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Linear and Nonlinear Programming with Maple: An Interactive, Applications-Based Approach
Helps Students Understand Mathematical Programming Principles and Solve Real-World Applications Supplies enough mathematical rigor yet accessible enough for undergraduatesIntegrating a hands-on learning approach, a strong linear algebra focus, Maple™ software, and real-world applications, Linear and Nonlinear Programming with Maple™: An Interactive, Applications-Based Approach introduces undergraduate students to the mathematical concepts and principles underlying linear and nonlinear programming. This text fills the gap between management science books lacking mathematical detail and rigor and graduate-level books on mathematical programming. Essential linear algebra toolsThroughout the text, topics from a first linear algebra course, such as the invertible matrix theorem, linear independence, transpose properties, and eigenvalues, play a prominent role in the discussion. The book emphasizes partitioned matrices and uses them to describe the simplex algorithm in terms of matrix multiplication. This perspective leads to streamlined approaches for constructing the revised simplex method, developing duality theory, and approaching the process of sensitivity analysis. The book also discusses some intermediate linear algebra topics, including the spectral theorem and matrix norms. Maple enhances conceptual understanding and helps tackle problemsAssuming no prior experience with Maple, the author provides a sufficient amount of instruction for students unfamiliar with the software. He also includes a summary of Maple commands as well as Maple worksheets in the text and online. By using Maple's symbolic computing components, numeric capabilities, graphical versatility, and intuitive programming structures, students will acquire a deep conceptual understanding of major mathematical programming principles, along with the ability to solve moderately sized real-world applications. Hands-on activities that engage studentsThroughout the book, student understanding is evaluated through "waypoints" that involve basic computations or short questions. Some problems require paper-and-pencil calculations; others involve more lengthy calculations better suited for performing with Maple. Many sections contain exercises that are conceptual in nature and/or involve writing proofs. In addition, six substantial projects in one of the appendices enable students to solve challenging real-world probl | 677.169 | 1 |
Community Reviews
whe where that new formula all of a sudden came from?!
So most books I've read didn't teach me anything. The Cartoon guide to statistics was a partial exception since they finally managed to explain soms ideas understandably. And the drawings weren't annoying, sometimes they even helped the text!
On the other hand, this book also made unexplained logical jumps and sometimes changed notational systems from one page to another. I guess it's that twist in the brain.
This is by far the most accessible book I have found on this subject so far. But that still doesn't mean that it is understandable....more
When a man lives alone, he invents ways to keep busy. One summer I decided I was going to become an AP statistics professor. Yes, I know. Anyhow, I spent that summer studying like you would not believe. This was one of the books I read.
This was a rather straightforward approach to statistics. While it has cartoons, there is real mathematics here. I swear!
many students, the problem with statistics isn't the algebra or computations, which are straightforward if tedious. It is grasping the concepts of statistics -- the notions underlying probability, sampling, distributions, the central limit theorem and Gaussians, how they relate to estimates of error.
This is precisely where most college level statistics texts fall short. They may well present all of the equations needed (and then some). They may well derive them and present examples of their use. However, to my experience they do a fairly mediocre job of simply conveying the idea that underlies all of the algebra and computation. A student is left memorizing dozens of equations and relations without ever gaining a deep understanding of what they mean (or really how to properly use them).
Using the Cartoon Guide to Statistics as a supplement to a college text, however, gives the weaker student a conceptual bridge. Best of all, it is a bridge that they can cross in one sitting! They can read it, cover to cover, in a matter of a few hours, and then refer back to it when an idea confuses them for the rest of the term.
The other group the book is ideal for is younger students in high school (or even middle school), ones whose attention span is not yet up to the task of slogging through a serious book in statistics written in small print by a humourless author. Again such a student would be well-served by having another more mathematical reference handy, but if one's goal is just to convey the idea and methodology underlying the ideas of: distribution, mean of the distribution, variance of the distribution, standard deviation, and the central limit theorem (and what else is there, really?) you can hardly do better....more
"chicken soup" for people who have to endure courses in statistics, it failed to make me fall in love with the subject, but enough to make we stood in awe with the level of obsessions some people have to measure our lives with numbers. Hooray to all statisticians who provide guidance to understand the world we're living in - but everybody need to remind themselves that we need further look into each phenomenon lest we get disoriented - don't blame the statistics for misunderstanding!
There are many math courses that I should have skipped & just read this book instead. Needs Excel worksheets & macros to help one enjoy/experiment. Might make them myself. I bought much of the whole Cartoon Guide/History series & leave them near the kids.
A nice basic review of statistics. I read through this with my stats text from college, looking up derivations/proofs of the important results discussed more intuitively in this book. I think this was a valuable approach, I certainly got more mileage out of this book with this kind of reading. c(4.0) Actually not a bad refresher reference concepts, I might actually consider starting here.
Might be a good companion to business school or something like that. It would've been inadequate to get me ramped up for grad school though (had to do that the hard way). I know it's not Gonick's style, but including some exercises would've been really cool/helpful. Perhaps he could consider some companion books. Making 'homework' fun with cartoons could be a winning combination....more
Thought this one was great. It does a great job of gradual learning curve mixed with an emphasis on real world application but it is also unafraid to toss a little math your way. Not to mention it is really funny at times. I've been chewing up stat books lately as an attempt to refresh on these concepts for work. This is a great refresher and then I'd add Data Smart as a good extension to more modern issues (like clustering, and social graph stuff).
Very good concept with impressive cartoon display. Basic concepts of statistics are explained well with a good examples and real life problems. You will not get bored while reading this book unlike other stereotypical statistics books. Authors' sense of humor is great and how they tried to incorporate it into the 'mathematical' cartoons is brilliant. Overall good book for the beginner students who have just started studying statistics.
Borrowed from a friend as a quick reference to my statistics homework. While some of the illustrations are helpful, it's far easier to sit down with a tutor for explanations, than to try to understand the illustrations. This is better suited as a gift to someone who already has a rudimentary knowledge of statistics.
Oh, CGS. Apparently the only statistics textbook out there without terrible, glaring errors, according to my professor. I really can see the utility of this teaching method, especially for high school - but it doesn't have any practice problems! (what, me, complain about having no homework?) Good as a reference, but it should not be your only source for learning stats.
going to make it any easier.
I love the fact that so many of the topics were covered in my Six Sigma Black Belt training. From hypothesis testing and anova to simple experimentation they cover a lot. Nice thing is that in many cases it goes into the topic but stops before going too deep.
My favorite part of the book actually came in the conclusion when they are discussing data quality. In a couple paragraphs they discuss how in the early days of statistical sampling statisticians actually did hands on data collection verses today's statisticians who rely on somebody collecting data so they can manipulate it in a computer. The authors stress that you lose some of that connection to the data by not directly being involved with its collection. My master black belt stressed the very same sentiments.
All in all a decent reference book that will become part of my Reference Library....more
T The deranged math-phobe popped up several more times throughout the book, typically when the author presented more formulas. If the message is that math is confusing, this is an opportunity for the author to explain the math in an intuitive way. If the concept can't be explained in a non-scary way, it should probably be left out of a book like this.
There were a couple good discussions, in particular the medical-test example on Bayesian probabilities, and the elements of regression. In general, I would prefer to see the author do a better job presenting a more limited selection of topics. The table of contents as-it could fill up two books of this size....more
computer program mentioned was MINITAB (which usually gets a reaction, a negative one [it's still used by companies that practice six sigma process:]). I'm moving on to the Manga Guide to Statistics in a week or so. Overall, I think it was a good offer and a helpful read. Even if I do not feel like I "completely get it" - the book was a pleasure to read, rare for a text covering math (in my opinion). ...more
th thing for sure - things that it did cover, it covered them well. yes, my print had a few typos & the book did take some logical leaps (especially in the later chapters where the author assumed that we have absorbed everything) but I learned a lot more than my school (& even college!) teachers ever taught. All things said, the book brings together 2 of my favorites. Maths & Cartoons. Good enough for me....more
A quick, fun way to learn statistics. The concepts are explained using some funny illustrations. This concept itself earns the book a great review. The only drawback is that at times the explanations are hurried and it hard to follow the authors on those bits.
But overall, a great alternative to hugely verbose and boring stats books.
who can't quite wrap their heads some what a confidence interval is, or hypothesis testing, or proper sampl who can't quite wrap their heads some what a confidence interval is, or hypothesis testing, or proper sampling and the need for it. Great book...more
A wonderful book that graphically illustrates the basics or statistics and probability. It is humorous and well written. My only qualm is that it isn't very rigorous. I don't think that rigor would be appropriate for this kind of book, but it would be interesting to see someone try. Great for reviewing your stats knowledge!
Gonick as always finds a clear, clever way to present the material he's given. Somehow, this ended up being even more mathematical than the Cartoon Guide to Physics, but suffered from typos and occasionally inconsistent notation. Still, a good way to get familiar with the basic terms & reasoning involved in the statistical methods described.
The book is not light on equations. Half concept, half equations, but I feel that the points get lost in the details. It's not as well organized as I would have liked and the illustrations are mere flourish rather than really helpful to the understanding.
Meh...the concept is sound, the cartoons are okay, and the examples are clever, but the definitions leave much to be desired for a book that is trying to make statistics easy for newbies. This book is used by my company for its Six Sigma training, and my thinking is that folks experiencing statistics for the first time, or even those with a little bit of knowledge, will struggle with this book.
Simply epic! This book is a must-read for anyone trying to understand statistical concepts intuitively. The hilarious examples and awesome cartoons make it impossible to put this book down, which you would never think can really happen for a stats book! Definitely recommend reading it!
diversity of his interests, and the success with which his books have met, have together earned Gonick the distinction of being "the most well-known and respected of cartoonists who have applied their craft to unravelling the mysteries of science" (Drug Discovery Today, March 2005)....more | 677.169 | 1 |
Appropriate for one- or two-semester Advanced Engineering Mathematics courses in departments of Mathematics and Engineering.
This clear, pedagogically rich book develops a strong understanding of the mathematical principles and practices that today's engineers and scientists need to know. Equally effective as either a textbook or reference manual, it approaches mathematical concepts from a practical-use perspective making physical applications more vivid and substantial. Its comprehensive instructional framework supports a conversational, down-to-earth narrative style offering easy accessibility and frequent opportunities for application and reinforcement. | 677.169 | 1 |
Mathematics for Multimedia explains the mathematics behind multimedia applications such as compression, signal processing, and image/video processing. This timely and thoroughly modern text is a rigorous survey of selected results from algebra and analysis, requiring only undergraduate math skills. More specifically, it focuses on when and why modern robust methods provide more accurate results. The topics are 'gems' chosen for their usefulness in understanding and creating application software for multimedia signal processing and communication.
The course Mathematics for Multimedia has been offered biennially at Washington University since 1997. It has proved popular with students from sophomores to beginning graduate students who seek an innovative course of rigorous contemporary mathematics with practical applications. For the instructor, the material is divided into six chapters that may be presented in six lecture hours each. Thus, the entire text may be covered in one semester, with time left for examinations and student projects. For the student, there are more than 100 exercises with complete solutions, and numerous example programs in Standard C. Each chapter ends with suggestions for further reading. The book also describes and illustrates easy-to-use software for applying cutting-edge techniques.
This book is aimed at a wide audience, including computer science and multimedia students and professors as well as those interested in employing mathematics in multimedia design and implementation.
Explains when and why modern robust methods provide more accurate results
REVIEWS for Mathematics for Multimedia | 677.169 | 1 |
Complex Variables and Applications, 9e will serve, just as the earlier editions did, as a textbook for an introductory course in the theory and application of functions of a complex variable. This new edition preserves the basic content and style of the earlier editions. The text is designed to develop the theory that is prominent in applications of the subject. You will find a special emphasis given to the application of residues and conformal mappings. To accommodate the different calculus backgrounds of students, footnotes are given with references to other texts that contain proofs and discussions of the more delicate results in advanced calculus. Improvements in the text include extended explanations of theorems, greater detail in arguments, and the separation of topics into their own sections. | 677.169 | 1 |
tenth edition of this bestselling text includes examples in more detail and more applied exercises; both changes are aimed at making the material more relevant and accessible to readers. Kreyszig introduces engineers and computer scientists toMore...
In books such as Introductory Functional Analysis with Applications and Advanced Engineering Mathematics, Erwin Kreyszig attempts to relate the changing character and content of mathematics to practical | 677.169 | 1 |
Linear Algebra With and accessible book from one of the leading figures in the field of linear algebra provides readers with both a challenging and broad understanding of linear algebra. The author infuses key concepts with their modern practicalMore...
This thorough and accessible book from one of the leading figures in the field of linear algebra provides readers with both a challenging and broad understanding of linear algebra. The author infuses key concepts with their modern practical applications to offer readers examples of how mathematics is used in the real world. Topics such as linear systems theory, matrix theory, and vector space theory are integrated with real world applications to give a clear understanding of the material and the application of the concepts to solve real world problems. Each chapter contains integrated worked examples and chapter tests. The book stresses the important role geometry and visualization play in understanding linear algebra. For anyone interested in the application of linear algebra theories to solve real world | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
This new title in Barron's E-Z Series contains everything students need to prepare themselves for an algebra class. Separate chapters focus on fractions, integers, ratios, proportions, expressions, equations, inequalities, graphing, statistics and probability basics, word problems, and more. Review questions and chapter reviews all have answers. The fast-growing E-Z Series presents new, updated, and improved versions of Barron's longtime popular Easy Way books. New cover designs, new interior layouts, and more graphic material than ever make these books ideal as self-teaching manuals. Teachers have discovered that E-Z titles also make excellent supplements to classroom textbooks. Skill levels range between senior high school and college-101 standards. All titles present detailed reviews of the target subject plus short quizzes and longer tests to help students assess their learning progress. All exercises and tests come with answers.
Review
"Looking for a comprehensive pre-algebra review? Then this may be just the book you want. It builds the foundation of mathematics and geometry through eighteen chapters of mathematical knowledge needed to understand the language of math. Although this book is meant as a self-teaching manual, it could also be incorporated into a mathematics program as a stand-alone text or as a review for students enrolling in a high school or college algebra course. A notable feature of the book is its flexibility. It provides helpful preparation for anyone planning to take an algebra course." Recommended -- Mathematics Teacher
Synopsis
(back cover)
Pre-Algebra makes sense when you approach it the E-Z way! Open this book for a clear, concise, step-by-step review of:
Fractions
Integers
Ratios
Proportions
Equations
Inequalities
Statistics and Probability Basics . . . and more
Measure your progress as you learn Pre-Algebra:
Take time to review and understand each important concept
Solve each chapter's problems and check your answers And discover that learning Pre-Algebra can be E-Z! | 677.169 | 1 |
For Sale is a brand new version of Student Solutions Manual For Beginning And ...For Sale is a brand new version of Student Solutions Manual For Beginning And Intermediate Algebra by Elayn Martin-Gay and this book is ready for immediate shipment.more
In this engaging book, the concept of the soliton is traced from the beginning...In this engaging book, the concept of the soliton is traced from the beginning of the last century to modern times with its recent applications.more
The Bittinger Graphs and Models Series helps readers learn algebra by making c...The Bittinger Graphs and Models Series helps readers learn algebra by making connections between mathematical concepts and their real-world applications. Abundant applications, many of which use real data, offer students a context for learning the...more
In mathematics we are interested in why a particular formula is true. Intuitio...In mathematics we are interested in why a particular formula is true. Intuition and statistical evidence are insufficient, so we need to construct a formal logical proof. The purpose of this book is to describe why such proofs are important, what they...more
"Approximation by Multivariate Singular Integrals "is the first monograph to i..."Approximation by Multivariate Singular Integrals "is the first monograph to illustrate the approximation of multivariate singular integrals to the identity-unit operator. The basic approximation properties of the general multivariate singular integral...more
"John Bird's approach to mathematics, based on numerous worked examples and in..."John Bird's approach to mathematics, based on numerous worked examples and interactive problems, is ideal for vocational students who require an entry-level textbook.Theory is kept to a minimum, with the emphasis firmly placed on problem-solving...more
Do you think it makes sense to split a day into twenty-four hours? Would anoth...Do you think it makes sense to split a day into twenty-four hours? Would another number have been a better choice? Good questions promote students mathematical thinking and understanding. These best-selling books offer a wealth of sample questions and...more
Presents short rhymes about numbers of objects from one through one hundred an...Presents short rhymes about numbers of objects from one through one hundred and provides information about the New Hampshire natural history and social studies topics that the objects represent.more
An illustrated collection of jump rope rhymes arranged according to the type o...An illustrated collection of jump rope rhymes arranged according to the type of jumping they are meant to accompany.more
"X and the City," a book of diverse and accessible math-based topics, uses bas..."X and the City," a book of diverse and accessible math-based topics, uses basic modeling to explore a wide range of entertaining questions about urban life. How do you estimate the number of dental or doctor's offices, gas stations, restaurants, or...more
Nowadays we are facing numerous and important imaging problems: nondestructive...Nowadays we are facing numerous and important imaging problems: nondestructive testing of materials, monitoring of industrial processes, enhancement of oil production by efficient reservoir characterization, emerging developments in noninvasive imaging...more | 677.169 | 1 |
9780131410855
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Summary
Algebra for College Students is typically used in a very comprehensive 1-semester&Intermediate Algebra course serving as a gateway course to other college-level mathematics courses.& The goal&of&the Intermediate Algebra course&is to provide students with the mathematical skills that are prerequisites for courses such as College Algebra, Elementary Statistics, Liberal-Arts Math and Mathematics for Teachers.& This Algebra for College students text may also be used in a 1-semester, lower-level College Algebra course as a prerequisite to Precalculus.
Table of Contents
Preface To the Student
Basic Concepts
Study Skills for Success in Mathematics, and Using a Calculator
Sets and Other Basic Concepts
Properties of and Operations with Real Numbers
Order of Operations Mid-Chapter Test: Sections 1.1
Exponents
Scientific Notation
Equations and Inequalities
Solving Linear Equations
Problem Solving and Using Formulas
Applications of Algebra Mid-Chapter Test: Sections 2.1
Additional Application Problems
Solving Linear Inequalities
Solving Equations and Inequalities Containing Absolute Values
Graphs and Functions
Graphs
Functions
Linear Functions: Graphs and Applications
The Slope-Intercept Form of a Linear Equation Mid-Chapter Test: Sections 3.1
Factoring a Monomial from a Polynomial and Factoring by Grouping Mid-Chapter Test: Sections 5.1
Factoring Trinomials
Special Factoring Formulas
A General Review of Factoring
Polynomial Equations
Rational Expressions and Equations
The Domains of Rational Functions and Multiplication and Division of Rational Expressions
Addition and Subtraction of Rational Expressions
Complex Fractions
Solving Rational Equations Mid-Chapter Test: Sections 6.1
Rational Equations: Applications and Problem Solving
Variation
Roots, Radicals, and Complex Numbers
Roots and Radicals
Rational Exponents
Simplifying Radicals
Table of Contents provided by Publisher. All Rights Reserved.
Excerpts
This book was written for college students who have successfully completed a first course in elementary algebra. My primary goal was to write a book that students can read, understand, and enjoy. To achieve this goal I have used short sentences, clear explanations, and many detailed, worked-out examples. I have tried to make the book relevant to college students by using practical applications of algebra throughout the text. Features of the Text Full-Color Format.Color is used pedagogically in the following ways: Important definitions and procedures are color screened. Color screening or color type is used to make other important items stand out. Artwork is enhanced and clarified with use of multiple colors. The full-color format allows for easy identification of important features by students. The full-color format makes the text more appealing and interesting to students. Readability.One of the most important features of the text is its readability. The book is very readable, even for those with weak reading skills. Short, clear sentences are used and more easily recognized, and easy-to-understand language is used whenever possible. Accuracy.Accuracy in a mathematics text is essential. To ensure accuracy in this book, mathematicians from around the country have read the pages carefully for typographical errors and have checked all the answers. Connections.Many of our students do not thoroughly grasp new concepts the first time they are presented. In this text, we encourage students to make connections. That is, we introduce a concept, then later in the text briefly reintroduce it and build upon it. Often an important concept is used in many sections of the text. Students are reminded where the material was seen before, or where it will be used again. This also serves to emphasize the importance of the concept. Important concepts are also reinforced throughout the text in the Cumulative Review Exercises and Cumulative Review Tests. Chapter Opening Application.Each chapter begins with a real-life application related to the material covered in the chapter. By the time students complete the chapter, they should have the knowledge to work the problem. A Look Ahead.This feature at the beginning of each chapter gives students a preview of the chapter. This feature also indicates where this material will be used again in other chapters of the book. This material helps students see the connections between various topics in the book and the connection to real-world situations. The Use of Icons.At the beginning of each chapter and of each section, a variety of icons are illustrated. These icons are provided to tell students where they may be able to get extra help if needed. There are icons for theStudent's Solution Manual;theStudent's Study Guide; CDs and videotapes; Math Pro 4/5 Software;thePrentice Hall Tutor Center;and theAngel Website.Each of these items will be discussed shortly. Keyed Section Objectives.Each section opens with a list of skills that the student should learn in that section.The objectives are then keyed to the appropriate portions of the sections with red numbers such as1. Problem Solving.George Polya's five-step problem-solving procedure is discussed in Section 1.2. Throughout the book problem solving and Polya's problem-solving procedure are emphasized. Practical Applications.Practical applications of algebra are stressed throughout the text. Students need to learn how to translate application problems into algebraic symbols. The problem-solving approach used throughout this text gives students ample practice in setting up and solving application problems. Detailed, Worked-Out Examples.A wealth of examples have been worked out in a step-by-step, detailed manner. Important steps | 677.169 | 1 |
Understand Electrical and Electronics Maths
Understand Electrical and Electronics Maths covers elementary maths and the aspects of electronics. The book discusses basic maths including quotients, algebraic fractions, logarithms, types of equations and balancing of equations. The text also describes the main features and functions of graphs and the solutions to simpler types of electronics problems. The book then tackles the applications of polar coordinates in electronics, limits, differentiation and integration, and the applications of maths of rates of change in electronics. The activities of an electronic circuit; techniques of mathematical modeling; systematic techniques for dealing with the more difficult sets of simultaneous equations; alternating currents and voltages; and analysis of waveforms are also considered. The book provides answers to exercises for each chapter. Students taking electronics and courses related to electrical engineering at levels up to and including higher national certificate and diploma will find the | 677.169 | 1 |
Precalculus A Problems-OrientedMore...
Get understanding of college algebra and trigonometry, this text provides you with the tools you need to be successful in this course. Preparing for exams is made easy with iLrn, an online tutorial resource, that gives you access to text-specific tutorials, step-by-step explanations, exercises, quizzes, and one-on-one online help from a tutor. Examples, exercises, applications, and real-life data found throughout the text will help you become a successful mathematics student!
David Cohen is a British joutnalist who has written for The Independent, The Guardian, and British GQ, as well as The New York Times. In 1997, he was the recipient of a harkness fellowship hosted by Columbia University, which enabled him to write this book. He lives in london with his wife and two daughters.
Fundamentals
Sets of Real Numbers
Absolute Value
Solving Equations (Review and Preview)
Rectangular Coordinates
Visualizing Data
Graphs and Graphing Utilities
Equations of Lines
Symmetry and Graphs
Circles
Equations And Inequalities
Quadratic Equations: Theory and Examples
Other Types of Equations
Inequalities
More on Inequalities
Functions
The Definition of a Function
The Graph of a Function
Shapes of Graphs
Average Rate of Change
Techniques in Graphing
Methods of Combining Function
Iteration
Inverse Functions
Polynomial And Rational Functions
Applications to Optimization
Linear Functions
Quadratic Functions
Using Iteration to Model Population Growth (Optional Section)
Setting up Equations That Define Functions
Maximum and Minimum Problems
Polynomial Functions
Rational Functions
Exponential And Logarithmic Functions
Exponential Functions
The Exponential Function y = ex
Logarithmic Functions
Properties of Logarithms
Equations and Inequalities with Logs and Exponents
Compound Interest
Exponential Growth and Decay
Trigonometric Functions Of Angles
Trigonometric Functions of Acute Angles
Algebra and the Trigonometric Functions
Right-Triangle Applications
Trigonometric Functions of Angles
Trigonometric Identities
Trigonometric Functions Of Real Numbers
Radian Measure
Radian Measure and Geometry
Trigonometric Functions of Real Numbers
Graphs of the Sine and the Cosine Functions
Graphs of y = A sin(Bx-C) and y = A cos(Bx - C)
Simple Harmonic Motion
Graphs of the Tangent and the Reciprocal Functions
Analytical Trigonometry
The Addition Formulas
The Double-Angle Formulas
The Product-to-Sum and Sum-to-Product Formulas
Trigonometric Equations
The Inverse Trigonometric Functions
Additional Topics In Trigonometry
The Law of Sines and the Law of Cosines
Vectors in the Plane, a Geometric Approach
Vectors in the Plane, an Algebraic Approach
Parametric Equations
Introduction to Polar Coordinates
Curves in Polar Coordinates
Systems Of Equations
Systems of Two Linear Equations in Two Unknowns
Gaussian Elimination
Matrices
The Inverse of a Square Matrix
Determinants and Cramer's Rule
Nonlinear Systems of Equations
Systems of Inequalities
Analytic Geometry
The Basic Equations
The Parabola
Tangents to Parabolas (Optional)
The Ellipse
The Hyperbola
The Focus-Directrix Property of Conics
The Conics in Polar Coordinates
Rotation of Axes
Roots Of Polynomial Equations
The Complex Number System
Division of Polynomials
Roots of Polynomial Equations: The Remainder Theorem and the Factor Theorem | 677.169 | 1 |
Vector Analysis for Mathematicians, Scientists and Engineers: The Commonwealth and International Library: Physics Division
Vector Analysis for Mathematicians, Scientists and Engineers, Second Edition, provides an understanding of the methods of vector algebra and calculus to the extent that the student will readily follow those works which make use of them, and further, will be able to employ them himself in his own branch of science. New concepts and methods introduced are illustrated by examples drawn from fields with which the student is familiar, and a large number of both worked and unworked exercises are provided. The book begins with an introduction to vectors, covering their representation, addition, geometrical applications, and components. Separate chapters discuss the products of vectors; the products of three or four vectors; the differentiation of vectors; gradient, divergence, and curl; line, surface, and volume integrals; theorems of vector integration; and orthogonal curvilinear coordinates. The final chapter presents an application of vector analysis. Answers to odd-numbered exercises are provided as the end | 677.169 | 1 |
4
Algebra When Ready Only when students exhibit demonstrable success with prerequisite skillsnot at a prescribed grade levelshould they focus explicitly and extensively on algebra, whether in a course titled Algebra 1 or within an integrated mathematics curriculum. Exposing students to such coursework before they are ready often leads to frustration, failure, and negative attitudes toward mathematics and learning. NCTM Position : Algebra: What, When, and for Whom (September 2011)
5
Major Themes that Start in PreK and Go all the Way through Grade 12 Exploring and extending patterns Representing mathematical ideas with symbols and objects Using mathematical models to represent quantitative relationships Analyzing change in various contexts
8
The Far Too Typical Experience! 1.Here is an equation: y = 3x + 4 2.Make a table of x and y values using whole number values of x and then find the y values, 3.Plot the points on a Cartesian coordinate system. 4.Connect the points with a line. Opinion: In a students first experience, the equation should come last, not first.
13
Piles of Tiles A table can help communicate the number of tiles that must be added to form each successive pile? (the recursive rule) ? Pile 12345678.. Tiles 47101316192225.. Pile 1 Pile 2 Pile 3 Pile 4 Pile 5
18
Piles of Tiles How is the change, add 3 tiles, from one pile to the next (recursive form) reflected in the graph? Explain. How is the term 3n and the value 1 (explicit form) reflected in the graph? Explain. Y = 3n + 1
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Piles of Tiles The recursive rule Add 3 tiles reflects the constant rate of change of the linear function. The 3n term of the explicit formula is the repeated addition of add 3 Y = 3n + 1
28
Equations Arise From Physical Situations What is the perimeter of shape 6?
29
Find the Perimeter Shape12345678.. Perimeter A table can help communicate the length of the sides that must be added to a shape to find the perimeter of the next shape? (the recursive rule)
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Find the Perimeter A table can help communicate the length of the sides that must be added to a shape to find the perimeter of the next shape? (the recursive rule) Shape12345678.. Perimeter4681012141618
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Find the Perimeter Can we find the perimeter of shape N without using the recursive rule? (the explicit rule) Shape123456..N Perimeter468101214.. 2N + 2
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Equations Arise From Physical Situations What is the perimeter of shape 6?
33
Find the Perimeter Shape12345678.. Perimeter A table can help communicate the length of the sides that must be added to a shape to find the perimeter of the next shape? (the recursive rule)
34
Shape12345678.. Perimeter6111621263136 A table can help communicate the length of the sides that must be added to a shape to find the perimeter of the next shape? (the recursive rule) Find the Perimeter
35
Can we find the perimeter of shape N without using the recursive rule? (the explicit rule) Shape123456..N Perimeter51116212631.. 5N + 1 Find the Perimeter
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How many beams are needed to build a bridge of length n? Bridge of length 6
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Multiple Representations The understanding of mathematics is advanced when concepts are explored in a variety of forms including symbols, graphs, tables, physical models, as well as spoken and written words. | 677.169 | 1 |
Mathematics
DT6248
Description
The programme is designed for those with an interest in the mathematical sciences or who require a better understanding of mathematics for their jobs and wish to improve their mathematical skills for their future career. It is a single-stage, add-on programme and provides a solid background in the fundamentals of mathematics. The programme is two years in duration and requires the successful completion of four modules, two of which are taken each year. The modules are delivered in the evenings, making the programme attractive to those in full-time employment. The programme is suitable for individuals who have already studied some mathematics at third level within another discipline. Graduates of the programme are eligible to enter the part-time ordinary degree programme in Mathematics (DT7248) or transfer to the full-time honours degree programmes in the SchoolDiscrete Mathematics
Algebra 1
Year 2 consists of two modules:
Calculus /Analysis 2
Numerical Methods 1
Students may enter either Year 1 or Year 2 which are offered on alternate years. (The order in which the two years are completed is not relevant.) The Year 1 programme is scheduled to run in 2015/2016.
Entry Requirements
Completion of a suitable number of third-level mathematical modules
or
A level 6 or level 7 qualification in a mathematically related discipline.
or
any other such qualification that the Institute may deem equivalent.
Examinations / Assessments
Continuous assessment counts for 30% of final mark. Examinations are held at the end of Semester 2.
Progression
Students who have successfully completed the programme are eligible to enter the part-time Ordinary Degree programme in Mathematics or stage 3 of the full-time Honours Degree programme in Mathematical Sciences or Industrial Mathematics.
Learning Outcome
On successfully completing the Higher Certificate, the learner will:
have an appreciation of the fundamental concepts of Mathematics,
be able to solve a wide range of basic problems in Calculus, Algebra and Numerical Methods
be able to use software packages to assist in the modelling and solution of mathematical problems
Career Opportunities
Mathematical, analytical and problem-solving skills are required in the modern workplace and career opportunities abound for numerate graduates. A mathematical qualification is the key to a wide range of employment opportunities and this flexibility allows graduates to enter whichever sector is thriving and advance their careers quickly. A mathematics qualification is ideal for upskilling and recent graduates have found employment in the ICT sector and the financial services. Graduates of this programme will develop analytical and problem solving skills and be in a good position to embark on high achieving careers in industry, commerce, the teaching and other professions, and the public sector. The School of Mathematical Sciences offers a ladder of qualifications, both part time and full time, and graduates can take steps towards an honours degree, postgraduate study and degrees by research.
Unable to print? Email part.time@dit.ie with your full name & address and we'll post it out to you | 677.169 | 1 |
Using Geometer's Sketchpad
Geometry is one of the most visual courses in high school mathematics. The Geometer's Sketchpad is an electronic ruler and compass for today's technological geometry classroom. This module is designed to familiarize teachers with this new tool through interactive tutorials and then to demonstrate how it can be used effectively in the teaching of geometry. Credit: 1 grad. sem. hr.
Common Core Standardsfor Mathemtical Practice that are emphasized include:
4. Model with mathematics.
5. Use appropriate tools strategically.
7. Look for and make use of structure.
Using the Geometer's Sketchpad in the Mathematics Classroom was written by Jeremy Bartusch of the University of Illinois in November 1996. It was revised by Tony Peressini in January 1998, and by Tom Anderson in March 2004 and again in October 2011 for Sketchpad Version 5. | 677.169 | 1 |
Product Description:
'This is a very interesting book and one that can only help our battle to make algebra come alive and help our students see the useful and interesting problems that algebra can help us to deal with' - Peter Hall, Imberhorne School, East Grinstead
Each activity is presented as a reproducible student investigation. It is followed by guidelines and notes for the teacher. Each activity is keyed to the National Council of Teachers of Mathematics (NCTM) Standards, Revised. This link to the NCTM standards allows teachers to facilitate linking classroom activities to specific state and school district content standards.
First and foremost, the activities are meant to be motivational. As much as possible, we want this book to achieve the goal of being attractive to people who thought they didn't like mathematics. To accomplish this, it is necessary for the activities to be quite different from what students encounter in their basal texts-different in both substance and form. This seems especially critical; no matter how excellent a basal text is being used, nearly every class experiences the "blahs." Unfortunately, this sort of boredom is often well entrenched long before the teacher and perhaps even the students are aware of it. Presenting activities on a regular basis gives the variety and change of pace needed to sustain interest in any subject.
REVIEWS for Making Algebra Come Alive | 677.169 | 1 |
Performance task for Logarithms on Population Growth
836 Downloads
Word Document File
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0.03 MB | 3 pages
PRODUCT DESCRIPTION
This is a common core "type" of performance task looking at both exponential functions and linear functions when it comes to population growth. It asks numerous reasoning questions with data and math questions in between. Students are asked to come up with a linear model in slope intercept form looking at a population in 1980 and then again in 2000. It also asks students to come up with an exponential model using that same data. The students will also need to graph (on a separate sheet of paper both models to try and determine which model of better use to predict future populations. It then asks to predict what the population will be in 2011, and asks students to use both models when calculating. After they have calculated the population using both models, the worksheet gives what the actual population was in the year 2011 and asks various questions on which model works better, why it works better, what factors affect population growth and so on and so forth.
As common core is among us this is a higher level of critical thinking that is both mathematically difficult (uses various mathematical skills) and has high application | 677.169 | 1 |
Introduction Recap Introduction To the topic Consolidation Development
• • •
introduce the idea of a complex number and explain the Argand diagram. demonstration of the algebra of complex numbers in Cartesian form. introduce the polar form of a complex number and demonstrate the algebra of complex numbers in polar form. | 677.169 | 1 |
lab: An Introduction with Applications
Overview
More college students use Amos Gilat's" MATLAB: An Introduction with Applications" than any other MATLAB textbook. This concise book is known for its just-in-time learning approach that gives students information when they need it. The new edition gradually presents the latest MATLAB functionality in detail. Equally effective as a freshmen-level text, self-study tool, or course reference, the book is generously illustrated through computer screen shots and step-by-step tutorials, with abundant and motivating applications to problems in mathematics, science, and engineering. | 677.169 | 1 |
Topic Topics covered include spectral theory of elliptic differential operators, the theory of scattering of waves by obstacles, index theory for Dirac operators, and Brownian motion and diffusion....more | 677.169 | 1 |
MAT 121
MATHEMATICS FOR BUSINESS AND INFORMATION SCIENCE
LECTURE 12
WING HONG TONY WONG
2.5 Multiplication of matrices
Just like matrix addition, where A + B can be dened if and only if A and B have the
same sizes, matrix multiplication AB can be dened on
MAT 121
MATHEMATICS FOR BUSINESS AND INFORMATION SCIENCE
LECTURE 16
WING HONG TONY WONG
3.1 Graphing systems of linear inequalities in two variables
Recall that if we are given a linear equation
ax + by = h,
the way to graph it is to nd the slope a and th
MAT 121
MATHEMATICS FOR BUSINESS AND INFORMATION SCIENCE
LECTURE 3
WING HONG TONY WONG
2.1 Systems of linear equations: an introduction
Example 1. Determine whether each system of linear equations has one and only one
solution, innitely many solutions, or
MAT 121
MATHEMATICS FOR BUSINESS AND INFORMATION SCIENCE
LECTURE 1
WING HONG TONY WONG
2.1 Systems of linear equations: an introduction
Recall from our algebra knowledge that ax + by = h, where a and b are not both 0,
represents a straight line on a recta | 677.169 | 1 |
7th Grade Math Final Exam and Study Guide~Common Core
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1.41 MB | 12 pages
PRODUCT DESCRIPTION
Comprehensive 7th grade math final exam covering all major concepts taught during the year. Directly aligns to common core. This is a great way to assess your students' understanding of the skills they should acquire during 7th grade math.
The study guide is designed to prepare students to be successful on the exam, but does not give them questions that are exactly like the exam. There are a wide variety of types of questions and difficulty level.
Topics:
Rational Numbers
Statistics and Probability
3-Dimensional Geometric Figures
Area and Perimeter of 2-D Figures
Proportional Reasoning
Angle Relationships
Equations, Inequalities and Expressions
Answer key is provided for both the exam and the study guide
Check out my Exam Bundle for $8.50. Includes an exam for each of the 7th grade math units, and a study guide that is aligned to each exam. Also includes a comprehensive final exam, and mid-term | 677.169 | 1 |
Product Description:
This expanded sixth edition of the Angel/Porter text contains two additional chapters, Graph Theory and Voting and Apportionment. This best-selling text balances solid mathematical coverage with a comprehensive overview of mathematical ideas as they relate to varied disciplines. This text provides an appreciation of mathematics, highlighting mathematical history, applications of mathematics to the arts and sciences across cultures and introduces students to the uses of technology in mathematics. It is an ideal book for students who require a general overview of mathematics, especially those majoring in the liberal arts, elementary education, the social sciences, business, nursing and allied health fields.
REVIEWS for A Survey of Mathematics with Applications | 677.169 | 1 |
Discussion Question 2
• Due Date: Day 4 [Main] forum
• Post your response to the following: The introduction of calculators and computers into the mathematics classroom has made it faster and easier for students to complete difficult problems. Technology should not, however, replace the students' understanding of basic mathematical operations. What are some appropriate and effective ways to use technology in the mathematics classroom? How can you ensure that students do not depend too heavily on technology to... | 677.169 | 1 |
Nirmala Kashyap
Math Instructor
Department(s):
Contact Info:
Course(s):
Course Notes:
This class is taught online using MyStatLab. Students will need to buy the software by the end of first day of classes to avoid being dropped from class. MyStatLab comes with e-textbook (students are NOT required to buy printed textbook) and costs approximately $65 online (much cheaper than printed textbook). Students study on their own (can get help at Math Learning Center on campus) using e-material from MyStatLab and videos posted online and turn in homework and quizzes online. Two mandatory paper/pencil exams will be given on campus. Enrolled students will be emailed with further instructions (including exam dates, and how to buy MyStatLab etc.) once the class is set up on blackboard and MyStatLab (approximately 2 weeks before the start of semester).
Please do NOT take this class if you think it will be easier than a traditional, on-campus class. Math 160 is an intensive transfer-level class. To succeed in this challenging class you must have strong Algebra and Arithmetic skills, stay focused, take charge of your learning, work extremely hard (15-18 hours per week), be a self-motivated learners who does not need prompting in order to complete assignments, and able to balance personal responsibilities with class requirements, You must be proficient in computer skills and good at understanding and following written directions. For more info. contact Nirmala Kashyap at Nirmala.Kashyap@gcccd.edu
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Math 160 Online FAQ
Q&A For Prospective Students
1. What is an online course?
An online course is a course that is taught over the Internet. The student enrolled in an online course does not attend class and thus does not have the advantage of hearing the instructor explain the material or of participating in classroom discussions of the material. Instead, the student learns the material by using the resources provided by the teacher, and in general, questions are answered via email and discussions take place on electronic discussion boards. Tests are given on campus in a proctored environment.
2. Is an online mathematics course a wise choice for you?
Online mathematics courses require self-discipline and a solid background in the prerequisite course. If you feel that your preparation is not adequate, you should consider taking a traditional classroom version of Math 160.
3. Is this online course a self-paced course?
No. This online course, like its traditional classroom counterpart, has schedules, deadlines, assignment due dates, and scheduled tests.
4. How do I learn the material?
All course material will be available in MyStatLab. There you will find an e-text, lectures, videos, guided tutorials, homework assignments, quizzes, test reviews and paper-and-pencil chapter reviews to download and work out. To complete the course, you will work your way through seven sequenced lessons that will guide you through the course.
5. How can I learn how to use the MyStatLab software?
There are directions on MyStatLab welcome page to help you learn this information. If you need in person help, please email me and I will be happy to meet with you at Grossmont College Campus.
6. If I take the online course, will I be required to come to the Grossmont College campus?
Yes, you must come to Grossmont for midterm and the Final Exam.
7. Are there any circumstances under which I can take my tests at another community college, college or university?
NO, All exams are given on Grossmont college campus.
8. What are the required materials for this class?
The required material for this class is the MyStatLab Student Access Kit. This kit contains the MyStatLab software as well as an electronic version of the text. You may purchase the kit online at using a credit card, or may purchase an access code from the College Bookstore and then log on using the access code. Detailed instructions about logging in to MyStatLab will be posted on Blackboard and will be available for those who will be registered for the class.
9. What resources are available to assist me in learning the material?
Here are some of the resources available for you
Ø MyStatLab: Use the software to access videos, example problems, step-by-step solutions, the electronic text, and your homework, quizzes and other stuff.
Ø MathLearningCenter Assistance: Instructors and aides in the Math Learning Center are available to answer homework questions that you may have.
Ø Tutoring: Free tutoring is available in the Tutoring Center. To find hours and locations go to the Tutoring Center website.
Ø Office hours and Email Communication: I have online office hours on Mondays from 6:00 pm - 7:00 pm. I also check and respond to emails within 24 hours during the weekdays. I do not check email on the weekends. I check only Grossmont and MyStatLab emails. Please don't use any emails other than these two.
Ø The Discussion Board on MyMathLab: Use the discussion board to ask questions and to answer questions of other students in the class. I do monitor these discussions and post responses to unanswered questions. Please remember in using the Board to be kind, generous, and patient with your classmates. All interactions must be appropriate in content and language, as they would be in an actual classroom.
10. How do I get started on the class?
To get started in the class, register with Admissions and Records. You will be emailed further instruction two weeks before the class starts. | 677.169 | 1 |
'A' Level Further Mathematics
Examination Board & Code : to be decided, see below
Length of Course : 2 Years to full A level
What is the new Further Maths A Level about?
Further Mathematics is a challenging qualification, which both extends and deepens your knowledge and understanding beyond the standard A level Mathematics.
New AS and A levels in Mathematics and Further Mathematics are being introduced in England for first teaching from September 2017. The changes include:
New linear structure: AS will be decoupled from A level, and all assessment will take place at the end of the course. Exam questions may draw on the content of the whole A level.
New emphasis: There will be more emphasis on problem solving, reasoning and modelling, and a requirement for the use of technology to permeate teaching and learning.
New content: The content of AS and A level Mathematics will be fixed. It will include pure mathematics, mechanics and statistics (including analysis of large data sets). There will be some choice in content for AS and A level Further Mathematics.
Is this course for me?
In addition to the minimum academic level specified in our sixth form entry requirements, students will be required to pass the entry level algebra competency test and to complete the summer homework that is set during the Induction week.
Where could it lead?
If you are planning to take a degree such as Engineering, Sciences, Computing, Finance/Economics, etc., or perhaps Mathematics itself, you will benefit enormously from taking Further Mathematics. Further Mathematics introduces new topics such as matrices and complex numbers that are vital in many STEM degrees. Students who have studied Further Mathematics find the transition to such degrees far more straightforward.
Further Mathematics qualifications are highly regarded and are warmly welcomed by universities. Students who take Further Mathematics are regarded as demonstrating a strong commitment to their studies. Some prestigious university courses require students to have a Mathematics qualification and others may adjust their grade requirements more favourably to students with Further Mathematics.
What do other students say?
"Further Maths is challenging but the teachers and the help from the Mathematics Department makes it enjoyable & achievable."
"Further Mathematics is appealing to the country's top universities as it demonstrates the student's ability to succeed in challenging environment."
Draft structure of A level Further Mathematics
Please note: various Exam Boards have submitted their specifications to OFQUAL and are still awaiting accreditation. As a result we have been unable to decide upon the Exam Board. Please click to gain an idea of the structures proposed. | 677.169 | 1 |
Higher Engineering Mathematics practical introduction to the core mathematics principles required at higher engineering level
John Bird's approach to mathematics, based on numerous worked examples and interactive problems, is ideal for vocational students that require an advanced textbook.
Theory is kept to a minimum, with the emphasis firmly placed on problem-solving skills, making this a thoroughly practical introduction to the advanced mathematics engineering that students need to master. The extensive and thorough topic coverage makes this an ideal text for upper level vocational courses.
Now in its seventh edition, Engineering Mathematics has helped thousands of students to succeed in their exams. The new edition includes a section at the start of each chapter to explain why the content is important and how it relates to real life. It is also supported by a fully updated companion website with resources for both students and lecturers. It has full solutions to all 1900 further questions contained in the 269 practice exercises.
Recommendations:
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Prerequisite: MATH 206 or ENGR 102; MATH 325.
Note: Does not count towards the mathematics major or minor. Credit may be applied toward the MAT degree but not towards any other graduate degree in mathematics. Credit not allowed for both MATH 387 and MATH 587.
Topics include: Pigeon-hole principle, counting techniques, binominal coefficients, generating functions, stirling and catalan numbers, permutations and graphs | 677.169 | 1 |
Instead of using a simple lifetime average, Udemy calculates a course's star rating by considering a number of different factors such as the number of ratings, the age of ratings, and the likelihood of fraudulent ratings.
Master the Fundamentals of Math
Learn everything from the Fundamentals, then test your knowledge on 90+ quiz questions
Bestselling
4.6
(259 ratings)
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HOW MASTER THE FUNDAMENTALS OF MATH IS SET UP TO MAKE COMPLICATED MATH EASY
This 162-lesson course includes video and text explanations of everything from the Fundamentals, and it includes more than 40 quizzes (with solutions!) to help you test your understanding along the way. Master the Fundamentals of Math is organized into the following sections:
Numbers
Negative numbers
Factors and multiples
Decimals
Fractions
Mixed numbers
Ratio and proportion
Exponents
Radicals
Scientific notation
And here's what you get inside of every lesson:
Videos: Watch over my shoulder as I solve problems for every single math issue you'll encounter in class. We start from the beginning... I explain the problem setup and why I set it up that way, the steps I take and why I take them, how to work through the yucky, fuzzy middle parts, and how to simplify the answer when you get it.
Notes: The notes section of each lesson is where you find the most important things to remember. It's like Cliff Notes for books, but for math. Everything you need to know to pass your class and nothing you don't.
Quizzes: When you think you've got a good grasp on a topic within a course, you can test your knowledge by taking one of our quizzes. If you pass, wonderful. If not, you can review the videos and notes again or ask me for help in the Q&A section.
HERE'S WHAT SOME STUDENTS OF MASTER THE FUNDAMENTALS OF MATH HAVE TOLD ME:
"Wonderful course so far, great first step in refreshing my knowledge of math in preparation for college calculus come fall after 2 years off." - Blake L.
"I swear I was math illiterate before, but no longer! Krista makes most things understandable, and if I don't understand something the first time, I just replay it another time or two and I get it! You have NO idea what this means to me. While I may not be ready for more advanced math like algebra or calculus (yet), I feel just a tiny bit smarter. Thanks!" - Tracy B.
"This is a really awesome course that takes you step by step through each topic explaining all the pertinent rules and how they apply. Each quiz fortifies what you learned in the lectures as well as testing your knowledge of the material covered. This is a very well thought out course and I enjoyed it immensely!" - William O.
"The instructor is great, very good and easy to understand instructions." - John C.
"After trying a few of the other Math Fundamental courses and a GED class just to brush up on math. This is by far the best class I have found. The instructor speaks in an easy and clear English voice. If you are looking to brush up on you math skill then this is the class you are looking for." - Maxx Z.
"Taking this course as a refresher, I was as struck by Ms. King's firm grasp of the fundamentals as I was by her engaging presentation. Too many math teachers breeze over key concepts which leaves many students puzzled and eventually lost. This course is exceptional in that each concept is thoroughly explained, with no steps skipped." - Ben S.
Who is the target audience?
Current middle school/junior high students, or students about to start middle school who are looking to get ahead
Homeschool parents looking for extra support with the fundamentals
Anyone who wants to study math for fun after being away from school for a while
I'd go to a class, spend hours on homework, and three days later have an "Ah-ha!" moment about how the problems worked that could have slashed my homework time in half.
I'd think, "WHY didn't my teacher just tell me this in the first place?!"
So I started tutoring to keep others out of that aggravating, time-sucking cycle. Since then, I've recorded tons of videos and written out cheat-sheet style notes and formula sheets to help every math student—from basic middle school classes to advanced college calculus—figure out what's going on, understand the important concepts, and pass their classes, once and for all. | 677.169 | 1 |
Find a West University Place, TX SAT...Thomas and received an A in the course. Linear Algebra is the study of matrices and their properties. The applications for linear algebra are far reaching whether you want to continue studying advanced algebra or computer science. | 677.169 | 1 |
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Unformatted text preview: w a must be an integer multiple of 9. Our last concern is end behavior.
Since the leading term of p(x) is ax4 , we need a < 0 to get p(x) → −∞ as x → ±∞. Hence, if we
choose x = −9, we get p(x) = −9x4 + 6x3 − 82x2 + 54x − 9. We can verify our handiwork using
the techniques developed in this chapter. 226 Polynomial Functions This example concludes our study of polynomial functions.9 The last few sections have contained
what is considered by many to be 'heavy' mathematics. Like a heavy meal, heavy mathematics
takes time to digest. Don't be overly concerned if it doesn't seem to sink in all at once, and pace
yourself on the exercises or you're liable to get mental cramps. But before we get to the exercises,
we'd like to offer a bit of an epilogue.
Our main goal in presenting the material on the complex zeros of a polynomial was to give the
chapter a sense of completeness. Given that it can be shown that some polynomials have real zeros
which cannot be expressed using the usu...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School. | 677.169 | 1 |
Circuit Training - The Quadratic Formula (algebra)
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2.61 MB | 4 pages
PRODUCT DESCRIPTION
Engage your students with these twelve problems to help them practice applying the Quadratic Formula to solve equations! Students check their work as they go, advancing in the circuit by hunting for their answer(s). When they find their answer(s), this becomes the next problem. I wrote this for algebra one students, though it would be a great review for algebra two students before they learn about imaginary numbers.
There is no answer key included in the circuit since the answers are imbedded in the circuit. The only thing the teacher needs to do is work the circuit ahead of the students to decide how to best use it in their classroom. Notes? Guided Practice? Independent Practice? Cooperative activity? Scavenger hunt?
If you ever get stuck or suspect there is an error, please do not hesitate to contact me at virginia.cornelius@gocommodores.org | 677.169 | 1 |
1 Page 1 1 Issues: (a) Representing problems (b) Methods & common flaws in problem solving (c) Expertise 1. Problems and problem representation – Well-structured vs. ill-structured problems – Stages in problem solving – The importance of problem representation 2. Common flaws in problem solving – Analogies – Hindrances to forming appropriate representations 3. Problem solving methods – Algorithms and heuristics – Heuristics: Hill climbing, means-ends analysis, working backward 4. Expertise – Very domain specific (chess study) – Power law of practice – Characteristics of expertise 2 • A Problem consists of some initial state in which a person begins and a goal state that is to be attained, plus a non-obvious way of getting from the first to the second. Initial State Goal State Methods 3 4 Polya (1957) • Form a representation • Construct a plan • Execute plan • Checking/Evaluation Reformulate 5 Problem Solving Concepts: Initial & Goal states. Intermediate States. Representation of problem. Operators: actions that move between states Problem Space: Whole range of possible states and operators, only some of which will lead to goal state 6 n = 4p, and p = n - 30 Initial state n = and p = Goal state Substitute for p Divide by 4 Divide by 30 Problem space whole range of possible states and operators, only some of which will lead to the goal state The price of a notebook is four times that of a pencil. The pencil costs $.30 less than the notebook. What is the price of each? Substitute for n Subtract 4p from both sides: p = 4p - 30 -3p = -30 Divide, substitute Operators Intermediate states 40 10
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2 Page 2 7 O O O O O O O O O There are 9 dots arranged in a square below. Using no more than 4 lines, connect all the dots without lifting your pencil. 8 • Initial & Goal States well defined • Operators: Four connected lines • Representation: Graphical layout • Problem space: all possible lines you can draw O O O O O O O O O 9 O O O O O O O O O 10 • For many problems, the representation may make it easier or harder to solve. – Algebra problems easier as equations – Geometry problems easier graphically – Decision problems easier when relevant information is laid out in a grid 11 One morning, exactly at sunrise, a Buddhist monk began to climb a tall mountain. A narrow path, no more than a foot or two wide, spiraled around the mountain to a glittering temple at the summit. The monk ascended at varying rates of speed, stopping many times along the way to rest and eat dried fruit he carried with him. He reached the temple shortly before sunset. After several days of fasting and meditation he began his journey back along the same path, starting at sunrise and again walking at varying speeds, with many pauses along the way. His average speed descending was, of course,
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This note was uploaded on 12/26/2009 for the course PSYCH 240 taught by Professor Gehring during the Fall '08 term at University of Michigan. | 677.169 | 1 |
Solving Quadratics Comic Book Project
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PRODUCT DESCRIPTION
Students create a comic book introducing advanced mathematical concepts to young adults. The comic book must include examples of solving quadratic equations using all different methods, explanations, and illustrations.
Project details and scoring rubric are included.
Note: If you'd like to purchase 10 of my projects as a bundle and save $20, please go to the following link:
00. | 677.169 | 1 |
In management, how important is it to learn to use mathematics to solve problems?
Mathematics is defined at freedictionary.com as the study of the measurement, relationships, and properties of quantities and sets, using numbers and symbols Mathematics is important to understand most subjects, including science, technology, medicine, economy, business, and finance. Besides solving and presenting equations, it focuses on careful analysis of data and its structure.
Mathematics allows for checks and balances, and interpretation of the economic status of the business; while enhancing reasoning and problem-solving skills.
Mathematical tools are used to map market trends and forecasting; and statistics and probability are used in everyday business and economics.
Financial and Business Mathematics are direct applications to business and economics; applied mathematics such as probability theory and management science, queuing theory, time-series analysis, and linear programming are vital for business.
While there are many programs and utilities, which will help in mathematic application to business, without solid mathematical knowledge, it is impossible to know if results are accurate.
Business and financial success is dependent on business strategies and operations policies, all of which are calculated; therefore, without the knowledge of how to calculate and interpret the information and apply it correctly, the business could be in jeopardy.
How valuable is an MBA degree, without the ability to identify and calculate the cost of a company's capital, its return on investment, gross margin percentage, break-even point, or what percent of a population would be willing to buy its product?
Since mathematics is a cornerstone in Business Management, an MBA degree would be useless without the ability to identify and calculate cost of company capital, return on investment, gross margin percentage, break-even point, and the percentage of the...
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When you have the right math teacher, learning math can be painless and even fun! Let Basic Math and Pre–Algebra Workbook For Dummies teach you how to overcome your fear of math and approach the subject correctly and directly.
Basic Math and Pre-Algebra Workbook For Dummies, 2nd Edition helps take the guesswork out of solving math equations and will have you unraveling the mystery of FOIL in no time. Whether you need to brush up on the basics of addition, subtraction, multiplication, and division or you're ready to tackle algebraic expressions and equations, this handy workbook will demystify math so you can get back to having fun in math class.
Factor fearlessly, conquer the quadratic formula, and solve linear equations There's no doubt that algebra can be easy to some while extremely challenging to others. If you're vexed by variables, Algebra I For Dummies, 2nd Edition provides the plain-English, easy-to-follow guidance you need to get the right solution every time! | 677.169 | 1 |
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Grade 6/7 Math (British Columbia, Canada Curriculum)
Straightforward video lessons explaining Grade 6/7 Math. Great for home school programs, tutoring or extra help at home.Hello and welcome to Math for Grade 6/7. Completing this course will provide you with the basic concepts and skills in preparation for Grade 8 Math. This course is based on the Provincial Curriculum for British Columbia, Canada. Please take the time to apply the skills covered in a way that is meaningful to you personally, by applying the skills in context.
Target audience:
·Home school students or students who are looking for extra support in math such as tutoring. Parents looking to provide homeschooling or tutoring support to their children will want to sign their children up for this course.
What is the course about?
·This course covers all major topics in the Grade 7 Math curriculum for British Columbia, Canada.
·Topics range from Data Analysis to Fractions to Percent to Algebra and Geometry.
Materials included:
·There are 29 video lessons that cover the British Columbia, Canada curriculum that are laid out in a logical and easy to follow format.
·There are 5 practice/sample questions with an answer key for each video lesson topic.
·One sample lesson as to how to apply the concepts taught in a personal way to yourself to make your learning more relevant.
How long to complete course?
·The course could be completed over the course of a week assuming one hour of study per day.
How course is structured:
·The course begins with an introduction explaining how to use the instructional videos and then is grouped into topical themes that progress sequentially.
·Each video lesson builds on the skills from the previous lesson.
Why take the course?
· This course is well suited to home school students or students looking for extra support with the basics of math for grade 6 or 7. Taking your time with this course and applying the skills covered in a relevant way to yourself will result in a strong understanding of the concepts covered in grade 7 math.
Hello and let me introduce myself. My name is Toby Beck and I currently am a teacher in the West Vancouver School District, in British Columbia, Canada. I have a Degree in English, a Degree in Education and a Graduate Diploma in Education Technology. I have taught in West Vancouver, for the past twelve years. I am an avid user of digital tools in my classroom and am excited to use such tools to reach a wider audience. I have presented at numerous Computer Using Educators conferences in the Pacific Northwest over the years. I am a constructivist teacher and often employ Challenge Based Learning or Self Directed Learning in my teaching practice. I am also intrigued as to how the push and pull of nature vs nurture plays out in the classroom and at home for students. Outside of the classroom I play soccer and am a huge Barcelona FC fan. | 677.169 | 1 |
Jessica Weaver
Math 135-24
The Fibonacci Numbers
The Fibonacci number sequence consists of the numbers that are derived by adding
the first two numbers to get a third number and so on. The ratio of any number to the next of
number to the following number
CHAPTER 8
Sequences, Induction,
and Probability
8.1 Sequences and Series
v
Objectives
Find particular terms of sequence from the general
term
Use recursion formulas
Use factorial notation
Use summation notation
Sequences and Series
An sequence is a functi
3.2 Polynomial Functions &
Their Graphs
Objectives
Identify
polynomial functions.
Recognize
characteristics of graphs of polynomials.
Determine
end behavior.
Use
factoring to find zeros of polynomials.
Identify
Use
zeros and their multiplicities.
In
Chapter 2
Functions and Graphs
2.1 Basics of Functions and
their Graphs
Objectivesrelation.
Find the domain and range of a
Determine whether a relations is a function.
Determine is an equation represents a function.
Evaluate a function.
Graph functions by
College Algebra Advice
Showing 1 to 3 of 16
I took this college course in my fall term of high school my senior year.
Course highlights:
Learned basic algebraic functions, the base of what you need for college.
Hours per week:
3-5 hours
Advice for students:
Make sure to keep up the work daily after each class so that you do not fall behind.
Course Term:Fall 2017
Professor:Ms. Pope
Course Tags:Math-heavyGreat Intro to the Subject
Mar 05, 2017
| No strong feelings either way.
This class was tough.
Course Overview:
good thing to have finished, needed for most majors. recommend knocking it out early instead of later. course hero made it a breeze and recommend taking a teacher that has documents on course hero.
Course highlights:
anything from linear equations to mapping just about any point on a graph through equation. math is math is math is math is math. numbers are numbers and variables are places for numbers.
Hours per week:
6-8 hours
Advice for students:
Take a good professor! look into who is going to be teaching you an extremely hard course. it will not be easy if you have a professor that doesn't speak fluent english, etc.
Course Term:Spring 2017
Professor:Brown
Course Required?Yes
Course Tags:Math-heavyMany Small AssignmentsCompetitive Classmates
Jan 10, 2017
| Would not recommend.
This class was tough.
Course Overview:
Linda Suns class is very difficult. If you miss a class you're basically missing a whole section which will affect your test grade. She is also from china and has a very strong accent so if you're not good with accents then I suggest finding a different professor to save your grade.
Course highlights:
I wouldn't say that there were really any highlights to this course and I didn't really learn any new information that I didn't already know.
Hours per week:
6-8 hours
Advice for students:
Find a different professor otherwise your GPA might be hurt by it. The actual course isn't bad but trying to understand what she's saying can be difficult at times. | 677.169 | 1 |
Product Description
▼▲
Jacobs Elementary Algebra is a one-year math curriculum for high school students. Easy to follow instruction will be helpful for both teachers and self-directed students. A simple introduction of new concepts is followed by worked examples and exercise sets. This student text is divided into 17 sections, covering functions and graphs, integers, rational numbers, exponents, polynomials, factoring, fractions, and more. Select solutions are provided in the text, with full answers available in the solutions manual. 380 | 677.169 | 1 |
Maths in Focus - 2 Unit Maths in Society: Book 1
The first book in a new two-volume series providing complete coverage of the new Preliminary course to be introduced in NSW in 2000. Encourages students to learn and practice mechanical skills as well as looking at the way that these skills may be used in everyday life | 677.169 | 1 |
Problems and Exercises in Discrete Mathematics
Many years of practical experience in teaching discrete mathematics form the basis of this text book. Part I contains problems on such topics as Boolean algebra, k-valued logics, graphs and networks, elements of coding theory, automata theory, algorithms theory, combinatorics, Boolean minimization and logical design. The exercises are preceded by ample theoretical background material. For further study the reader is referred to the extensive bibliography. Part II follows the same structure as Part I, and gives helpful hints and solutions. | 677.169 | 1 |
Maplesoft, leading developer of advanced mathematical and analytical software, has released an update to Maple 18.0, the mathematical computing software for education and research in mathematics, engineering, and the sciences.
With Maple 18, Maplesoft offers enhanced tools for developing interactive applications and quizzes, together with additional features to enrich and streamline the student experience. Maple supports the easy creation of interactive Math Apps for use in the classroom and through The M?�bius Project, an initiative from Maplesoft that supports the creation, sharing, and grading of Math Apps. With Maple 18, instructors can take advantage of increased flexibility in the one-step Math App creation tool to quickly create even more complex applications, and easily create randomly generated quizzes for their students. Maple 18.01 Update Details - Significantly enhanced efficiency for many numerical linear algebra computations - New keyboard shortcuts for ?�?Execute All?�? ([Ctrl or Cmd]+[Shift]+[Enter]) and for entering slideshow mode ([F11] or [Cmd]+[F11]) - Improved export of 2-D plots - PDF export improvements for documents that include code edit regions - Enhancements to the limit command About Maplesoft Maplesoft is a leading developer of advanced mathematical and analytical software. Its innovative suite of products harnesses the power of mathematics, providing industry and academia with the most advanced mathematical tools complete with fully integrated numerics and symbolics. If you touch math... you need Maple. Name: Maplesoft Maple Version: 18.01 Home: Interface: english OS: MacOsx Size: 756.0 mb | 677.169 | 1 |
Linear Equations & Inequalities in 1 Variable
This unit focuses on the algebra of equations and inequalities in one variable. Students deepen their understanding of the properties of equality and how the basic mathematical operations can be used to transform expressions, equations, and inequalities. These understandings form a basis for students to solve equations and inequalities in one variable and to rearrange formulas to isolate a specific quantity. Students use algebraic tools and concepts to investigate and interpret equations in a useful way. | 677.169 | 1 |
For a one- or two-term introductory course in discrete
mathematics.
Focused on helping students understand and construct proofs and
expanding their mathematical maturity, this best-selling text is
an accessible introduction to discrete mathematics.
Johnsonbaugh's algorithmic approach emphasizes problem-solving
techniques. The Seventh Edition reflects user and
reviewer feedback on both content and organization.
Longlisted for the National Book AwardNew York
Times Bestseller
A former Wall Street quant sounds an alarm on the mathematical
models that pervade modern life — and threaten to rip apart
our social fabric
We live in the age of the algorithm. Increasingly, the decisions
that affect our lives—where we go to school, whether we get a car
loan, how much we pay for health insurance—are being made not by
humans, but by mathematical models. In... more...
Embrace the new world of fiance by leveraging the power of
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Set up your own wallet, buy and sell Bitcoin, and execute
custom transactions on the Blockchain
Leverage the power of Bitcoin to reduce transaction costs and
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A practical step-by-step guide to break down the Bitcoin
technology to ensure safe transactions
Who This Book Is For... more...
Compelling tips and tricks to improve your mental skillsDon't you wish you were just a little smarter? Ron and Marty Hale-Evans can help with a vast array of witty, practical techniques that tune your brain to peak performance. Founded in current research, Mindhacker features 60 tips, tricks, and games to develop your mental potential. This accessible compilation helps improve memory, accelerate learning, manage time, spark creativity, hone math and... more...
Are you the innovative type, the cook who marches to a different drummer -- used to expressing your creativity instead of just following recipes? Are you interested in the science behind what happens to food while it's cooking? Do you want to learn what makes a recipe work so you can improvise and create your own unique dish? More than just a cookbook, Cooking for Geeks applies your curiosity to discovery, inspiration, and invention in the kitchen.... more...
Writing requirements is one of the core competencies for anyone
in an organization responsible for defining future Information
Technology (IT) applications. However, nearly every independently
executed, root-cause analysis of IT project problems and failures
in the past half-century have identified "misunderstood or
incomplete requirements" as the primary cause. This has made
writing requirements the bane of many projects. The real... more...
This valuable problem-solving guide puts in your hands the power
you need today to resolve faults in and coax peak performance from
new, experimental, or just plain temperamental circuits. Written by
one of the bestselling practical electronics authors of all
timeÑhis books have sold more than 2 million copies in 9 languages
worldwideÑThe Electronic Troubleshooting Handbook, Volume I, gives
you full descriptions of the operation of important... more...
Containing papers presented at COMPRAIL 2006, this book represents
the latest research, development and application of computers to
the management, design, manufacture and operations of railways and
other passenger, freight and transit systems. The conference
attracted a large number of papers, divided into the following
sections: Planning; Safety; Passenger interface systems; Decision
support systems; Computer techniques; Converting metros to... more... | 677.169 | 1 |
The best way to penetrate the subtleties of the theory of integration is by solving problems. This book, like its two predecessors, is a wonderful source of interesting and challenging problems. As a resource, it is unequaled. It offers a much richer selection than is found in any current textbook. Moreover, the book includes a complete set of solutions. This is the third volume of Problems in Mathematical Analysis. The topic here is integration for real functions of one real variable. The first chapter is devoted to the Riemann and the Riemann-Stieltjes integrals. Chapter 2 deals with Lebesgue measure and integration. The authors include some famous, and some not so famous, inequalities related to Riemann integration. Many of the problems for Lebesgue integration concern convergence theorems and the interchange of limits and integrals. The book closes with a section on Fourier series, with a concentration on Fourier coefficients of functions from particular classes and on basic theorems for convergence of Fourier series. The book is mainly geared toward students studying the basic principles of analysis. However, given its selection of problems, organization, and level, it would be an ideal choice for tutorial or problem-solving seminars, particularly those geared toward the Putnam exam. It is also suitable for self-study. The presentation of the material is designed to help student comprehension, to encourage them to ask their own questions, and to start research. The collection of problems will also help teachers who wish to incorporate problems into their lectures. The problems are grouped into sections according to the methods of solution. Solutions for the problems are provided. Problems in Mathematical Analysis I and II are available as Volumes 4 and 12 in the AMS series, Student Mathematical Library15633 27206821832981
Descripción American Mathematical Society. Estado de conservación: New. 082183298218329819901219990 | 677.169 | 1 |
A comprehensive set of computer exercises of varying levels of difficulty covering the fundamentals of signals and systems. The exercises require the reader to compare answers they compute in MATLAB #65533; with results and predictions made based on their understanding of material. Chapter covered include Signals and Systems; Linear Time-Invariant Systems; Fourier Series Representation of Periodic Signals; The Continuous-Time Fourier Transform; The Discrete-Time Fourier Transform; Time and Frequency Analysis of Signals and Systems; Sampling; Communications Systems; The Laplace Transform; The z-Transform; Feedback Systems. For readers interested in signals and linear systems.
21 Day Unconditional Guarantee
REVIEWS for Computer Explorations in Signals and Systems Using | 677.169 | 1 |
ISBN 9789384905125
ISBN-10
9384905127
Binding
Paperback
Language
(English)
Subject
Entrance Exam Preparation
This is a unique book that comprises all mathematical formulae at one place along with the study of all variety of Graphs encompassing the mathematics syllabus. The 1st part book covers 27 chapters just on formulae. The author has left no stone unturned in covering the exhaustive list of formulae. The second part of the book focuses on the concepts and numerical problems on Graphs. All the graphs have been covered and possible questions have been discussed. With 10 chapters on the concepts and numerical of Graphs, along with detailed graphical solutions, the book will not only help the young Engineering and Olympiad aspirants but shall also appeal to the Mathematics lovers and mentors of Engineering entrance examinations all over the country. It has been often found that many students struggle with formulae in Mathematics. On the other hand, there have been innumerable instances where just knowing a formula has triggered solution for a complex problem just in the blink of an eye. The book aims at a thorough understanding of the fundamentals of Graphs and learning useful problem solving techniques that will definitely see a candidate sail through tougher questions. A quick glimpse on the pattern of questions asked in IIT JEE and other leading exams like BITSAT, VITEEE, etc. will clearly show how tricky questions can become a problem if one is not well adept with the concept of Graphs. The book will be of immense help for all those students who are aiming for top ranks in various Engineering entrance examinations provided they solve each and every sum and go through all the formulae. | 677.169 | 1 |
Slide 3 3 In my view, (Vote for ONE:) In helping students succeed in first-year chem, the math background of entering students is: A. A major problem B. A minor problem C. Not a problem Good Afternoon. My slides today, with plenty of references, are posted on the web. I will cover just selected slides, but I will put up the web address at the end where you may view the slides with additional detail if you find any of them interesting. Could I ask: How many of you frequently teach any form of first year college chemistry? Anyone teach High School or AP chemistry? My focus today is going to be on calculations in those courses. I'd like to start by asking you to please read the question on this slide -- then be ready to vote for A, B, or C. Ready? How many of you would vote raise your hand. for A, ________ B? ______ C? _______ (Voting audience was 100% major)
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Slide 4 4 Vote for ONE: To solve calculations in first-year Chem, it is most important for students to have background knowledge in A. Use of a calculator B. The theory of mathematics C. Fundamentals of math computation Please read and be ready to vote. We ask students To be able to solve like THIS …..
Slide 5 5-- Zumdahl, 5 th edition Or THIS
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Slide 8 8 Vote for ONE: To solve calculations in General Chem, it is most important for students to have background knowledge in A. Use of a calculator B. The theory of mathematics C. Fundamentals of math computation All of these are important. But if you had to pick ONE, Which is most important? How many vote for A: ___________ B: ___________ C: _________ (Voting audience was 100% math computation)
Slide 10 10 Virginia Math Results: VA all students Grade 9 1998 1999 2000 2001 2002 Total Math 54 55 55 55 55 • Stanford 9 standardized test given statewide • National percentile average = 50 on 1995 norms My goal today is to look at the evidence. Is math preparation is a problem? If so, WHY? And how can we fix the problem? About 8 years ago, I was asked to represent my Virginia faculty organization On a task force looking at the issue Of why so many students entering college were needing remediation. Being a chem instructor, I thought the problem was the K-12 math programs But when I looked at the standardized tests, in "Total Math" Virginia scores were above average – average being the 50th percentile -- and steady. (point) But when I looked in the report detail, I found that on the test VA was using
Slide 11 11 Two sub tests were reported described as • "Math Problem Solving , which focuses on reasoning skills, and • Math Procedures , which measures the student's facility with computation ." Talk them
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GeoGebra Means No Excuses For Not Studying Maths This Summer
There are lots of summer bridging activities that promise basic remediation and skill building over the holiday months, but very few of them combine as many maths courses as GeoGebra. And thanks to its recent update, the software offers even more for students, especially since it includes statistics, graphing practice, spreadsheet building, and more.
Interesting, GeoGebra was designed and intended for advanced math users, not casual students, but the full complement of instructional and evaluational tools makes it a great summer practice software. Even better the tools are useful enough that students can actually work at their own pace and self-teach many new concepts.
Best of all, the geometry and algebra instruction is in line with the content students are already being introduced to in school and it's available open source for non-commercial use. That means classroom teachers aren't licensed to use it for whole group instruction, but teachers and parents can use it or recommend it for home use. Teachers who do choose to go for the license will find the authoring tool highly useful as it lets the user create interactive and scorable lessons.
"GeoGebra is really for math experts and is a complex application aimed at users who are comfortable with difficult math, but it does have advantages over other applications in that GeoGebra provides multiple representations of objects that are all dynamically linked. Basically, the idea is to connect geometric, algebraic, and numeric representations in an interactive way. This can be accomplished with points, vectors, lines, and conic sections. With GeoGebra you can directly enter and manipulate equations and coordinates, thereby enabling you to plot functions; work with sliders to investigate parameters; find symbolic derivatives; and use commands such as Root or Sequence."
To check out GeoGebra for yourself and see how it can boost a student's maths understanding, download the update from FileHippo by clicking HERE. | 677.169 | 1 |
TI-Interactive!
Students examine two applications of the TI Interacive! In this secondary mathematics lesson students examine the slope as a characteristic property of a line. In the second demonstration, students examine matrices and matrix operations in a problem solving context involving insect population. | 677.169 | 1 |
TI-84: Typing in Functions to Graph
Word Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.76 MB | 4 pages
PRODUCT DESCRIPTION
Students explore the TI-83/84 on their own by following the step-by-step instructions.
This intro to the graphing calculator focuses on typing in common functions correctly and exploring how to use the GRAPH screen effectively
Many of my students are intimidated by the many buttons of the TI-83/84, especially because it is so much less intuitive than their calculator apps on their phones. They've found this to be a very useful exercise and I have them keep it in the front of their binders | 677.169 | 1 |
for a one-term course in elementary algebra is intended for students with a firm background in arithmetic. Problem solving and applications are emphasized. Coverage progresses from algebraic expressions and equations and inequalities toMore...
This text for a one-term course in elementary algebra is intended for students with a firm background in arithmetic. Problem solving and applications are emphasized. Coverage progresses from algebraic expressions and equations and inequalities to quadratic equations. New to this edition are sections on connecting concepts, study tips, and exercises designed to foster intuitive problem solving for particular types of problems. Other new material includes a "gentle" introduction to interpolation and extrapolation, and expanded material on rates and units. Bittinger teaches at Indiana University. Ellenbogen teaches at Community College of Vermont. c. Book News Inc | 677.169 | 1 |
purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering.
Data science courses contain math—no avoiding that! This course is designed to teach learners the basic math you will need in order to be successful in almost any data science math course and was created for learners who have basic math skills but may not have taken algebra or pre-calculus. Data Science Math Skills introduces the core math that data science is built upon, with no extra complexity, introducing unfamiliar ideas and math symbols one-at-a-time.
In this course, you will learn the science behind how digital images and video are made, altered, stored, and used. We will look at the vast world of digital imaging, from how computers and digital cameras form images to how digital special effects are used in Hollywood movies to how the Mars Rover was able to send photographs across millions of miles of space.
This course is an introduction to the finite element method as applicable to a range of problems in physics and engineering sciences. The treatment is mathematical, but only for the purpose of clarifying the formulation. The emphasis is on coding up the formulations in a modern, open-source environment that can be expanded to other applications, subsequently | 677.169 | 1 |
Interests
Browse by
Delving deeper
Steve Phelps and Michael Todd Edwards
New Life for an Old Topic: Completing the Square Using Technology
M
athematics teaching has always been a curious blend of the old and the new. As the use of technology becomes more commonplace in school classrooms, this blend becomes even more pronounced. When teachers and students revisit traditional topics using technology, they are afforded opportunities to connect mathematical ideas in powerful, previously unimagined ways. The National Council of Teachers of Mathematics (NCTM) captures the importance of connections clearly in its Principles and Standards for School Mathematics (2000): "The notion that mathematical ideas are connected should permeate the school
mathematics experience at all levels. As students progress through their school mathematics experience, their ability to see the same mathematical structure in seemingly different settings should increase" (p. 64). A recent technology-oriented investigation led us to surprising connections among quadratics, mathematical envelopes, tangent lines, and tangent parabolas. In particular, our work with the TI-Nspire computer algebra system (CAS) enabled our students to generate conjectures and test hypotheses in ways not possible with pencil and paper alone while providing new insights into traditional, skill-oriented topics.
"Delving Deeper" offers a forum for classroom teachers to share the mathematics from their own work with the journal's readership; it appears in every issue of Mathematics Teacher. Manuscripts for the department should be submitted via For more background information on the department and guidelines for submitting a manuscript, visit content.aspx?id=10440#delving. Edited by J. Kevin Colligan, jkcolligan@verizon.net RABA Center of SRA International, Columbia, MD 20145 Dan Kalman, kalman@american.edu American University, Washington, DC 20016 Virginia Stallings, vstalli@american.edu American University, Washington, DC 20016 Jeffrey Wanko, wankojj@muohio.edu Miami University, Oxford, OH 45056
COMPLETING THE SQUARE WITH TILES: AN OVERVIEW
At a local teaching conference focusing on the use of interactive whiteboards (IWB), our interest was piqued by the speaker's discussion of a seemingly routine topic—completing the square. The presentation began innocently enough, with the speaker demonstrating the drawing capabilities of IWB software. As he represented algebraic expressions with virtual Algebra Tiles (Bruner 1966; Dienes 1960; ETA/Cuisenaire 2008; Picciotto 2008), the speaker used tiles of three different dimensions—with areas of 1, x, and x2 square units (Corn 2004). The shapes (see fig. 1) were familiar to many in the audience. The speaker then combined tiles to represent more complicated algebraic expressions. For
instance, the trinomial x2 + 6x + 1 was represented as shown in figure 2. By manipulating the tiles within the IWB environment, the presenter demonstrated several examples of completing the square, such as the one illustrated in figure 3. The arrangement of tiles in figure 3a suggests that 8 units are needed to complete the square for the expression x2 + 6x + 1 (Corn 2004). The expression that results from this addition—namely, x2 + 6x + 9—is a perfect square trinomial, expressible as the square of a binomial: (x + 3)2 (see fig. 3b). The speaker shared several more examples of this sort, each time completing the square by splitting the x tiles into two piles with the same number of pieces—placing one pile in a horizontal "stack" below the x2 piece and the other pile in a vertical "stack" to the right of the x2 piece. While discussing pedagogical advantages and limitations of such an approach, the presenter made an off-hand comment that intrigued us: He noted that tiles were not helpful when completing the square with trinomials containing odd linear coefficients (such as x2 + 5x + 1) because odd numbers of x tiles cannot be split into two piles with the same number of pieces. Although no one questioned this claim, the authors sat through the remainder of the session wondering whether this statement were true. At the conclusion of the presentation, we sat in the conference center lobby, constructing squares with presentation software on our laptops. Rather than splitting the x blocks into two equal piles, we added blocks of various dimensions onto our initial construction to complete the square. Within five minutes, we had generated a family of expressions that could be added to x2 + 5x + 1 to complete squares.
(a)
(b)
Fig. 4 There are many ways to complete a square begun with x2 + 5x + 1; two are shown here.
Two examples of completed squares are shown in figure 4, one completed by adding x + 8 and the other completed by adding 3x + 15. We now explore unexpected mathematical connections fostered by our new interpretation of completing a square, in particular the following: • The family of linear functions that complete squares for various trinomials (i.e., square completers) and the relationship of these functions to the original (i.e., seed) trinomial • Connections between square-completer families and the traditional completing-the-square algorithm Technology plays a central role throughout our exploration, informed by concrete examples. For instance, IWB and virtual Algebra Tiles motivated our initial questions about alternative conceptions of completing the square. Subsequent explorations involving more sophisticated mathematical
Vol. 104, No. 3 • October 2010 | Mathematics Teacher 231
simulation and data analysis are fostered using TINspire CAS and dynamic geometry software.
FAMILIES OF SQUARE COMPLETERS
Consider our previous example involving x2 + 5x + 1. Students are typically taught that exactly one expression will complete the square for such a trinomial when in fact any one of an entire family of linear expressions may be added to generate perfect square trinomials. Teachers (and their students) typically add only a constant term when given the choice because adding additional variables complicates algebraic expressions when solving equations. Restricting one's attention to square completers with no variables, although pedagogically justifiable, ignores rich and unexpected mathematical connections. By exploring the fam-
ily of expressions that, when added to the original expression, create a perfect square polynomial, we provide students with opportunities to reconsider traditional content using inquiry-oriented, technology-rich perspectives. For instance, figure 4 illustrates that adding x + 8 to x2 + 5x + 1 generates the perfect square trinomial x2 + 6x + 9 = (x + 3)2. Likewise, adding 3x + 15 yields x2 + 8x + 16 = (x + 4)2. In fact, an infinite number of different expressions may be added to x2 + 5x + 1 to generate perfect square trinomials. We define the nth square completer of p(x) as the expression one adds to p(x) to yield the perfect square trinomial (x + n)2. Table 1 illustrates family members for p(x) = x2 + 5x + 1 for various n. Note that the second differences from entries in the second column are constant, suggesting a quadratic relationship between n and the nth square completer. Later we discuss a CAS-assisted derivation of a general formula for the nth square completer for arbitrary p(x).
INITIAL INVESTIGATIONS WITH CAS
To gain a more thorough understanding of the family of square completers, we constructed a multipage TI-Nspire document to generate family members for a given trinomial p(x) automatically. On the first page of the document, students manipulated sliders to change the coefficients of the seed trinomial: ax2 + bx + c. On the second page, algebraic expressions for various family members were generated within a CAS-enhanced spreadsheet. Screen shots from the first two pages of the TI-Nspire document are highlighted in figure 5. As figure 5a suggests, coefficients of the seed trinomial (a, b, and c) controlled by the sliders in a TI-Nspire Graphs & Geometry window were linked to a CAS-enhanced spreadsheet, as shown in figure 5b. Algebraic expressions for square completers were calculated in column D of the spreadsheet and graphed back in the Graphs & Geometry window. In figure 5a, the envelope formed by family members appeared to be quadratic. To explore this conjecture more rigorously, students constructed intersection points among consecutive
family members and then generated a quadratic function to fit these ordered pairs (see fig. 6). The white dots in figure 6a depict intersections of nth and (n + 1)st square completer graphs. The bold curve was generated by fitting a quadratic function to these intersections. As students manipulated values of a, b, and c with sliders, the envelope and associated quadratic regression were updated dynamically. This feature enabled students to look for connections between various seed polynomials and corresponding envelopes of square completers such as those provided in table 2. Because of the amount of time required to generate such graphs by hand, such an exploration would have been highly impractical for our students without technology. Clearly, the entries in table 2 suggested a relationship between the equations of quadratic envelopes and seed polynomials—namely, seed p(x) = x2 + bx + c appeared to have envelope –x2 – bx – c + 0.25. Although we were not certain whether this conjecture held in general, the cases warranted further investigation.
Fig. 8 Students can calculate qn(x) using the TI-Nspire CAS.
CONNECTIONS WITH TANGENT CURVES
Because the sign of the coefficients of the seed and the envelope appeared to be opposites, we modified our TI-Nspire document to graph the opposites of square completers of p(x), defining qn(x) as the opposite of the nth square completer for a given p(x). Explicitly, qn(x) = p(x) – (x + n)2. The modified TINspire document simultaneously graphed qn(x) for various n and p(x). A sample graph for p(x) = x2 + 5x + 1 and qn(x) with –20 < n < 20 is shown in figure 7. By inspection, students conjectured that the graph of qn(x) was tangent to p(x) for all n. To determine whether such a hypothesis were true, we first constructed a formula for qn(x) given p(x) = x2 + bx + c: qn(x) = p(x) – (x + n)2 by definition = x2 + bx + c – (x2 + 2nx + n2) by substitution = (b – 2n)x + (c – n2) algebraic simplification
Several students performed similar calculations in a step-by-step fashion using CAS. Such an approach is highlighted in figure 8. Insight into the general formula qn(x) was provided through analysis of a concrete representation
Vol. 104, No. 3 • October 2010 | Mathematics Teacher 233
of (x – n)2 – (x2 + bx + c) with Algebra Tiles (see fig. 9). The gray and white regions—that is, (2n – b)x + n2 – c—represent the nth square completer of x2 + bx + c—that is, – qn(x). With qn(x) defined as –(2n – b)x – (n2 – c) for arbitrary p(x) = x2 + bx + c, students used CAS to verify that the graph of qn(x) was tangent to p(x) for all n. In using CAS to verify that each square-completer family member was tangent, we first defined p(x) as the general trinomial x2 + bx + c and qn(x) as –(2n – b)x – (n2 – c). Solving the equation p(x) = qn(x), we determined intersection points of x2 + bx + c and qn(x), the nth member of the square completers. Solutions to the equation, as calculated with TI-Nspire CAS, are shown in figure 10. From figure 10, we saw that there was exactly one solution to the equation—namely, x = –n. This result strongly suggested that every member of the square-completer family qn(x) was tangent to p(x) at x = –n. Establishing with certainty that qn(x) was tangent to p(x) required little more than elementary calculus. Noticing that the slope of qn(x) is –(2n – b), students noted that this is precisely the value of p′(x) = 2x + b evaluated at x = –n. Hence, they correctly concluded that qn(x) and the tangent line to
p(x) at x = –n were parallel. Because qn(x) and p(x) have only one point in common, students concluded that qn(x) was the tangent line to p(x) at x = –n.
GRAPHICAL INTERPRETATION OF SQUARE COMPLETERS
How do p(x), qn(x), and the completed square (x + n)2 fit together graphically? As an example, we asked students to consider p(x) = x2 + 5x + 1 and the specific square completers q-1(x) and q2(x) (see fig. 11). In figure 11a, the vertex of the completed square is below the point of tangency; in figure 11b, the vertex is above the point of tangency. Square completers q-1(x) and q2(x) are members of qn(x). Students were able to recognize that the plotted functions were tangent to p(x) at x = 1 and x = –2, respectively. Moreover, they noted that each member of qn(x) led to a completed square trinomial whose vertex was located on the x-axis directly above (or below) the point of tangency. By completing the square in this manner—that is, by adding on the appropriate Algebra Tiles—we effectively translated p(x) so that the vertex was vertically aligned with the point of tangency and on the x-axis.
COMPLETING THE SQUARE REVISITED
At this juncture, it was instructive to connect our new observation back to more traditional notions of completing the square—that is, taking half the linear coefficient, squaring, and adding. In the traditionally understood method of completing the square, students add nothing but unit blocks to the initial trinomial; they do not add x blocks. Algebraically, this approach implies that qn(x) = k, a constant function. Hence, when we complete the square in the conventional sense, the coefficient of the linear term of qn(x) is zero—in other words, –(2n – b) = 0, implying that n = b/2. We know that qn(x) is tangent to p(x) at x = –n, which implies that, in traditional completing-the-square situations, qn(x) is a horizontal line tangent to p(x) at the vertex of the trinomial. This observation led us to a novel interpretation of completing the square. Geometrically speaking, by completing the square, one translates p(x) vertically so that its vertex lies on the x-axis. We encouraged students to see that completing the square for p(x) = x2 + 5x + 1 involves steps similar to the following:
Step 1: p( x ) = x 2 + 5x + 1 2 2 2 Step 2: p( x ) + 2.5 − 1 = x + 5x + 1 + 2.5 − 1 Step 3: p( x ) + 5.25 25 = x 2 + 5x + 6.2 25 5
Fig. 10 Points of intersection for p(x) and qn(x) were determined using the CAS.
(
)
(a)
(b)
Step 4: p( x ) + 5.25 25 = x + 2.5
Fig. 11 Note the location of the vertex of the completed square trinomial in relation to the points of tangency of the square completers in both (a) and (b).
Step 5: p( x ) = x + 2. .5 5 − 5. .25
(
(
)
)
2
2
234 MATHEMATICS TEACHER | Vol. 104, No. 3 • October 2010
(a)
(b)
Fig. 12 The traditional completing-the-square algorithm can be given a graphical interpretation.
Fig. 13 Changing the coefficient of the leading term to a number other than 1 yields similar results.
In step 4 of the calculations above, the seed function p(x) is translated vertically +5.25 units. In particular, the vertex of p(x) is translated from (–2.5, –5.25) to (–2.5, 0). We represented this idea graphically by plotting p(x) and q2.5(x) simultaneously with the TI-Nspire CAS (see fig. 12). Recalling previous discussions of the square completers q-1(x) and q2(x), students viewed the traditional method of completing the square as a special case of a much larger completing-the-square theory.
FOUR SUGGESTIONS FOR FURTHER RESEARCH
There are many fruitful paths to explore when we combine the concrete visual clues of Algebra Tiles with the rich technology applications found in the TI-Nspire CAS. The following four investigations offer interested readers a glimpse of the ways a technologically enhanced Algebra Tiles view of polynomials can lead students to previously unforeseen connections. These investigations are readily accessible to students in first-year algebra through precalculus.
Fig. 14 Square completers for a linear function form a family of tangent parabolas.
1. Quadratics with Leading Coefficients Other Than 1
Consider p(x) = 2x2 + 5x + 2. The traditional algorithm for this case would have students first divide through by 2 (the leading coefficient) to make the leading coefficient 1 and then proceed as the traditional algorithm dictates—that is, take half the linear coefficient, square it, and add. Instead, we look for opportunities to add on tiles to complete a square. As we have established, many combinations of Algebra Tiles would suffice to complete a square; for instance, adding 2x2 + 3x + 2 would work, as would adding on 2x2 + 11x + 14. Graphing p(x) along with the family of the opposites of all possible square completers—that is, qn(x)—produces the graph in figure 13. (a) (b)
Fig. 15 3D Algebra Blocks (a) can be used to model completion of a cube; the graph (b) shows p(x) and some cube completers.
2. Completing the Square for Linear Equations
Begin with a single x block and 3 unit blocks and model x + 3. What Algebra Tiles would one need to add to complete the square? One could add x2 + 3x + 1 or even x2 + 9x + 22. Either way, if one examines the family of square completers, the result is a family of parabolas tangent to the line y = x + 3 (see fig. 14).
Vol. 104, No. 3 • October 2010 | Mathematics Teacher 235
and their manipulation more concretely. Moreover, the use of technology afforded both us and our students opportunities to connect mathematical ideas in powerful, previously unimagined ways. The process of completing the square is transformed from a series of algebraic steps to be memorized, executed, and quickly forgotten to a geometric process that reveals the vertex of a parabola through geometric translation.
Fig. 16 Completing the square for the equation of a circle produces graphs like this one.
3. Completing Cubes
Similarly, 3D Algebra Blocks can be used to model p(x) = x3 + 2x2 + x (see fig. 15). What would be required to complete the cube? As before, we could add x2 + 2x + 1, or 4x2 + 11x + 8, or 7x2 + 26x + 27, to name just a few cube completers. Graphing the opposite of these yields a family of parabolas in which some are tangent although each member intersects the cubic p(x) at only one point. Further investigation will yield connections between these cube completers and Taylor polynomials.
4. Squaring the Circle
Experienced Algebra Tiles users will recognize both x and y blocks; many also are comfortable with various methods of modeling negative integers. These methods can be used to model quadratics that represent conics. For instance, consider the collection of x2 + y2 – 4 tiles used to model the circle x2 + y2 = 4. In general, to complete the square (x + y + n)2, one could add 2xy + 2x + 2y + 5, or 2xy + 4x + 4y + 8, or 2xy + 16x + 16y + 68. Graphing the circle and the family of square completers for the circle produces a family of hyperbolas, some of which appear to be tangent to the circle (see fig. 16).
SUMMARY
We have explored alternative interpretations of completing the square along with the mathematical implications of such interpretations. Throughout our investigations, technology—specifically, Algebra Tiles—has played a central role. From generating initial questions with interactive whiteboards to constructing and testing hypotheses with students using TI-Nspire CAS and dynamic geometry software, technology has made various facets of this investigation accessible to secondary school students. In particular, the tools provide students with a means of considering abstract mathematical objects such as symbolic polynomials
236 Mathematics Teacher | Vol. 104, No. 3 • October 2010 STEVE PHELPS, sphelps@ madeiracityschools.org, teaches geometry at Madeira High School in Cincinnati, Ohio. MICHAEL TODD EDWARDS, edwardm2@muohio.edu, is an associate professor in the Department of Teacher Education at Miami University in Oxford, Ohio. Both are interested in the use of technology in teaching and learning mathematics, with particular emphasis on computer algebra systems, dynamic geometry software, and pencasting technology. | 677.169 | 1 |
Reviewed OER Library
CCGPS Algebra 1
Note that this resource was reviewed during the Spring 2013 review period. The resource may or may not have been updated since the review. Check with the content creator to see if there is a more recent version available.
Intended Audience
Grade: 9 Subject: Algebra Scope: full course Duration: school year
License
License: CC BY-NC-SA 3.0
Note: This course contains content produced by other organizations which may use a different open license. Please confirm the license status of these third-party resources before reusing them.
Access restrictions: no
Format and Features
Format: Web
Includes Video
Includes Interactives
Resource is Printable
Common Core
No standards correlations available.
Professional development
No professional development is available from the developer.
Review
Background from OER Project Review Team
Georgia Virtual Learning is part of the Georgia Department of Education. This self-directed online Algebra 1 course was designed to align with the Common Core Georgia Performance Standards. This should factor into the viewer's analysis of the review results. Existing OER resources are combined with created material into a structured course format. There are quite a few links to content which is limited to individual, not institutional, use on a free basis. Though designed to take advantage of digital media, a print option is available to reproduce much of the content.
Amount of work required to bring into CCSS alignment (average score): Extreme (0.8)
The Georgia Virtual Learning materials are well organized, clearly presented, and allow for students to successfully navigate an Algebra 1 course on their own. The inclusion of the unit projects increase the level of rigor and allows students an opportunity to persevere through more challenging problems while communicating their understanding. The materials still lean toward teacher centered instruction, but students are able to choose a variety of options in each lesson for assistance. There are videos they can watch, interactive practice problems, activities that allow for immediate feedback and formative assessments. Students can choose to watch the video or read the material on their own. They can do as many or as few practice problems within the lesson, and have the opportunity to retake their formative quizzes. The presentation is engaging.
These materials do not include adequate focus on quadratic functions and polynomials, as described by PARCC. There may not be enough independent practice for some students to achieve mastery, and there could be more opportunities for students to communicate their understanding. Students aren't allowed the opportunity to explore or discover the mathematics very often in this course. However, the explanations and examples are clear. This text could be used for independent study or as a remediation tool for students that are struggling in a math classroom.
I would use these materials in my classroom: Disagree (On a 4 point scale: Strongly Disagree, Disagree, Agree, or Strongly Agree)
The materials seem to be designed primarily for a student to interact with on the computer, rather than for a teacher to use to instruct a group of students. In every lesson, there is a 5-8 min video explanation of the content, as well as a written elaboration with examples. There are exercises and apps that students can work with, and check their understanding.
There is no support for a teacher to facilitate productive classroom activity and discourse. There is little attention to mathematical reasoning and no direct attention to the Standards for Mathematical Practice.
Seems to be an integrated curriculum even though it is labeled Algebra 1. Includes a unit on Transformations in the Coordinate Plane and one on Connecting Algebra and Geometry through the Coordinate Plane. The rate of alignment on the Algebra 1 CCSS Worksheet is 15/21 standards, and on the Integrated 1 CCSS Worksheet is 15/19 standards.
Missing from Algebra 1 standards: Use Properties of rational and irrational numbers, Write expressions in equivalent forms to solve problems, Arithmetic and zeros of polynomials, solve simple rational and radical equations in one variable, solve quadratic equations in one variable, limits Analyzing functions using different representations to linear and exponential functions only, and skips writing functions in different but equivalent forms to reveal and explain properties of the function.
The alignment to CCSS standards that are identified as present is extremely weak for a number of reasons.
The curriculum is organized to provide information to students, and does not provide opportunities or support for students to make sense of content through active engagement in thinking and talking about mathematics.
The course provides content aligned to grade 8 standards, and does not reach level of rigor for Alg1 (?)
Primarily asks students to produce answers to problems, and does not provide much opportunity to produce arguments, explanations, diagrams, models, etc.
There are errors and inconsistent use of notation and language. Example: In Unit 2, function notation is not introduced until lesson 11. However, in Lesson 6, students are asked to create equations using function notation.
Tricks! Example: Unit2, Lesson 7 "Finding Intercepts for Standard Form Equations Using the Cover- Up Method".
There is not a balance between conceptual understanding, procedural fluency and application. There is an emphasis on procedural fluency.
I would use these materials in my classroom: Disagree (On a 4 point scale: Strongly Disagree, Disagree, Agree, or Strongly Agree)
This curriculum is divided into units that attempt to provide a coherent framework for the course. However the design of the curriculum places the strongest focus on major content such that the coherence suffers. Coherence emerges as a strength in the lessons focused on below-grade content but a deficiency in lessons with at-grade-level content. This also originates from an unbalanced approach to rigor. Rigor in application is incorporated in most lessons while understanding is often overlooked. This is most apparent in the assessments with most mathematical problems focusing on rigor in fluency and application. I would use this curriculum when I am in need of a resource for a specific topic that provides my students with a skills-based learning experience involving context.
An example of the struggle between focus and coherence is the unit titled "Understanding Linear and Exponential Relationships" where the "essential questions" guiding the unit do not include the terms linear or exponential. Instead the unit's essential questions are focused on understanding functions and their characteristics. While the unit extensively addresses functions at a basic level, the unit contains only one lesson involving exponential relationships (or functions) and never makes a connection between linear and exponential relationships.
The mathematical practices are never identified and do not appear to have been included in the focus of the design of the curriculum.
The most well designed lessons have a combination of videos that serve to warm-up to, present, show examples, practice, and review the topic at hand. The lesson titled "Graphs of Equations and Functions" in the third unit is one such lesson. It is in these lessons that the students receive the most engaging of the various styles of direct instruction and have access to enough practice to encourage fluency. Less than one-third of lessons have this feature. In fact many lessons, particularly in the second half of the course, have little to no interactivity or practice/assessment. Examples include the "Histograms" lesson in the "Describing Data" unit and the "Reflections Are Isometries" lesson in the "Transformations in the Coordinate Plane" unit.
The most pressing area for improvement is the need for this curriculum to address all common core standards found in the Algebra 1 course, including the mathematical practices. For example, quadratic functions are completely absent from this course.
I would use these materials in my classroom: Agree (On a 4 point scale: Strongly Disagree, Disagree, Agree, or Strongly Agree)
The Georgia Virtual Algebra appears to be the work of a committee hastily convened and given a week during the summer to prepare an interactive algebra curriculum. There is no consistent layout of material on each page and no consistent type fonts are used. The video links have been assembled from multiple places. Khan Academy and the Monterey Institute for Technology and Education lessons are used heavily. And the videos in the Monterey Institute's lessons are from Khan Academy. This is not necessarily a bad thing but care needs to be taken that more than just the titles are aligned. The approach in the lesson, in the examples, in the video, in the problems and in the solutions needs to be consistent. Students cannot be expected to learn independently, or in class for that matter, if the approaches, the techniques, vary widely. Apparently there was not sufficient time for proofreading or Beta testing as links are missing, some links don't work, and solutions are missing, or in the worst case, are incorrect. I did not find copyright dates so I can only suspect that the alignment to CCSS was done after the fact. Titles may match but content does not align with the CCSS. The level of rigor is inconsistent. Students are asked to mainly demonstrate their understanding in a series of multiple choice problems or short-answer problems. This is not interactive work. Most work was of the "plug and chug" variety. Incorrect answers simply direct the student to go back and work the same problem over. There is no diagnostic component. I did not see any opportunity for students to defend their own work or critique the work of others. Rather than rework this material it should be completely written with the CCSS in mind. I would hope that far better lessons could be developed in house rather than relying on mostly web-based materials that were written and developed before the CCSS existed.
I would use these materials in my classroom: Strongly Disagree (On a 4 point scale: Strongly Disagree, Disagree, Agree, or Strongly Agree)
It is my understanding that the Georgia Virtual curriculum is intended more for credit recovery, with a focus on independent student work, rather than a comprehensive teacher-led Algebra course. As credit recovery, I feel the curriculum could be reasonably effective. However, such is not the case for implementation in an Algebra classroom.
The strength of the Georgia Virtual curriculum is its technology integration – there are numerous presentations and links to help develop student understanding of the content standards. However, there are serious gaps in the content covered – there are no modules covering quadratics, polynomials, or rational expressions. The lesson content is also not exemplary: some topics are developed in context, but the majority of material is developed through procedural examples. Furthermore, the procedural examples seem particularly wordy for an Algebra 1 student. Finally, the coherence within the modules is lacking as the topics are sometimes seamless then make huge leaps. For instance, the module titled "Understanding Linear and Exponential Relationships" has 14 solid lessons on linear relationships then suddenly there are three lessons that aren't well tied to the linear relationships. These less coherent lessons include maximum and minimum values with non-linear functions, increasing and decreasing intervals for various functions (written in interval notation), and exponential functions. With one lesson on exponential functions the unit title seems a misnomer.
Modules are broken down into sections that are visible as different pages. There are breadcrumbs at the top of the page that allow a student or teacher to jump to different modules. The navigation through each of the lessons is very straight-forward.
The structure for each module above is well-developed. While the content as a whole needs much work for use in an Algebra course, the module projects pose thought-provoking and rigorous questions. Thus, I would consider using some of the material as a supplement to my classroom content or as a resource for struggling students.
I would use these materials in my classroom: Disagree (On a 4 point scale: Strongly Disagree, Disagree, Agree, or Strongly Agree) | 677.169 | 1 |
Math Summer Assignments
The following courses have a recommended summer homework packet available. Each packet contains solutions so that students can check their work as they progress.
Purpose: Review previous content in preparation for the upcoming school year.
Usage in the Classroom: Teachers will review content in the packet throughout the beginning of the school year to help ensure that students have the necessary background for their current course.
Grading: The packet itself will not be collected and graded. However, teachers will review the content of the packet throughout the beginning of the school year. As the material is reviewed, it is subject to being included in homework/classwork, tests or quizzes | 677.169 | 1 |
Introduction•YSTEMS:mbination of components acting together toSYSTEMS: combination of components acting together to perform a task.Componentis a single unit of a system.Dynamic vs staticsystem•MATHEMATICAL MODEL: mathematical description of the system's dynamic characteristics.Linear differential equationsdiffti lti22()510()()dxtxtftdtNon-linear differential equations222(1)0xxtPrepared by R. A. BurdissoME3514 System Dynamics | 677.169 | 1 |
Graphs Before Algebra: an Exercise in Imaginary Curriculum
Abstract
The technology that we use to present a topic to our students, and the technology that we expect our students to use in class, can have profound effects on what is taught and the way in which it is taught. Sometimes this does not happen for a number of reasons, including concerns as to how much the students will "understand" using the new technology, and the development of new curriculum and new methods can be slowed or stopped altogether. One way to work through these difficulties is to present blocks of "imaginary curriculum" as a basis for discussion among teachers and authorities, working through the sorts of understanding we would expect students to develop and identifying points where something might be gained or lost. A specimen topic is developed in this paper.
In this country in the middle years of secondary school students typically learn to work algebraically with quadratic functions, first multiplying out products of terms and then reversing the process to factorize quadratics. Later this is extended into completing the square and solving quadratic equations by factorizing, by completing the square or by use of the standard formula. Later again students use these algebraic skills in constructing graphs of quadratic functions.
In this paper the author demonstrates how the process can be completely reversed, using modern hand-held graphing calculators, by studying graphs first and then deducing all the standard algebraic processes. Examination of the underlying processes suggests that this method is at least as easy to understand, if not clearer; that students will "understand" at least as much as they do by learning the traditional processes; and that they will be more able to extend their skills into more advanced problems. In an ideal curriculum students would probably learn both approaches and the connections between them, to enable them to check their work and reinforce their understanding in multiple ways. | 677.169 | 1 |
Mathematical Physics
Quaternion Dynamics, Part 2 - Identities, Octonions, and Pentuples
This text develops various identities for Hamilton's quaternions. The results are presented in order of difficulty. Results are organized as Axioms, Vectors, Quaternions, and Matrices. There are also sections for Octonions and Pentuples. Axioms are presented first and are largely without rigorous proof. Subsequent identities are constructed from prior identities. When complex conjugates are discussed, the author's thinking is biased towards the original quaternion having a positive vector portion and the conjugate having a negative vector portion. To genuinely understand what is presented, it is recommended that the reader should visualize the concepts in addition to manipulating them algebraically. The algebra is certainly true, but the visual understanding is more elegant and intuitive. This text will likely be updated occasionally | 677.169 | 1 |
Teaching
Math Studies Algebra, Fall 2016
When:
Mondays, Wednesdays, Fridays 13:30
Where:
Wean Hall 8427
What:
Algebra is the art of changing the perspective. The change is mainly achieved through abstraction, which strips the irrelevant details and brings the important to the forefront. The extra generality also enables the connections between far-flung mathematical concepts.
The aim of this course is both to introduce the algebraic way of thinking, and to convey the basic language of algebra.
That language is the language of groups, rings, modules, fields. We shall see, for example, how the group theory unifies such topics
as integer arithmetic, tessellations, solubility of polynomial equations, and counting holes in a pretzel. We shall also learn and use
some category theory, which is a higher-level abstraction that unifies different algebraic notions.
Resources:
Not all topics that we cover are in the book, and some topics we will cover differently.
Links to additional resources will be posted as the course progresses.
More fun:
More fun can be had at my office hours on Wednesdays 2:30–3:30pm and Thursdays 9:30–10:30am in Wean 6202. I am also available by appointment.
Course activities:
Mastery of any subject The main advantage of a class over just
reading a textbook is the ability to ask questions, propose ideas, and interact in other ways.
In particular, discussions during the lectures are encouraged.
In the fall semester, the homework will count for 25% of the grade. During the semester there will be three in-class tests (on September 23, October 19, December 2).
Each test will count for 15% of the grade. A take-home final exam will count for 30%. No collaboration or use of external resources is allowed on tests or on the final.
Homework must be submitted in LaTeX via e-mail. I want both the LaTeX file and the PDF that is produced from it. The filenames must be of the form lastname_alg_homeworknumber.tex
and lastname_alg_homeworknumber.pdf respectively. Pictures do not have to be typeset; a legible photograph of a hand-drawn picture is acceptable.
The homework must be submitted by 1:30pm of the day it is due. For each minute that it is late, the grade will be reduced by 10%.
Lectures:
August 29: Introduction. Groups. Associativity as a consequence of function composition. Examples of groups. Dihedral groups.
August 31: Fields. GLn(R). Generators for the dihedral groups. Normal form for the dihedral group. Group presentations.
Symmetric groups. Cyclic permutations. Homework #1 | 677.169 | 1 |
The Math Handbook: Everyday Math Made Simple
9781848661653
184866165724.01
More Prices
Summary
This is the perfect introduction for those who have a lingering fear of math. If you think that math is difficult, confusing, dull or just plain scary, then The Math Handbookis your ideal companion . Covering all the basics including fractions, equations, primes, squares and square roots, geometry and fractals, Dr. Richard Elwes will lead you gently towards a greater understanding of this fascinating subject. Even apparently daunting concepts will be explained simply, with the assistance of useful diagrams, and with a refreshing lack of jargon. So whether you're an adult or a student, whether you're the sort of person who does Sudoku puzzles, crosswords, or has always been daunted by numbers at work, school or in everyday life, you won't find a better way of overcoming your nervousness about math and learning to enjoy this most amazing of human discoveries. | 677.169 | 1 |
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Unformatted text preview: GATEWAY EXAMS IN CALCULUS at Michigan Stole University Richard 0. Hill
(rev. 7/2/97) Gateway Exams are designed to test basic skills and knowledge that are needed later in a
course or in ensuing courses. They in no way test understanding of concepts, which is done in
other ways. The choice of problems is meant to be representative, not comprehensive. The problems on such exams have surely been seen by students before and are of the type
that students should have some facility with. Ifthe students find there are certain types of
problems they have trouble with, they are responsible for practicing sufficiently, on their own or with others. to gain the requisite skills. Suggested sources of explanations and practice problems
are available. The algebra exam (Calculus Gateway I) covers ten types of algebra and trigonometry
problems that arise in calculus. For emphasis, these ten types do not entail a comprehensive
coverage of all such skills; it is hoped facility with these will lead to facility with the rest.
Following is a description of the types of problems and some hints about what we are expecting. 2 2 2 2
1. Typical question: Expand [sz - or [3x2 — 73-13" + . Comments: You should have memorized (u + v)2 = u2 + Zuv + v2 and (u — v)(u + v) = u2 - v2 . (In particular, doing these problems by "FOIL" is mathematically
correct but much slower. In calculus, such a computation would be a small part of a big problem
that you definitely do not want to make longer than you have to.) For this problem, 2 z 2
(3x — = (341:2)z —2(3xz)[33-]+[3%) = 91:22.2 ~12xy +4-E-2- and 2 2
[3x2 - 33-13". + = (3xz)z - = 91:92.2 - 4%. Finally. you should be encouraged to do the middle steps in your head. 2. Typical question: Simplify {In/x3 . Comments: Rational exponents come up often enough in calculus that you need facility with them. Here, {In/F =[g(x3)m]lls = [ms/2]" = [J's/2]"s = xuz = 3. Typical question: Factor 1:2 - 6x - 7. Solve: J:2 = 6x +7 .
Comments: So many calculus, engineering. economics, etc.. problems have simple
quadratic problems embedded in them that you should be able to do simple ones by hand. Here, ,, ,t:2 ~6x-7=(x-7)(x+l) and x2 -—--6x~+-7,x2 -6x--7=O,(x-7)(x+l)=0,x=-l,7. .- 4. Typical question: Sketch the graphs 0f y = e: and y = 1-23 +1 '
x Comments: Even though you have a calculator handy, you should have memorized
addition and multiplication tables. For similar reasons. even though you have a graphing
calculator handy. there are several functions which are so common that you should have images in
your mind of the basic shapes of their graphs. and also know simple translations and expansions.
You should know the graphs of: ' 1 l
y=mx+b yzxz y=xJ yzlxl y=~/-; y=-; y=-—2- y=sinx,y=cosx ' 5. Typical question: Simplify as much as possible: bf—f- . b a
Comments: One way to simplify fractions like this is to multiply numerator and
denominator by the lowest common denominator. a bab aab bfz 37.1-3771 _ '¢b'_(a-b)(a+b)_ +b
1 lab-lab tab" an, ' a_"b"""" balblal 6. Typical question: Simplify as much as possible: a — (1 - {a -[a - (a - l)]}). I
Comments: Be careful of negative signs as you clear parentheses and brackets. Here.
a-(l- {a—[d-(a- 1)") =a-l+{a -[a-a+ 1]} =a— Hot-[1] =2a-2 7. Typical question: Find the point-slope equation for the following straight lines:
(a) The line through the points (-2. 3) and (4, 1).
(b) The line through the point (~2, 3) and parallel to the line y = -3x +7.
(c) The line through the point (~2, 3) and perpendicular to the line y = ~3x +7. Comments: You shduld know y - yo = m(x - x0) is the point-slope equation for a
change in y change in x ' I. straight line. Here, (x0, yo) is any point on the line and m :- slope :- F ( ) change in y [.3 1 d '
a Y m = m = = _~ ' .
or Change in x 4 _ (4) 3 an you can use either paint. Thus two possxble answers are y *3 = --§-(x +2) or y -l = -l~(x - 4). For (1") and (c) you should know the
slopes of parallel lines are equal and the slopes of perpendicular lines are negative reciprocals.
Thus the answer for (b) is y — 3 = —-3(x + 2) and the answer for (c) is y - 3 =J1.(x + 2). 8. Typical question: The graph below represents a straight line with equation
y = mx + b. Estimate m and b. change in y
change in x the calculus problems that you will see will require you to estimate slopes and sometimes also
estimate 1:. Here, for (i) m =1 3 and b =5 -l and for (ii) m = -l/3 andb ~ 1. 9. Typical question: Sketch a 30° - 60° right triangle and label its sides. Use it to find
It 7:: Comments: You should know m = slope = and b = y-intercept. Many of cos-g and csc-g. 79 x I
Comments: You should know the triangles / C i 3
' C
. a 7: 3 2r 2
and that in a right triangle '4 sm9 = etc. Thus cos; = cscz = - = 2.
s 1 10. Typical question: Sketch the graph of y = sin x over the interval = {-n S x 5 iv: .
Find sin(- #72) and sin%7r. Comments: You should know the graphs of y s sin x; y = cos x, y = tan x (as well as y a
cotx, y a sec x, y = csc 2:) over intervals bigger than just [0, 27: 1. Using these graphs is one way
of easily finding some values of the trigonometric functions. For example. you can easily see from
the graph that sin(--§1r) 5 —1 and sinivt a -l ...
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I use this worksheet to help breakdown how to solve algebraic equations for my 8th graders. This worksheet assumes prior knowledge in solving algebraic equations. Each step (and page) adds an additional level of difficulty which is helpful for both students and teachers to determine where their student is in experience/confidence with algebraic equations.
I often use this as a pre-test for my 8th graders to determine how comfortable they are manipulating algebra expressions.
The math in this worksheet definitely gets complicated, so I wouldn't assign this to a new-to-algebra student | 677.169 | 1 |
Introduction
Matrix Algebra underlies many of the current tools for experimental design and the analysis of high-dimensional data. In this introductory data analysis course, we will use matrix algebra to represent the linear models that commonly used to model differences between experimental units. We perform statistical inference on these differences. Throughout the course we will use the R programming language.
What you'll learn:
Matrix algebra notation
Matrix algebra operations
Application of matrix algebra to data analysis
Linear models
Brief introduction to the QR decomposition | 677.169 | 1 |
Product Description:
Geared toward helping students visualize and apply mathematics, Elementary Algebra: Graphs and Models uses illustrations, graphs, and graphing technology to enhance students' mathematical skills. This is accomplished through Interactive Discoveries, Algebraic/Graphical Side-by-Sides, and the incorporation of real-data applications. In addition, students are taught problem-solving skills using the Bittinger hallmark five-step problem-solving process coupled with Connecting the Concepts and Aha exercises. And, as you have come to expect with any Bittinger text, we bring you a complete supplements package that now includes an Annotated Instructor's Edition and MyMathLab, Addison-Wesley's online course solution.
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Answers : (1)
I will suggest you to go for multiple books.because generally a single books doesnot have all the chapter well explained.
For vector and 3D geometry refer IIT MATHA by M L KHANNA.
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. | 677.169 | 1 |
Applied Combinatorics with Problem Solving
by Jackson and Thoro. This book is out of print and copies can be
obtained at Gray's Bookstore only.
Objectives:
In this class, we will study the fundamentals of
discrete mathematics, including deductive proof, inductive proof,
counting techniques, binomial coefficients, the pigeonhole
principle, the inclusion-exclusing principle, recurrence
relations, generating functions, and graphs.
Responsibilities:
You are responsible for attending class on a
regular basis and maintaining comprehension of the scheduled class
objectives for each day. You are expected to be active
participants in class, to turn in assignments promptly, and to
attend examinations. Examinations will be on Friday, February 22 and
Friday, April 5 during class time, and Friday, April 26
at 8AM. Except in case of emergency, late assignments will
only be accepted for half credit within one week of the original
due date. It is important that you show your work or outline the
process of discovery for each problem on the homework
assignments. No credit will be given for answers which do not
include work, unless otherwise stated.
Special needs:
Any scheduled absence during a quiz or examination,
or any other special needs, must be brought to my attention
before the end of the second week of class. Unscheduled absences
will be handled on a case-by-case basis, with exceptions generally
made only for documented emergencies.
Honesty:
There are many resources available to help you
succeed in this class, including consultation during office hours,
secondary texts, and cooperation with other students. It is
important, however, that all papers handed in be the result of
your individual comprehension of the course material. Duplication
of others' work is both a disservice to your own education and a
serious violation of the university's academic honesty
policy.
Grades:
Homework problems account for 40% of your
grade. Each of the two midterm examinations will be 15% of your
grade, and the final examination will contribute 30%. A 90%
overall guarantees a grade of A–, 80% guarantees a B–,
and 70% guarantees a C–.
Changes:
The syllabus is subject to change. Changes will be
announced in class and updated online. | 677.169 | 1 |
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MATH TREK Algebra 1
04/01/04
For curriculum-based algebra instruction, teachers and students can use MATH TREK Algebra 1. The multimedia program includes tutorials, assessments and student tracking. Students can use the program's scientific calculator, glossary and journal to help them complete the various exercises and activities. The assessment and student-tracking features provide immediate feedback to students so that they can stay on top of their progress. This engaging program, complete with sound, animation and graphics, can be used on stand-alone computers or a network. NECTAR Foundation, (613) 224-3031,
This article originally appeared in the 04 | 677.169 | 1 |
Algebra 2 / Trig - We discussed Sec. 9.1, focusing on various ways (table, graph, equation) to represent a particular set of data. We also reviewed the method necessary for using our graphing calculator to identify the "regression" equation associated with a particular data set.
College Algebra - We completed our review ahead of tomorrow's 2nd semester final exam. Click on the link at the top of the page for more information regarding the exam.
Frosh Geometry - We completed a retake of the "Arcs/Angles of a Circle" quiz. HW for Thursday - Read Sec. 10.6, watch all lesson videos provided with the online text, and complete the Sec. 10.6 worksheet packet provided in class.
Tuesday, May 10, 2011
Monday, May 9, 2011
Algebra 2 / Trig - We reviewed Sec. 14.3 - 14.5 in preparation for tomorrow's quiz covering Sec. 14.3 - 14.6. No formal HW was assigned for Tuesday; however, students should come to class on Tuesday prepared to take their quiz.
Thursday, May 5, 2011
Algebra 2 / Trig - No class today, due to Schedule A. Complete assignment that was given in class on Wednesday.
College Algebra - We completed our discussion of basic probability. HW for Friday - complete the Sec. 13.3 - 13.5 worksheet packet that was distributed in class on Wednesday. We will take a quiz over Sec. 13.3 - 13.5 in class on Friday.
College Algebra - We continued our discussion of probability, focusing on inclusive and mutually exclusive events (Sec. 13.4 in our text). HW for Wednesday - Complete the Practice side of the Lesson 12-6 worksheet.
Frosh Geometry - We continued our discussion of angles of a circle, focusing on interior angles (Sec. 11.5 in our text). We worked on a sketchpad file in class to help us determine the formula for finding the measure of an interior angle. HW for Wednesday - Complete the sketchpad file that we worked on in class.
College Algebra - We continued our discussion of probability, focusing on calculating probabilities for dependent and independent events. HW for Tuesday - Complete the Practice side of the Sec. 12-5 worksheet distributed in class. Additionally, we collected the take-home quiz covering Sec. 13.1 - 13.2 - Permutations and Combinations | 677.169 | 1 |
Interests
Browse by
A Survey of
Geometric Algebra
and
Geometric Calculus
c _
Alan Macdonald
Luther College, Decorah, IA 52101 USA
macdonal@luther.edu
October 15, 2012
(The current version is always available at my web page.)
"The principal argument for the adoption of geometric algebra is that it provides a single,
simple mathematical framework which eliminates the plethora of diverse mathematical de-
scriptions and techniques it would otherwise be necessary to learn." [10]
1 Foundations
1.1 Introduction
Geometric algebra and its extension to geometric calculus unify, simplify, and gen-
eralize many areas of mathematics that involve geometric ideas. They also provide a
unified mathematical language for physics, engineering, and the geometrical aspects of
computer science (e.g., graphics, robotics, computer vision).
This paper is an introduction to geometric algebra and geometric calculus, presented
in the simplest way I could manage, without worrying too much about completeness
or rigor. The only prerequisite is an understanding of undergraduate mathematics. In
a few inessential sections some knowledge of physics is helpful.
My purpose is to demonstrate some of the scope and power of geometric algebra and
geometric calculus. I will illustrate this for linear algebra, multivariable calculus, real
analysis, complex analysis, and several geometries: euclidean, noneuclidean, projective,
and conformal. I will also outline several applications.
Geometric algebra is nothing less than a new approach to geometry. Geometric
objects (points, lines, planes, circles, ... ) are represented by members of an algebra, a
geometric algebra, rather than by equations relating coordinates. Geometric operations
on the objects (rotate, translate, intersect, project, construct the circle through three
points, ... ) are then represented by algebraic operations on the objects.
Geometric algebra is coordinate-free: coordinates are needed only when specific
objects or operations are under consideration.
All this has significant advantages over traditional approaches such as synthetic
and analytic geometry and vector, tensor, exterior, and spinor algebra and calculus.
The advantages are similar to those of elementary algebra over arithmetic: elementary
algebra manipulates symbolic representations of numbers independently of their val-
ues, and geometric algebra manipulates symbolic representations of geometric objects
independently of their coordinates in some coordinate system.
Efficient algorithms have recently been developed for implementing geometric alge-
bra on computers [6].
At first you will likely find the novelty and scope of the mathematics presented here
overwhelming. This is to be expected: it takes years of serious study to understand the
standard approaches to the mathematics discussed here. But after some study I hope
that you find, with me, great unity, simplicity, and elegance in geometric algebra.
Readers who want to know more can consult the last section, Further Study, of this
paper. It includes a listing of many papers on available on the web.
The American physicist and mathematician David Hestenes initiated the modern
development of geometric algebra in the 1960's. He built on the work of Hamilton,
Grassmann, and Clifford a century or so earlier. After a slow start, geometric algebra
has today attracted many workers in many disciplines.
Hestenes was awarded the 2002 Oersted Medal, the American Association of Physics
Teachers "most prestigious award". His medal lecture, "Reforming the Mathematical
Language of Physics" [5], was published in The American Journal of Physics, which
has published several papers on geometric algebra. In his lecture, Hestenes claims that
"geometric algebra simplifies and clarifies the structure of physics, and . . . [thus has]
immense implications for physics instruction at all grade levels." I believe that this is
equally true of mathematics.
2
1.2 The Geometric Algebra
The most popular algebraic structure today for Euclidean n-space is the inner prod-
uct space R
n
. This section presents a powerful extension of this structure, the geomet-
ric algebra G
n
. In subsequent sections, armed with this algebra, we will unify, simplify,
and generalize many areas of mathematics, and give several applications.
1.2.1. The geometric algebra G
n
. I first present the structure of G
n
concisely,
and then, in the next subsection, elaborate.
The geometric algebra G
n
is an extension of the inner product space R
n
. First, it is
an associative algebra with one. That is, it is a vector space with a product satisfying
properties G1-G4 for all scalars a and A, B, C ∈ G
n
:
G1. A(B + C) = AB + AC, (B + C)A = BA + CA.
G2. (aA)B = A(aB) = a(AB).
G3. (AB)C = A(BC).
G4. 1A = A1 = A.
The product is called the geometric product. Members of G
n
are called multivec-
tors. This allows us to reserve the term "vector" for vectors in R
n
. (They are also
multivectors.) This is a convenient terminology. We list three more properties.
G5. The geometric product of G
n
is linked to the algebraic structure of R
n
by
uu = u u = [u[
2
for all u ∈ R
n
. (1.1)
G6. Every orthonormal basis for R
n
determines a standard basis (defined below)
for the vector space G
n
.
That's it! That's the geometric algebra. We have not proved that the mathematical
structure just described exists. For that, see [9].
1.2.2. Elaboration. Equation (1.1) shows that nonzero vectors have an inverse in
G
n
: u
−1
= u/[u[
2
.
Use a polarization identity, Eq. (1.1), and distributivity:
u v =
1
2
((u +v) (u +v) −u u −v v)
=
1
2
_
(u +v)
2
−u
2
−v
2
_
=
1
2
(uv +vu) .
If u and v are orthogonal, then this equation gives the important
vu = −uv. (u, v orthogonal) (1.2)
Example: (1 +e
1
e
2
)(e
1
−2e
2
) = −e
1
−3e
2
.
If u and v are orthogonal and nonzero, then from Eq. (1.1), (uv)
2
= uvuv =
−uuvv = −[u[
2
[v[
2
< 0. Therefore uv is not a scalar or a vector. It is something new,
a 2-vector, or bivector.
3
Standard basis. Let ¦e
1
, e
2
, e
3
, e
4
¦ be an orthonormal basis for R
4
. This example
of a standard basis for the vector space G
4
suffices to understand the concept:
1 basis for 0-vectors (scalars)
e
1
e
2
e
3
e
4
basis for 1-vectors (vectors)
e
1
e
2
e
1
e
3
e
1
e
4
e
2
e
3
e
2
e
4
e
3
e
4
basis for 2-vectors (bivectors)
e
1
e
2
e
3
e
1
e
2
e
4
e
1
e
3
e
4
e
2
e
3
e
4
basis for 3-vectors (trivectors)
e
1
e
2
e
3
e
4
basis for 4-vectors.
The subscripts on the products of e's are increasing from left to right, and all such
products are in the basis.
According to Eq. (1.2) rearranging the order of the e's in member of the basis at
most changes its sign. Thus the original product and its rearrangement are linearly
dependent. We cannot use both in a basis. The standard basis uses the arrangement
with the subscripts increasing from left to right.
Since vectors in R
n
are in G
n
and since G
n
is closed under the geometric product,
every linear combination of geometric products of vectors fromR
n
is in G
n
. A standard
basis shows that all multivectors are of this form.
Products of k different e's span the subspace of k-vectors. Properties G1-G6 imply
that k-vectors are independent of the orthonormal basis ¦e
1
, e
2
, . . . , e
n
¦.
The (one and only) zero is a k-vector for all k. This is necessary if k-vectors are to
form a subspace of G
n
.
We might try to form a 5-vector in G
4
, e.g., e
1
e
2
e
3
e
4
e
2
. But by the product rules,
Eqs. (1.1) and (1.2), this is equal to e
1
e
3
e
4
. There are no 5-vectors in G
4
. More
generally, there are no m-vectors in G
n
with m > n.
Each member of a standard basis contains a given e or it does not. Thus G
n
has
dimension 2
n
.
How Geometric Algebra Works
Geometric algebra represents geometric objects in R
n
with members of G
n
.
Geometric algebra represents geometric operations
on these objects with algebraic operations in G
n
.
Coordinates are not used in these representations.
Geometric objects include points, lines, planes, circles, and spheres. Geometric oper-
ations include rotations, projections, constructions of lines between two points, con-
structions of circles through three points, determining areas of parallelepipeds, and
determining angles between subspaces.
Abbreviate "geometric algebra" to GA and "vector algebra" to VA.
Let's see what we can do with GA.
4
1.3 The Inner and Outer Products
We investigate the geometric product of two given vectors u and v in R
n
. Let
¦e
1
, e
2
¦ be an orthonormal basis for a plane containing the vectors. Let u = a e
1
+b e
2
and v = c e
1
+ d e
2
. Then from the product rules, Eqs. (1.1) and (1.2),
uv = (ac + bd) + (ad −bc) e
1
e
2
. (1.3)
1.3.1. The inner product. The first term on the right side of Eq. (1.3), ac +bd,
is the usual inner product of u and v: u v = [u[ [v[ cos θ .
Fig. 1: The bivector
u ∧ v .
1.3.2. The outer product. The second term on the right
side of Eq. (1.3) is the outer product of u and v. This bivector
is denoted u ∧ v. Just as u represents an oriented length, u ∧ v
represents an oriented area. For the factor ad −bc is the signed
area of the parallelogram with sides u and v: [u[ [v[ sin θ. And
i ≡ e
1
e
2
specifies the plane in which the area resides. Thus
u ∧ v = (ad −bc) e
1
e
2
= [ u[ [ v [ sin θ i . (1.4)
See Fig. 1. (If you want to know why e
1
e
2
is denoted i, square it. We will have much
more to say about this.)
The inner product u v = [u[ [v[ cos θ is intrinsic to the vectors, not depending on
the basis ¦e
1
, e
2
¦. So too with the outer product. For [ u[ [ v[ sin θ is intrinsic. And if
we rotate to another basis
_
e
1
e
2
_
=
_
cos α sin α
−sin α cos α
__
e
1
e
2
_
, then i
= e
1
e
2
= i.
In 3D VA, the oriented area u ∧ v is represented by the cross product u v. But VA
does not generalize to higher dimensions. Moreover, we shall see that even in 3D, ∧ is
superior to in several respects. Thus the cross product plays only a minor role in GA.
1.3.3. The fundamental identity. Rewrite Eq. (1.3) in terms of the inner and
outer products to obtain the fundamental identity
uv = u v +u ∧ v. (1.5)
Forget the coordinates of u and v which led to this equation. Remember that the
geometric product of two vectors is the sum of a scalar and a bivector, both of which
have a simple geometric interpretation.
There is no hint in VA that and (reformulated to ∧) are parts of a whole: the
geometric product, an associative product in which nonzero vectors have an inverse.
1
1.3.4. Important miscellaneous facts. We will use them without comment.
• u ∧ v , unlike uv, is always a bivector.
• uv is a bivector ⇔ u v = 0 ⇔ u ⊥ v ⇔ uv = u ∧ v ⇔ uv = −vu.
In particular, for i ,= j, e
i
e
j
= e
i
∧ e
j
.
• uv is a scalar ⇔ u ∧ v = 0 ⇔ u | v ⇔ uv = u v ⇔ uv = vu.
• v ∧ u = −(u ∧ v).
• The inner and outer products are the symmetric and antisymmetric parts of the
geometric product: u v =
1
2
(uv +vu) and u ∧ v =
1
2
(uv −vu) .
1
Not every nonzero multivector has an inverse. Example: Let |u| = 1. If 1 − u had an inverse,
then right multiply (1 +u)(1 −u) = 0 by the inverse to obtain 1 +u = 0, a contradiction.
5
1.4 Represent Subspaces
A blade B for a k-dimensional subspace of R
n
is a product of members of an
orthogonal basis for the subspace: B = b
1
b
2
b
k
. We call B a k-blade, or a blade of
grade k. Nonzero scalars are 0–blades.
Geometric algebra represents subspaces with their blades.
A positive multiple of a blade represents the same subspace as the blade. A negative
multiple represents the same subspace but with the opposite orientation. For example,
6e
1
e
3
and −e
1
e
3
= e
3
e
1
represent opposite orientations of a 2D subspace.
The inverse of B is B
−1
= b
k
b
2
b
1
/[b
k
[
2
[b
2
[
2
[b
1
[
2
.
We use bold to designate blades, with lower case reserved for vectors. (Exception:
i.) We use upper case italic (A, B, C, . . . ) to denote general multivectors.
We shorten "the subspace represented by blade B" to "the subspace B", or simply
"B". For example, set theoretic relationships between blades, e.g., "A ⊆ B", refer to
the subspaces they represent.
1.4.1. The pseudoscalar. The (unit) pseudoscalar I = e
1
e
2
e
n
is an impor-
tant member of G
n
. (We use lower case in 2D: i = e
1
e
2
, as in Sec. 1.3.2.) Since
n-vectors form a 1-dimensional subspace of G
n
, every n-vector is a scalar multiple of
I. If e
1
e
2
e
n
= I
for another orthonormal basis, then I
is a unit n-vector. Thus
I
= ±I. We say that the orthonormal bases have the same orientation if I
B is a (k−1)-blade in B.
b. If a
⊥
⊥ B, then a
⊥
∧ B = a
⊥
B and a
⊥
B = 0.
Also, a
⊥
∧ B is a (k+1)-blade representing span(a
⊥
, B).
1.5.8. Theorem. (Extend the fundamental identity.) Let B be a k-blade. Then
for every vector a,
aB = a B+a ∧ B. (1.13)
In this equation,
a B is a (k −1) - blade in B (or 0),
a ∧ B is a (k + 1) - blade representing span(a, B) (or 0).
Proof. Using the lemma four times in Step (3),
aB = (a
+a
⊥
)B = a
B+a
⊥
B = a
B+a
⊥
B+a
⊥
∧ B+a
∧ B
= (a
+a
⊥
) B+ (a
⊥
+a
) ∧ B = a B+a ∧ B.
For the rest of the theorem, apply the lemma to a B = a
B and a∧B = a
⊥
∧B.
In general, AB ,= A B+A∧ B. Example: A = e
1
e
2
and B = e
2
e
3
.
8
1.5.9. Theorem. (Projections and rejections.) Let a be a vector and B a blade.
Decompose a with respect to B: a = a
∧ B = a ∧ B. This gives Eq. (1.15).
VA has no analogs of these simple formulas, except when B = b, a vector, in the
projection formula Eq. (1.14).
1.5.10. Theorem. (Subspace membership test.) For every vector a and blade B,
a ∈ B ⇔ a ∧ B = 0, (1.16)
a ∈ B ⇔ a B
∗
= 0. (1.17)
Proof. From a = a
+a
⊥
, a ∈ B ⇔ a
⊥
= 0. This, with Eq. (1.15), gives Eq. (1.16).
From (a B)
∗
= a ∧B
∗
, Eq. (1.12), applied to B
∗
, a B
∗
= 0 ⇔ a ∧B = 0. This,
with Eq. (1.16), gives Eq. (1.17).
We say that B is a direct representation of the subspace and B
∗
is a dual represen-
tation. Both are useful.
The distance from the endpoint of a to B is [ a
⊥
[ = [ (a ∧ B)/B[ . If B is a
hyperplane, e.g., a plane in R
3
, then it divides R
n
into two sides. Then a ∧ B is an
n-vector. The scalar (a ∧ B)
∗
/ [B[ is a signed distance from the endpoint of a to B.
1.5.11. Theorem. (Reflections.) Let F
B
(a) be the reflection of a vector a in a
blade B
a. If B is a k-blade, then F
B
(a) = (−1)
k+1
BaB
−1
.
b. If B is a hyperplane and b = B
∗
is a vector normal to B, then F
B
(a) = −bab
−1
.
Proof. a. We prove only the case B = b, a vector. From Figure 2, F
B
(a) = a
= ( a B)/B. The result of an operation can be substituted in other
expressions, which can then be manipulated algebraically.
All this without using coordinates. Geometric algebra is coordinate-free: coordinates
are needed only when specific objects or operations are under consideration.
These features of GA fit naturally into the modern techniques of object oriented com-
puter programming.
3
Note, for example, that Pe
1
e
2
(e
2
e
3
) = 0. This tells us that areas in the plane e
2
e
3
projected to
the plane e
1
e
2
have area 0 there.
4
|(A··· B)/B| = |A··· B|/|B| because B
−1
is a blade.
5
To compute b
3
, use (u
1
∧ u
2
)
−1
= (b
1
b
2
)
−1
= b
−1
2
b
−1
1
= (b
2
/|b
2
|
2
)(b
1
/|b
1
|
2
).
10
2 Algebra
2.1 Complex Numbers
2.1.1. Complex numbers. Let i be the pseudoscalar of a plane in R
n
. Then
a + bi, scalar + bivector, is a complex number. Since i
2
= −1, GA complex numbers
are isomorphic to the usual complex numbers. But be careful: GA complex numbers
are not points in a plane or 2D vectors.
Let θ be the angle between vectors u and v. The fundamental identity, Eq. (1.5),
shows that the product of two vectors is a complex number:
uv = u v +u ∧ v = [u[ [v[ cos θ +i [u[ [v[ sin θ.
Define e
i θ
= cos θ +i sin θ. Write the complex number uv in polar form:
uv = [u[ [v[ e
i θ
= r e
i θ
. (2.1)
Every complex number a+ib can be put in the polar form re
i θ
by setting r =
√
a
2
+ b
2
,
cos θ = a/r, and sin θ = b/r . Note the familiar e
i π/2
= i.
Traditional complex numbers use geometrically irrelevant real and imaginary axes. In-
troducing them breaks the rotational symmetry of the plane. This makes it impossible to
implement traditional complex numbers coherently in different planes in higher dimensions.
With GA complex numbers, the a and b in a + bi and the θ in e
iθ
have a geometric
meaning independent of any particular coordinate system. This is different from traditional
complex numbers, where, for example, θ is an angle with respect to the real axis.
The usual complex number i is not needed. It is not part of geometric algebra.
2.1.2. Theorem. The set of complex numbers in G
3
is a subalgebra of G
3
.
Proof. Let ¦i
1
= e
3
e
2
, i
2
= e
1
e
3
, i
3
= e
2
e
1
¦ be a basis for bivectors in G
3
. You can
check that
i
2
1
= i
2
2
= i
2
3
= −1 and i
1
i
2
= i
3
, i
2
i
3
= i
1
, i
3
i
1
= i
2
. (2.2)
Every complex number a+bi in G
3
can be put in the form a+b
1
i
1
+b
2
i
2
+b
3
i
3
. From
Eq. (2.2), the product of such multivectors is another. By Corollary 1.4.4, the product
is a complex number. This is sufficient to prove the theorem.
The theorem fails in G
4
: (e
1
e
2
)(e
3
e
4
) is not a complex number.
The identities in Eq. (2.2) characterize the quaternions. Thus the complex numbers
in G
3
form the quaternion algebra.
Traditionally, quaternions have been considered as scalar + vector. So considered, they
have not been satisfactorily united with vectors in a common mathematical system [3, 12].
Considered here as scalar + bivector, they are united with vectors in GA.
For many, quaternions are a 19
th
century mathematical curiosity. But many roboticists,
aerospace engineers, and gamers know better: quaternions are the best way to represent
rotations in 3D, as we will see in Sec. 2.2.1.
Complex numbers and quaternions are only two of the many algebraic systems embedded
in GA. We shall see several more examples. Having these systems embedded in a common
structure reveals and clarifies relationships between them. For example, we have seen that
the product of two vectors is a complex number.
11
The exterior (Grassmann) algebra is another system embedded in GA. It consists
of outer products of nonzero vectors. It is thus only part of GA, not using the inner or
geometric products. Taking advantage of these products simplifies the exterior algebra.
For example, the GA dual, whose definition uses the geometric product, is the Hodge
star dual, up to a sign. The GA definition allows easier manipulation.
2.2 Rotations
2.2.1. Rotations in R
3
. Orthogonal matrices and Euler angles are among the
many representations of rotations in R
3
in use today. GA provides a better represen-
tation.
In GA the rotation of a vector u by an angle θ around an axis n is given by
R
iθ
(u) = e
−nIθ/2
ue
nIθ/2
. (2.3)
(An underbar denotes a linear transformation.) We prove this in Sec. 2.2.3. The
formula expresses the rotation simply and directly in terms of θ and n. We write
R
iθ
(u) = RuR
−1
, where the unit complex number R = e
−nIθ/2
represents the rotation.
Thus in R
3
, unit quaternions represent rotations.
2.2.2. Rotations compose simply. Follow the rotation R
1
with the rotation R
2
.
Then their composition is represented by the product R = R
2
R
1
:
M → R
2
(R
1
MR
−1
1
)R
−1
2
= RMR
−1
.
Since the product of unit quaternions is another, we have a simple proof of the
important fact that the composition of 3D rotations is a rotation.
As an example, a rotation by 90
◦
around the e
1
axis followed by a rotation of 90
◦
around the e
2
axis is a rotation of 120
◦
around e
1
+e
2
−e
3
:
(cos 45
◦
−e
2
I sin 45
◦
) (cos 45
◦
−e
1
I sin 45
◦
)
= cos 60
◦
−
e
1
+e
2
−e
3
√
3
I sin 60
◦
.
In the general case, the composite quaternion Q = cos(θ/2) +nI sin(θ/2), where n
is a unit vector, the axis of the composite rotation.
Quaternions are superior to orthogonal matrices in representing 3D rotations: (i) It
is easier to determine the quaternion representation of a rotation than the matrix repre-
sentation,
6
(ii) It is more efficient to multiply quaternions than matrices, and (iii) If a
quaternion product is not quite normalized due to rounding errors, then divide by its norm
to make it so; if a product of orthogonal matrices is not orthogonal, then use Gram-Schmidt
orthonormalization, which is expensive and not canonical.
6
Rodrigues' rotation formula gives the matrix representation of a 3D rotation [11].
12
2.2.3. Rotations in R
n
. In GA, an angle is a bivector iθ: i specifies the plane in
which it resides and θ specifies its size.
7
In R
n
, an angle iθ specifies a rotation: i specifies the plane of the rotation and θ
specifies the amount of rotation. (Only in 3D does a rotation have an axis, the unique
direction normal to the plane of the rotation.)
Let the rotation carry the vector u to the vector v. First suppose that u ∈ i .
Left multiply uv = [u[[v[ e
iθ
, Eq. (2.1), by u and use u
2
= [u[
2
= [u[ [v[ to obtain
v = ue
i θ
; e
i θ
rotates u to v. The analog in standard complex algebra is familiar.
Now consider a general u. Decompose u with respect to i : u = u
⊥
+u
(u
⊥
⊥ i ,
u
∈ i). The rotation rotates u
as above but does not affect u
⊥
. Thus
v = u
⊥
+u
e
i θ
= u
⊥
e
−i θ/2
e
i θ/2
+u
e
i θ/2
e
i θ/2
= e
−i θ/2
u
⊥
e
i θ/2
+ e
−i θ/2
u
e
i θ/2
(since u
⊥
i = iu
⊥
and u
i = −iu
)
= e
−i θ/2
ue
i θ/2
. (2.4)
In 3D this reduces to Eq. (2.3). For in G
3
, n = i
∗
= i/I.
Rotate rotations. Let R
1
represent a rotation by angle i
1
θ
1
. Rotate the rotation:
rotate the plane i
1
of this rotation with a rotation represented by R
2
. It is easy to
show that the rotated rotation is represented by R
2
R
1
R
−1
2
. Rotations rotate just as
vectors do!
2.2.4. The special orthogonal group. In nD, n > 3, rotations by angles iθ
are not closed under composition. They generate the orientation and inner product
preserving special orthogonal group SO(n). Given e
−i θ/2
, choose, using Eq. (2.1), unit
vectors b
1
and b
2
so that e
−i θ/2
= b
1
b
2
. Then the rotation Eq. (2.4) can be written
u → (b
1
b
2
)u(b
1
b
2
)
−1
. We may drop the normalization [b
i
[ = 1 in this formula. Thus
a member of SO(n) can be represented by a product of an even number of nonzero
vectors B = b
1
b
2
b
2k
: u → BuB
−1
(Sec. 2.2.2).
Eq. (2.4) represents a rotation of 2π by e
−i 2π/2
= −1. The products B above
represent the simply connected double covering group of SO(n), with ±B representing
the same element of SO(n). This is another algebraic system embedded in GA. Matrices
do not represent the double cover.
"A particular area where geometric algebra provides a unifying language is
in the description of rotations. The most fundamental modern treatments
such as those of Abraham, Marsden, Arnol'd and Simo use differential topol-
ogy and describe rotations in terms of the Lie group SO(3). A rotation is
thus an element of a differentiable manifold, and combinations of rotations
are described using the group action. Infinitesimal rotations and rotational
velocities live in the tangent bundle TSO(3), a differentiable manifold with
distinctly non-trivial topology, from where they can be transported to the
tangent space at the identity, identifiable with the Lie algebra so(3). As
throughout all of the differentiable topology formulation of mechanics, a
proliferation of manifolds occurs. ... In geometric algebra there is no such
proliferation of manifolds: the mathematical arena consists only of elements
of the algebra and nothing more." [10]
7
As an aside, let iθ be the angle between a vector a and a blade B. (See Fig. 2.) Then
i tan(θ) = a
⊥
a
−1
(opposite over adjacent). To see this, note that a
⊥
a
−1
, the product of orthogonal
vectors, is a bivector in the plane i they span. And |a
⊥
a
−1
| = |a
⊥
||a
|
−1
= tan θ, where θ is the
scalar angle between a and B.
13
2.2.5. Pauli's electron theory. In 1927 Wolfgang Pauli published a quantum
theory of an electron interacting with an electromagnetic field. Pauli's theory does not
take Einstein's special relativity theory into account. A year later Paul Dirac published
a relativistic quantum theory of the electron. We compare the VA and GA formulations
of Pauli's theory in this section and of Dirac's theory in Sec. 2.4.5.
An electron has a property called spin, with a fixed value
1
2
. For our purposes think
of the electron as spinning about an axis s. The Pauli and Dirac theories describe the
position and spin axis of spin-
1
2
particles in an electromagnetic field. The field can
change the position and the axis. We consider here only the case of a particle at rest
in a uniform but time varying magnetic field. This case is important, for example, in
the theory of nuclear magnetic resonance.
In the VA formulation of classical electromagnetism, a magnetic field is represented
by a vector b ∈ R
3
. In the G
3
formulation, the field is represented by the bivector
B = −b
∗
, in the plane orthogonal to b. We saw in Sec. 2.2.3 that B and its scalar
multiples specify rotations.
The basic physical fact is that from time t to t + dt, s rotates through the angle
γB(t) dt, where γ is a constant. (Thus if B is constant in time, then s precesses with
respect to the plane Bwith constant angular speed γ[B[ .) From Sec. 2.2.3, the rotation
is represented by the unit quaternion e
−
1
2
γB(t)dt
.
In Pauli's theory the spin is not represented by s(t), but by the unit quaternion
ψ(t) which rotates s(0) to s(t) . From Sec. 2.2.2 we have the composition
ψ(t + dt) = e
−
1
2
γB(t)dt
ψ(t) ≈
_
1 −
γ
2
B(t) dt
_
ψ(t) .
This gives the GA version of Pauli's equation (for the spin): ψ
J
¸ e
J
A¸
0
e
−1
J
.
16
2.4 The Spacetime Algebra
2.4.1. Indefinite metrics. GA readily extends from R
n
to vector spaces with
an indefinite metric. In R
p,q
every orthonormal basis has p e
i
's with e
i
e
i
= 1 and
q e
i
's with e
i
e
i
= −1 . Correspondingly, in G
p,q
there are p e
i
's with e
i
e
i
= e
2
i
= 1
and q e
i
's with e
i
e
i
= e
2
i
= −1 . Many properties of G
n
remain valid in G
p,q
. In
particular, we still have Eq. (1.2): vu = −uv for orthogonal vectors u and v.
2.4.2. The spacetime algebra. Special relativity assigns spacetime coordinates
(t, x, y, z) to a definite time and place. In VA, spacetime is represented by R
1,3
. An
orthonormal basis has a time direction e
0
with e
0
e
0
= 1 and three space directions
e
i
with e
i
e
i
= −1. In GA, spacetime is represented by the spacetime algebra G
1,3
.
To get very far in 4D relativistic physics, methods beyond VA must be employed, usually
the tensor algebra or exterior algebra of R
1,3
, which are very different from anything a
student has seen before in VA. The transition from G
3
to G
1,3
is easier.
2.4.3. Boosts. Consider a second coordinate system (t
, x
, y
, z
) whose spatial
points (x
, y
, z
) move with respect to those of the first with velocity v. The origins
(0, 0, 0, 0) of the two systems coincide. And their spatial axes are parallel.
Fig. 7: An active "boost"
p → p
in a plane.
You can think of the ¦e
i
¦ basis as moving with the
new coordinate system. However, translating a vector
does not change it, and the e
i
are, with the e
i
, fixed
vectors in R
1,3
.
The transformation p=
3
i=0
a
i
e
i
→ p
=
3
i=0
a
i
e
i
maps p to the vector p
which "looks the same" in the
moving system. This active transformation is called a
boost, or Lorentz transformation. Fig. 7 illustrates an
analogy to a rotation in a plane.
The analogy is appropriate. For it can be shown
that the boost is a rotation in the e
0
v plane: p →
e
−e0
ˆ vα/2
pe
e0
ˆ vα/2
. (C. f., Eq. (2.4).) Here we have set v = tanh(α)ˆ v, where ˆ v
2
= −1.
8
And the exponential function is defined for all multivectors M by e
M
=
∞
i=0
M
i
/i! .
You can verify that e
e0
ˆ vα/2
= cosh(α/2) +e
0
ˆ v sinh(α/2).
VA represents boosts with 4 4 orthogonal matrices on R
1,3
. The GA exponential
function representation has advantages similar to the GA exponential function representa-
tion of rotations in R
3
. (See Sec. 2.2.2.) We will see an example in the next section,
where we compute the composition of two boosts. The results obtained there in a few lines
are much more difficult to obtain with VA, so difficult that it is a rare relativity text which
derives them.
8
We use units of space and time in which the speed of light c = 1. For example, time might be
measured in seconds and distance in light-seconds. Since speeds of material objects are less than c,
−1 < v
2
≤ 0. Thus tanh(α) = (−v
2
)
1
2 .
17
2.4.4. Composition of boosts. By definition, a boost is free of any spatial
rotation. Perhaps surprisingly, a composition of boosts is not. The composition of
boosts e
−e0
ˆ vβ/2
and e
−e0
ˆ uα/2
can be written as a spatial rotation followed by a boost:
e
−e0
ˆ vβ/2
e
−e0
ˆ uα/2
= e
−e0
ˆ wδ/2
e
−iθ/2
. (2.6)
Active transformations do not change reference frames, so we can express the boosts
in Eq. (2.6) using the common basis vector e
0
and vectors orthogonal to it (ˆ u, ˆ v, ˆ w).
The rotation e
−iθ/2
is called a Thomas rotation. It is an important physical effect.
We prove Eq. (2.6) by solving it uniquely for ˆ w, δ, i, θ on the right side. Expand
the exponentials and equate the terms with an e
0
component and those without:
s
α
c
β
ˆ u + c
α
s
β
ˆ v = s
δ
ˆ we
−iθ/2
,
c
α
c
β
−s
α
s
β
ˆ vˆ u = c
δ
e
−iθ/2
, (2.7)
where s
α
= sinh(α/2), c
β
= cosh(β/2), etc. Divide to obtain ˆ w and δ:
9
tanh(δ/2) ˆ w =
s
α
c
β
ˆ u + c
α
s
β
ˆ v
c
α
c
β
−s
α
s
β
ˆ vˆ u
=
tanh(α/2)ˆ u + tanh(β/2)ˆ v
1 −tanh(α/2) tanh(β/2)ˆ vˆ u
. (2.8)
When the boosts are parallel, this reduces to the familiar "addition of velocities" for-
mula in special relativity.
Equate the bivector parts and the scalar parts of Eq. (2.7):
s
α
s
β
ˆ v ∧ ˆ u = c
δ
sin(θ/2) i,
c
α
c
β
−s
α
s
β
ˆ v ˆ u = c
δ
cos(θ/2).
Divide to obtain i and θ:
tan(θ/2) i =
s
α
s
β
ˆ v ∧ ˆ u
c
α
c
β
−s
α
s
β
ˆ v ˆ u
=
tanh(α/2) tanh(β/2) ˆ v ∧ ˆ u
1 −tanh(α/2) tanh(β/2) ˆ v ˆ u
.
The rotation plane i is ˆ v ∧ ˆ u. To obtain a scalar expression for tan(θ/2), substitute
ˆ v ∧ ˆ u = sin φi and ˆ v ˆ u = −cos φ, where φ is the scalar angle from ˆ v to ˆ u.
2.4.5. Dirac's electron theory. The most elegant formulation of Dirac's rela-
tivistic quantum theory of the electron is in the spacetime algebra.
Recall from Sec. 2.2.5 that Pauli's theory represents spins by 3D rotations, i.e., by
members of SO(3). Dirac's theory represents spins by members of SO(1, 3).
The VA version of Dirac's equation uses the four 4 4 complex Dirac matrices
γ
0
, γ
1
, γ
2
, γ
3
associated with (but not identified with) orthogonal directions in space-
time. These matrices generate the Dirac algebra. The Dirac algebra is isomorphic to
G
1,3
, with γ
j
↔ e
j
.
Both Pauli and Dirac invented the geometric algebra for the space(time) in which they
were working out of necessity, without realizing that the algebras are not special to quantum
theory, but have deep geometric significance and wide applicability. From the perspective
of GA, the Pauli and Dirac algebras are uninteresting matrix representations of a geometric
algebra, which obscure the physical content of their theories.
9
To see that the right side of Eq. (2.8) is in fact a vector, multiply its numerator and denominator
by 1 − tanh(α/2) tanh(β/2)ˆ uˆ v. Then use Theorem 1.5.11a and ˆ u··· ˆ v =
1
2
(ˆ uˆ v + ˆ vˆ u).
18
3 Geometric Calculus
3.1 The Derivative
3.1.1. The norm. Expand a multivector A with respect to a standard basis:
A =
J
a
J
e
J
. Then the norm of A is defined by [A[
2
=
J
[a
J
[
2
. Limits in geometric
calculus are with respect to this norm.
Of course [A+B[ ≤ [A[ +[B[. It is easy to see that if A ⊆ B then [AB[ = [A[ [B[.
3.1.2. The vector derivative. In vector calculus ∇ =
j
e
j
∂
ej
, where ∂
ej
=
∂/∂x
j
is the directional derivative in the direction e
j
. We call ∇ the vector derivative
because it acts algebraically as a vector: we can multiply it by a scalar field f, giving
the vector field ∇f; dot it with a vector field f , giving the scalar field ∇ f ; and cross it
with f , giving the vector field ∇f . But ∇f , a product of vectors, cannot be formed
in vector calculus.
In geometric calculus ∇f does make sense. This product of vectors is scalar +
bivector: ∇f = ∇ f +∇∧ f , just as with our fundamental identity, Eq. (1.5). In this
way ∇ unifies the divergence and curl and generalizes the curl to nD.
The geometric calculus identity ∇
2
f = ∇(∇f ) cannot be written in vector calculus.
Instead, we must resort to ∇
2
f = ∇(∇ f ) −∇(∇f ) – and this only in R
3
.
3.1.3. Analytic functions. Let f(x, y) = u(x, y) + v(x, y) i , with u and v real
valued. Then
∇f = e
1
(u
x
+ v
x
i) +e
2
(u
y
+ v
y
i) = e
1
(u
x
−v
y
) +e
2
(v
x
+ u
y
) .
From the Cauchy-Riemann equations, ∇f = 0 ⇔ f is analytic. Generalizing, we call a
multivector field F on R
n
analytic if ∇F = 0.
This definition leads to a generalization of standard complex analysis to n (real)
dimensions. Many standard results generalize. A simple example: since ∇F = 0 ⇒
∇
2
F = ∇(∇(F)) = 0, analytic functions are harmonic functions. Most important,
Cauchy's theorem and Cauchy's integral formula generalize, as we shall see.
3.1.4. The derivative. It violates the spirit of GA to write f above as a function
of the coordinates (x, y). Henceforth we shall think of it, equivalently, as a function of
the vector x = e
1
x +e
2
y, and similarly in higher dimensions. As an example, you can
verify that ∇(xa) = na.
With this change of viewpoint we can generalize the derivative from functions of
a vector to functions of a multivector (and taking multivector values). First, the
directional derivative of F in the "direction" A is
∂
A
F(X) = lim
τ→0
F(X + τA) −F(X)
τ
.
If A contains grades for which F is not defined, then define ∂
A
F(X) = 0 . For example,
if F is a function of a vector x, then ∂
e1e2
F(x) = 0 .
The geometric calculus (multivector) derivative can now be defined as a generaliza-
tion of the vector derivative ∇ =
j
v
j
∧ (Ru
j
R
−1
) = 0 . The right side has an agreeable
geometric interpretation: the (bivector) sum of the parallelograms spanned by the v
j
and the rotated u
j
is zero. The equation can be solved for R.
Geometric calculus uses only one derivative, ∇φ, to solve this problem. Vector calculus
must break ∇φ into its four components ∂φ/∂r
i
. Is generally best not to break a multivector
into its components, just as it is generally best not to break a complex number into its real
and imaginary parts or a vector into its components.
3.1.6. Electromagnetism. Elementary electromagnetic theory is usually formu-
lated in 3D vector calculus. Two vector fields, the electric field e and the magnetic field
b, represent the electromagnetic field. The charge density scalar field ρ and the cur-
rent density vector field j represent the distribution and motion of charges. Maxwell's
equations are the heart of the theory:
∇ e = 4πρ, ∇e = −
∂b
∂t
, ∇ b = 0, ∇b = 4πj +
∂e
∂t
. (3.1)
The spacetime algebra G
1,3
of Sec. 2.4.2 provides a more elegant formulation. A
spacetime bivector field F unifying e and b represents the electromagnetic field. A
spacetime vector field J unifying ρ and j represents the distribution and motion of
charges. Maxwell's four equations become a single equation: ∇F = J. What a simple
equation: the derivative of one single grade field is another.
Multiplying ∇F = J by e
0
and equating the 0-, 1-, 2-, and 3-vector parts yields
the standard Maxwell equations, Eqs. (3.1).
Calculations using the G
1,3
formulation of Maxwell's equations are often easier, and
sometimes much easier, than using the R
3
formulation. This is in part due to the fact that
the GA derivative ∇, unlike the divergence and curl in Eqs. (3.1), is invertible, as we will
see in Sec. 3.2.4.
In geometric calculus the same derivative ∇ is used in the definition of an analytic
function, the minimization example, Maxwell's theory, the full Pauli theory, and the full
Dirac theory. That's unification.
20
3.2 The Integral
3.2.1. The integral. Let S be a compact oriented s-dimensional manifold in R
n
.
Let F(x) be a multivector valued field on S. Then we can form the integral
_
S
dSF. (3.2)
Here dS = [dS(x)[ I(x), where [dS(x)[ is an element of s-volume of S at x and I(x) is
the pseudoscalar of the tangent space to S at x. For example, if S is a surface in R
3
,
then [ dS[ = dA is an element of area of S and I(x) is the pseudoscalar of the tangent
plane to S at x (a bivector). If S is a volume in R
3
, then [dS[ = dV is an element of
volume of V and I(x) ≡ I is the pseudoscalar of R
3
. Note that the order of the factors
in the integrand is important, as the geometric product is not commutative.
Eq. (3.2) is a directed integral, using the directed measure dS. The integral
_
C
f(z) dz from complex analysis is a directed integral, a special case of Eq. (3.2).
3.2.2. The fundamental theorem. Let F be a multivector valued function
defined on S. The tangential derivative ∇
S
F of F can be thought of as the projection
of ∇ on S.
The Fundamental Theorem of (Geometric) Calculus
_
S
dS ∇
S
F =
_
∂S
dC F. (3.3)
This is a marvelous theorem. Its scalar part is equivalent to Stokes' theorem for
differential forms. Thus the divergence and Stokes' theorems of vector calculus are
special cases of Eq. (3.3). A generalization of Cauchy's theorem to a manifold is an
obvious special case: If F is analytic on S, i.e., if ∇
S
F = 0, then
_
∂S
dCF = 0.
The fundamental theorem also generalizes the residue theorem. Let Ω
n
be the
(n−1)-volume of the boundary of the unit ball in R
n
(e.g., Ω
2
= 2π). Let δ be Dirac's
delta function. Then we say that F has a pole at x
k
with residue the multivector R
k
if ∇
S
F(x) = Ω
n
R
k
δ(x −x
k
) near x
k
. Eq. (3.3) holds if F is analytic in S except at a
finite number of poles at the points x
k
. Thus
_
∂S
dC F =
_
S
dS ∇
S
F =
_
S
[ dS[I(x)
_
k
Ω
n
R
k
δ(x −x
k
)
_
= Ω
n
k
I(x
k
)R
k
.
If S is a region of a plane and the R
k
are complex numbers, then this reduces to
the usual residue theorem.
With directed integrals, complex analysis becomes a subdiscipline of real analysis: it
is the study of functions F with ∇F = 0. Traditional real analysis does not use directed
integrals. Unification with complex analysis cannot be achieved without them. For example,
consider Cauchy's theorem:
_
C
f(z) dz = 0. Green's theorem gives the real and imaginary
parts of the theorem separately. But the theorem cannot be written as a single formula in
vector calculus or with differential forms.
21
The fundamental theorem can be generalized, with several important corollaries.
Eqs. (3.4) and (3.5) below are examples.
3.2.3. F from ∇F and boundary values. Let F be a multivector valued function
defined in a region V of R
n
. Then for x
0
∈ V ,
F(x
0
) =
(−1)
n
Ω
n
I
__
V
x −x
0
[x −x
0
[
n
dV∇F(x) −
_
∂V
x −x
0
[x −x
0
[
n
dSF(x)
_
. (3.4)
In particular, F[
V
is determined by ∇F[
V
and F[
∂ V
. This generalizes Helmholtz's
theorem from 3D vector calculus.
If F is analytic, then Eq. (3.4) becomes
F(x
0
) = −
(−1)
n
Ω
n
I
_
∂V
x −x
0
[x −x
0
[
n
dSF(x) .
This is a generalization of Cauchy's integral formula. Geometric calculus enables us to
see better how Cauchy's formula fits into the scheme of things: it is a very special case of
Eq. (3.4), which applies to all functions in all dimensions.
3.2.4. Antiderivatives. Eq. (3.4) can be used to prove that F has an antideriva-
tive P : ∇P = F. In other words, ∇ is invertible. Clearly, two antiderivatives of F
differ by an analytic function. This generalizes the n = 1 case, where the analytic
functions are the constant functions.
For n > 2 an antiderivative is given by
P(x
0
) =
1
(n −2) Ω
n
__
V
[ dV[
[x −x
0
[
n−2
∇F(x) −
_
∂V
[ dS[
[x −x
0
[
n−2
n(x) F(x)
_
, (3.5)
where n(x) = ±I(x)
∗
is the unit outward normal to ∂V . (In fact, n = I
−1
I(x), where
I is the pseudoscalar of R
n
.)
If F is analytic, then the first integral is zero.
If [ ∇F(x) [ = O( [ x[
−2
) and lim
x→∞
xF(x) = 0 , then
P(x
0
) =
1
(n −2) Ω
n
_
R
n
[ dV[
[x −x
0
[
n−2
∇F(x) .
3.2.5. Potentials, Fields, Sources. Three fields P, F, and S with ∇P =
F and ∇F = S are called a potential, a field, and a source, respectively. Given a
source S with suitable boundedness, P and F always exist. A common situation in R
3
is S = s − s
∗
, where s is a scalar field and s is a vector field with ∇ s = 0. Then we
can take P = p − p
∗
, where p is a scalar field, p is a vector field with ∇ p = 0, and
F = f , a vector field. Vector calculus cannot form S or P. In particular, it cannot use
the simple formulas F = ∇P and S = ∇F. Instead, f = ∇p − ∇ p, s = ∇ f , and
s = ∇f .
22
4 Other Models
4.1 The Homogeneous Model
We have represented subspaces of R
n
(points, lines, planes, ... through the origin)
with blades of G
n
. This is the vector space model. But we have not represented
translated subspaces (points, lines, planes, ... not through the origin). Yet there is
nothing special geometrically about the origin.
4.1.1. The homogeneous model. The homogeneous model represents and ma-
nipulates all points, lines, planes, ... much as the vector space model represents and
manipulates those through the origin. It is the coordinate-free geometric algebra ver-
sion of homogeneous coordinates used in computer graphics and projective geometry.
The homogeneous model represents a translated k-dimensional subspace of
R
n
, and k-volumes in it, with a (k + 1)–blade of G
n+1
.
In particular, a point, which is a translation of the 0-dimensional subspace ¦0¦, is
represented by a 1-blade, i.e., a vector. To see how this works, extend R
n
with a unit
vector e orthogonal to R
n
. Then we have R
n+1
, and with it G
n+1
.
The homogeneous model represents a point P at the end of the vector
p ∈ R
n
with the vector p = e + p ∈ R
n+1
. (Only vectors in R
n
will be
denoted in bold.) Fig. 8 shows a useful way to visualize this for R
2
.
Fig. 8: Vector p = e + p repre-
sents the point P. Fig. 9: Bivector p∧q represents
the oriented segment PQ.
The vector p is normalized in the sense that p e = e e + p e = 1. However, the
representation is homogeneous: for scalars λ ,= 0, λ(e +p) also represents P .
The homogeneous model represents the oriented length PQ with the bivec-
tor p ∧ q. See Fig. 9.
Let v = q − p and v
be the vector with endpoint on the line PQ
and perpendicular to it. See the figure at right. Then p ∧ q determines,
and is determined by, v and v
) ∧ V shows that p ∧ q ∧ r does not determine p, q, or
r.
23
In VA and the vector space model of GA, vectors do double duty, representing both
oriented line segments and the points at their ends. In the homogeneous model oriented
line segments and points have different representations.
We now have two geometric algebras for Euclidean n-space: the vector space model G
n
and the homogeneous model G
n+1
. In both, blades represent geometric objects. In the
vector space model a vector represents an oriented line segment. In the homogeneous model
it represents a point. In the vector space model an outer product of vectors represents an
oriented area. In the homogeneous model it represents an oriented length. Oriented areas
and oriented lengths are different, yet they share a common algebraic structure. We have
to learn geometric algebra only once to work with both.
4.1.2. The Euclidean group. Rotations and reflections generate the orthogonal
group O(n). Include translations to generate the distance preserving Euclidean group.
In R
n
translations are not linear. This is a problem: "The orthogonal group is mul-
tiplicative while the translation group is additive, so combining the two destroys the
simplicity of both."[4]
The homogeneous model solves the problem. The subgroup of O(n+1) which fixes
e is isomorphic to O(n). For members of this subgroup map R
n
, the subspace of R
n+1
orthogonal to e, to R
n
. And since p q = 1 +p q, the map is also orthogonal on R
n
.
For a fixed a ∈ R
n
, consider the linear transformation x → x+(x e)a of R
n+1
. In
particular, e +p → e + (p +a). This represents a translation by a in R
n
. In this way
translations are linearized in the homogeneous model.
4.1.3. Join and meet. Geometric objects join to form higher dimensional objects.
For example, the join of two points is the line between them; the join of intersecting
lines is the plane containing them; and the join of a line and a point not on it is the
plane containing them.
Geometric objects meet in lower dimensional objects, their intersection. Thus the
meet of an intersecting line and plane is their point or line of intersection, and the meet
of two intersecting planes is their line of intersection.
GA defines the join and meet of two blades (only) to represent the join and meet
of the geometric objects that they represent.
The join of blades A and B is the span of their subspaces. From Eq. (1.11), if
A ∩ B = ¦0¦, then their join J = A ∧ B. However, there is no general formula for
the join in terms of the geometric product, as there is for the inner and outer products
(Eqs. (1.7) and (1.9)).
The meet of A and B is the intersection of their subspaces. In this section, let
X
∗
= X/J, the dual of X in join J of A and B. The meet of A and B is given by
A∨ B = A
∗
B. (4.1)
Before proving this we give examples of the join and meet.
In the vector space model, two lines through the origin join to form a plane. Two
examples are e
1
∧ e
2
= e
1
e
2
and e
2
∧ e
3
= e
2
e
3
. The two planes meet in a line:
(e
1
e
2
) ∨ (e
2
e
3
) = (e
1
e
2
)
∗
(e
2
e
3
) = −e
2
.
In the homogeneous model, two points join to form a line. Two examples are
e
∞
⇔ x =
a
a + b
p +
b
a + b
q,
a
a + b
+
b
a + b
= 1.
The representation does not determine p or q. But it does determine the distance
between them: (p ∧ q ∧ e
∞
)
2
= [p −q[
2
.
26
Planes. The representation p ∧ q ∧ e
∞
of lines gives a test: m, p, q collinear
⇔ m ∧ (p ∧ q ∧ e
∞
) = 0. If they are not collinear, then they determine a plane. The
direct representation of the plane is m∧ p ∧ q ∧ e
∞
, a "sphere through infinity". The
area A of the triangle with vertices m, p, and q is given by (m∧ p ∧ q ∧ e
∞
)
2
= 4A
2
.
A dual representation of the plane through the point p and orthogonal to n is the
vector π = n +(p n)e
∞
. This follows from x π = (x −p) n and the VA equation of
the plane (x −p) n = 0. A different form follows from the identity used in Eq. (4.5):
π = n + (p n)e
∞
= −(p e
∞
) n + (p n) e
∞
= p (n ∧ e
∞
). (4.6)
If instead we are given n and the distance d of the plane to the origin, then the VA
equation x n = d gives the dual representation π = n+de
∞
. From this, d = −π e
0
.
Circles. A direct representation of the circle through p, q, and r is p ∧ q ∧ r. We
can think of the lines p ∧ q ∧ e
∞
above as circles through infinity.
We determine the dual representation of the circle with center c, radius ρ, and
normal n. The circle is the intersection of the dual sphere c −
1
2
ρ
2
e
∞
(Eq. (4.5)) and
the dual plane c (n ∧ e
∞
) (Eq. (4.6)). From the dual representation of the meet, Eq.
(4.2), the dual representation of the circle is the bivector (c −
1
2
ρ
2
e
∞
) ∧( c (n∧e
∞
) ).
4.2.3. Representing geometric operations. Now that we know how to rep-
resent Euclidean objects in the conformal model, we turn to representing geometric
operations on the objects.
Let O be an orthogonal transformation of R
n+1,1
. Then pp = 0 ⇒ O(p)O(p) = 0.
It is not difficult to show that every null vector of G
n+1,1
is of the form λe
∞
or λp for
some p ∈ R
n
. Thus O induces a map of
¯
R
n
to itself. Such maps form the conformal
group of
¯
R
n
. (They preserve angles, but not necessarily orientations.)
The conformal group is generated (redundantly) by the following transformations:
• Rotations. A rotation around the origin by angle iθ is represented in the con-
formal model just as in the vector space model: p → e
−iθ/2
p e
iθ/2
. (Eq. (2.4).)
• Translations. The translation p → p+a is represented by p → e
−ae∞/2
pe
ae∞/2
.
Note that since e
∞
a = −ae
∞
and e
2
∞
= 0, e
±ae∞/2
= 1 ±ae
∞
/2 (exactly).
• Reflections. A reflection of the point p in a hyperplane with normal vector n
is represented in the conformal model just as in the vector space model: p →
−np n
−1
. (Theorem 1.5.11b.)
• Inversions. The inversion p → p
−1
is represented in the conformal model by
a reflection in the hyperplane normal to e
+
: p → −e
+
p e
+
. To normalize the
result, divide by the coefficient of e
0
(which is p
2
).
• Dilations. The dilation p → αp is represented in the conformal model by
p → e
−ln(α)E/2
p e
ln(α)E/2
, where E = e
∞
∧e
o
. Note that e
βE
= cosh β+E sinh β.
To normalize the result, divide by the coefficient of e
0
(which is 1/α).
The subgroup of the conformal group fixing e
∞
is the Euclidean group of R
n
. For
rotations, translations, and reflections fix e
∞
, while inversions and dilations do not.
27
4.2.4. Conformal transformations as automorphisms. As we have seen,
the representation of a conformal transformation of R
n
is constructed simply from
its geometric description. The representations are of the form p → ±V p V
−1
. They
extend to automorphisms of the entire geometric algebra. The automorphisms are of
the form M → ±V MV
−1
. The automorphisms preserve the geometric, inner, and
outer products and grades. They compose simply, as do rotations in the vector space
model. (Section 2.2.2). The following examples show how all this can be used to
advantage.
The rotations in Sec. 4.2.3 are around the origin. To rotate an object around a,
translate it by −a, rotate it around the origin, and translate it back by a. Thus a
rotation around a is represented by T
a
e
−iθ/2
T
−1
a
, where T
a
= e
−ae∞/2
. This says that
rotations translate in the same way as geometric objects. There is more. You can
quickly check that T
a
e
−iθ/2
T
−1
a
= e
−Ta(iθ)T
−1
a
/2
using the power series expansion of the
exponential. This says that the angle T
a
(iθ)T
−1
a
specifies the rotation around a. It is
the translation by a of the angle iθ specifying the rotation around 0.
We compute the angle θ between intersecting lines
1
and
2
. Translate the lines so
that their intersection is at the origin. They now have representations | 677.169 | 1 |
introductory text acts as a singular resource for
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Chapters discuss: function spaces and functionals, extension of
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elements of the theory of real functions on R.
The main goal of this third edition is to realign with the
changes in the Advanced Placement (AP ) calculus syllabus and the
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This book provides a self-contained and rigorous introduction to calculus of functions of one variable, in a presentation which emphasizes the structural development of calculus. Throughout, the authors highlight the fact that calculus provides a firm foundation to concepts and results that are generally encountered in high school and accepted on faith; for example, the classical result that the ratio of circumference to diameter is the same for all... more...
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be
found,
and
The
which
It is usual
first letters
of
the
or
bers
num-
therefore
are
represent known
to
alphabet,
a,
b, c, "c,
by
necessary
be strictly
not
rule, and so need
be either whole
fractional.
The
or
a
obeyed.
word
quantity
numbers,
numbers
Numbers
number.
either
numbers.
unknown
however
be
may
may
is often
word
the
used
last letters
"vvith the
this
z;
meaning
same
used
integer is often
x,y,
is
as
instea'l of whole
number.
of
himself to the use
beginner has to accustom
letters for representing numbers, and to leani the meaning
of the signs; we
portant
imshall begin by explaining the
most
shall assume
We
their use.
signs and illustrating
of Arithmetic,
that the student
has a knowledge of the elements
3.
The
and
that
he admits
truth
the
the
of
common
such
in all parts of mathematics,
to equals the wholes
are
equal, and
required
added
4.
The
number
sign + placed before
is to be
added.
represented by
T.
A.
Thus
6 is to be
number
a
a
+
added
as,
the
to
if equals be
like.
denotes
b denotes
notions
that
tho number
that
the
the
ber
num-
repi'e1
2
sented
by
6 is read
"
thus
a
number
number
number
sign placedbefore
a
-
representedby
a-b
sign,and
3, then
minus
plus b."
number
denotes that the
Thus a"b
denotes that tlio
is to be subtracted.
representedby b is to be subtixicted from the
The
5.
a
12.
represents
tj +
represent9 and h represent3,then a-\-h
The
sign+ is called the plttssign,and
If
a.
SIGNS.
PRINCIPAL
THE
a.
If
a
represents 6.
a
9 and
represent
The
sign
-
6 is read thus "a minus
"
denotes
is called the
b."
to add b to
that wo
c denotes
a, and then add c to the result ; a + b
b to a, and then subtract
the result;
to add
c from
are
denotes that we are to subtract b frc^n a, and tnen
a-b-^c
denotes
that we
tract
add c to the result; a"b"c
to subare
and
then
b from a,
subtract c from the result
6.
Similarlya
+
b
+
c
that
b represent
we
are
"
between
7. The
sign= denotes that the numbers
which it is placed are equal. Thus
b denotes that the
a =
number
representedby a is equal to the number
sented
repreAnd
of
the
b.
a-^b
denotes
that the sum
=
c
by
numbers
representedby a and b is equal to the nmnber
representedby c; so that if a represent 9, and b represent
3, then c must represent 12. The sign = is called the
is read thus "a equals b" or
sign of equality,and "="
"a is equal to b."
8.
The
9.
The
.
denotes
that the numbex^s between
which
it stands
to be multiplied together. Thus
are
that the number
a X " denotes
representedby " is to be
multipliedby the number representedhy b. U a represent
9, and b represent 3, then axb
represents27. The
signX is called the sign of midtiplication,and axb
m
read thus " a into b." Similarly
axbxc
denotes the product
of the numbers
and
c.
representedby a, b,
is however
often
signof multiplication
sake of brevity
"?" is used instead
; thus
for the
and has the
axbxc,
The
numbers
Thus 45
sign x
and
meaning; so also
has the same
meaning.
same
abc is used
omitted
of
a
x
5,
instead of
not be omitted when
must
sign of multiplication
are
exjjressedin the ordinary way by figures.
of 4 and
cannot be used to representthe produc-t
is adopted. Sometimes, how^ever,
which
way
a
of
instead
thus
the
x
is
used
is
used
4.5
sign
;
stead
inpoint
To prevent any confusion between
of 4x5.
the
of
and the point
multii^lication,
pointthus used as a sign
decimal
notation
for
used in the
it is advisable
fractions,
in
latter
the
the
to place
case
pomt
higher up; thus
is the
be
4*5 may
kept to
instead
not used
denote
4 +
"
.
But
in fact the point is
of the
signx except in cases where there
be no ambiguity. For example,1.2.3.4 may be put for
can
because
the pointshere will not be taken for
1x2x3x4
decimal points.
The
pointis
placed instead
sometimes
of the sign x
between two letters ; so that a fcis used instead oi axh.
But the point is here superfluous,
because,as we have
Nor is the point,
the
nor
said,ab is used instead oi axh.
dinary
sign X necessary between a number
expressedin the or.
by a figureand a number representedby a
letter; so that,for example,3a is used instead of 3xa,
and has the same
meaning.
way
which preThe sign -^ denotes that the number
cedes
it is to be divAded by the number which follows it.
Thus a-^h denotes that the number
by a is to
represented
be divided by the number
by h. \i a represent
represented
2
.
and
then
a-^h
h
represents The
represent4,
8,
sign is called the sign of division^and a^6 is read
10.
-=-
thus "a
"2/b."
There is also another way of denotingthat one
is to be divided by another; the dividend is
over
the
divisor with
used instead of
a -^
a
line between
6,and has the
same
them.
ber
num-
placed
Thus
v
is
meaning.
alphabet,and the signswhich
have alreadyexplained,
we
togetherwith those which may
occur
hereafter,arc called algebraicalsymbols, because
about which we
they are used to representthe numbers
the operations
performed on them, and
may be reasoning,
11.
The
letters of the
1"2
4
EXAMPLES.
L
their relations to each other.
symbols is called
ejc2)ression.
an
shall
We
12.
Any collection of Algebraical
algebraicalexpression,or briefly
an
give son;o
now
examples
as
Suppose
rt
l, h
=
%
=
7a
+
3"-2tZ+/=7
2ab
+
8bc-ac
If a=l, 6
2"
9a
3.
7ae
5.
abcd
+
+
d=5,
o,
6-10
+
e
130
+
9d-af.
4e-3a-3b
4.
8abc-bcd+9cde-de/.
Hcd
Ode
cde
5hcd
6.
+ 5c.
:^+
ab
2a
+
56
c^
3" + 2C
b+d
rt+c
"^
20c
be
cd
de
a +
d^-b
6~^c
b +
20
7e
+
,
e
^
bcd~-i;~
c +
,
6+
d
"
e +
a+b+c+d+e
c+e
"*"
^
2ac
'^~T~
c
'
j-~.
he
ae
d
c
Qb
??-^-.
fibde
,^
10.
,
"r+
+
"
\2a
6ade
-
numen-
2.
^
9.
6^46.
5,/=o, find the
following
expressions:
'
'
of
Then
"? =
abce + abde + acde + bcde.
8bc
values
10-3.
52-
=
b
4ac
explained;
G,f=0.
=
0=
+
48-6
d/=4:-\-
3c-2/
+
3bc +
+
=
2, c=:3, d=4,
=
cal values of the
L
+
c
exercise
an
the symbols v. Inch have been
in the use
these examples consist in findingthe numerical
certain algebraical
expressions.
of
e-d+c-b
+
a'
c +
3"
c-d'
COEFFICIENT.
FACTOR.
II.
Factor.
Wlien
13.
POWER.
TERMS.
Coefficient.Power.
5
Terms.
number
consists of the product of two
of the latter is called a factor of
one
numbers, each
the product. Thus, for example, 2 x 3 x 5
each
30 ; and
of the numbers
2, 3, and 5 is a factor of the product 30.
Or we
regard 30 as the product of the two factors,
may
factors 6 and 5,
2 and
15, or as the })roductof the two
or
more
=
or
as
the
productof
tlie two
factors 3 and 10. And
so, also,
the product of the two factors
consider
4ab as
may
4 and
"", or as the product of the two factors 4a and ",
the product of the two factors Ah and a; or we
or as
may
it
and
of
the
the
three
factors
and
4
h.
as
a
regard
product
we
"When a number
consists of the product of two
of the other
each factor is called the coefficient
factors,
factor: so that co^^c/"?"^
is equivalentto co-factor. Thus
14.
and
ab^ we call 4
the coeflicicnt of ah, and
ah the coefficient of 4; and
considering4.ah as the product of Aa and ", we call Aa
There will
the coefficient of 6, and h the coefficient of 4a.
the
Avord
in
coefficient
bo little occasion to use
practicein
in which
the
is
the
that
case
first,
except
any of these cases
4 is regarded as the coefficient of ah ; but for the sake of
of
distinctness we
coefficient
speak of 4 as the numerical
Thus
the numerical
ab in 4"6, or briefly
as
coefficient.
consideringAah
when
as
the
product
factor
a
letter
or
4
factor which
is
represented
other
a
figiu-e figures,and of anthat is by
,\'hichis representedalgebraically,
product consists of
thsit is by
arithmetically,
a
of
one
etters,the former
or
cal
factor is called the numeri-
coefficient.
all the factors of a product are
equal,the
product is called a poicer of that factor. Thus 7 x 7 is
called the second power
of 7 ; 7 x 7 x 7 is called the third
of 7 ;
of 7; 7 xl ^1 "il is called the fourth power
"power
and so on.
In like manner
ax
a is called the second power
of a; ax ax
of a; ax
ah
called the third power
ax
a
ax
is
is called tha fourth power
of a; and so on.
And a itself
15.
sometimes
Wlien
called thc^/v^ power
of
a.
6
FACTOR.
COEFFICIENT.
POIVER.
TERMS,
denoted
thus: instead of
briefly
power is more
v-c
expressingall the equal factors,
ex])ress the factor once,
it the number
and placeover
which
indicates how often it
is to be repeated. Thus "* is used to denote ay. a; a^ is
used to denote a x a x "; "* is used to denote a x " x a x ";
And
and so on.
a^ may be used to denote the first power
of a, that is a itself;
that a^ has the same
so
meaning as a.
16.
A
17.
A
number
placed
over
tunes the latter occurs
called an index of the poicer^
as
many
or,
an
index,
briefly,
or
or
an
another
to indicate how
factor in a poAver, is
a
exponent of the power
exponent.
Thus, for example,in a' the exponent
exponent is n.
18. The
student must
and an exponent. Thus 3"!means
coefficient
here 3 is a coefficient.
But d^ means
c
exponent.
is 3; in a" the
between
distinguish
very carefully
a
here 3 is an
;
three times c;
times c times c;
That is
(?
=
cy.cy.c.
of a, that is a\ is often called the
of ", that is
square of a, or a squared; and the third power
cubed.
There
cube
of
no
are
a^ is often called the
a, or a
inroad
"^
for
the
thus
words
in
such
use
higher powers;
*'
*'a to tJiefourth power," or briefly
a to the fourth.^'
19.
parts connected by the
it is called a simple expression. If an
parts connected by the signs 4- and
expressioncontain
an
Abe,and 5aV
are
"
"^
\Yhen
an
expressionconsists of two terms it is
it consists of three
called a binomial
: when
expression
terms it is called a trinomial
expression;any expression
consistingof several terms may be called a multinomial
21.
or
expression,
a
expression.
polynoinial
Thus
2a
3^
+
is
binomial
a
trinomial expression; and
multinomial
expressionor
Each
22.
the
of
a
"
expression;
b-"c
d
"
7
TERMS.
POWER.
COEFFICIENT.
FACTOR.
e
"
a
1h
"
+
be
miiy
is
called
5c
a
a
polynomial expression.
a
which
letters
of the
in
occur
is
term
a
term, and
the number
of the
called a dimension
Thus
a^l^c or
lettei*s is called the degree of tlie term.
is
said
of
be
six
aKaxbxb-xhxc
dimensions
of
to
or
coefficient is not counted ;
the sixth degi-ee. A numencal
a^b'^ are
thus
of the same
i)d'b* and
dimensions,
namely
dimensions.
seven
word
refers
ditnensicms
to
algebraicalmultiplicationsinvolved in the
of its
tcnn; that is,tlie degree of a term, or the number
dimensions, is the sum
of the ej-ponents of its algebraical
remember
that if no
factors, provided we
exponent be
expressed the exponent 1 must be understood, as indicated
the
of a and h
that the sum
denote
it thus {a-\-h)'"c
denote
or
is to be multipHed by c\ we
that
mean
+ h] X c, or simply {a^'b)cox {a + h\c; here we
the wlioU of ct + " is to be nmltiplied
by c. Now if we omit
and this denotes that 6 only
have a + lie,
the brackets we
Thus, suppose
to
\a
multipliedby c and the result added to a. Sunilarly,[a-\-'b c)d denotes that the result expressed by
a-vh
c is to be
multipliedby d^ or that the whole of
be multipliedby d\ but if we
omit
the
c is to
a + "
have a + b
brackets we
cd, and this denotes that c only
is to be multipUed by d and the result subtracted from
is to be
"
"
"
"
a
+
b.
So
also
by
{a"b
"
hy d
(a
"
"
+
"
-
+
c)x{d-\-e)denotes
+
+
c
is to
This
e.
+ e);just as
c)((/
So also sj(a+ b +
result expressed by
of this result
bo
multiphed by
30.
drawn
a
as
{a
is called
denoted
a
"
"
c
b
b+c)x{d
"
a
thus
"
and
d
We
have
+
?
e\"
pressed
ex-
the
result
the result
ab.
ex-
expressed
a
as
line is
forming
is used with, the same
e). A line used for this pose
purSo also {a + b
c)-^{d+ e')
may
cxd+e
+
+
vincidum.
d+e
a-\-b
that
ahxabx
Sometimes
instead of using brackets
the numbers
which are to be treated
over
meaning
be
denotes
{ab)^
"
Thus
pressed
ex-
that we
to obtain the
are
-i-b+ c, and then take the square root
{a+ b-c)-^{d-he) denotes
by a + 6 "; is to be divided by
number.
one
result
c) denotes
also
Eressed
y d+e.
the
aLo be denoted
simiDlythus
may
'tis shortened into ab.
ax
So also {ab)*
denotes dbxah\ and
So
that the result
"
;
reallya
and
here
cinculum
the
used
line between
in
a
particular
sense.
31.
explainedoil
the signswh'!ch are
used
in algebra. Wo
the
cases
may observe tliat in some
word
to the two signs + and
sign is appliedspeciallj'
;
thus in the Fwule for SuDtractiou wo shall speak of chonging
now
-
11
III.
EXAMPLES.
and
the signs,meaning the signs +
; and
"
in
multiplication
ing
of SignSjmean-
and division we shall speak of the Rule
to the signs+ and
a iiile relating
"
all the terms of an expression
connected
are
is
in
what
indifferent
it
order
the
+
by
si^
they are
and
thus
tlie
5
5
+
7
+
7
same
give
result,
placed;
namely,
12; and so also a + h and h + a givethe same
namoly,
result,
the sum
of the numbers which are representedby a and 6.
We may express this fact algebraically
thus,
33.
a +
h=^h
+ a.
+ c +
h=b
'b
Similarly,
When
34.
by
the
a
+
+ c=a
+ c + a.
consists of some
terms
expression
and
terms
some
sign +
preceded by
an
ceded
pre-
the
sign we may %vrite the former terms first in any order
and the latter terms after them in any order we
we
please,
notions of arithmetic.
please. This is obvious from the common
Thus, for example,
"
,
7+8-2-3=8+7-2-3=7+8-3-2=8+7-3-2,
'b
a +
"
c-e=b-\-a-c"e=^a-\-b
In some
35.
terms further,
by
"
e
c=h^a-e-c.
"
change the order of the
may
mixingup the terms which are preceded
by the sign witli those which are precededby the sign+
Thus, for example,suppose that a represents
presents
10,and b reand
c represents5, then
6,
cases
we
"
.
a-"b"c=a
for we
arrive ^vithout any
all the cases.
"
c +
b=b-c
+ a)
at 11
difficulty
as
the result in
Suppose however that a represents2, b represents6,
and c represents
+ b presentsa
a-c
5, then the expression
bocAuse we are thus apparently
difficulty,
requiredto take
from a let-s,
a greater number
namely,5 from 2, It "nll
be convenient to agree that such an expression
c + b,
us a
"
when
same
an
c
is
greaterthan a, shall bo understood
At presentwe
thingus a + b-c.
as
a + b"c
expression
except when
to
shall not
c
mean
use
is less than
the
such
a +
b;
that a + b
shall consider
so
"
c
"
13
TERMS.
LIKE
will not causg any
the
6 + a to mean
difficulty.
Similarly,
we
6.
thingas a
same
"
value of an expressionremains
Thus the numerical
be the order of the terms
the same, whatever
which
may
have
it.
follows
as
we
from
This,
partly
our
seen,
compose
and subtraction,and partlyfi'om an
notions of addition
agreement as to the meaning which we ascribe to an expression
when
our
ordinary arithmetical notions are not
36.
applicable.Such
strictly
a
conveniiony
and
agreement is called in algebi-a
an
cojiventioiicd is the
jective.
correspondingad-
ffe shall often,as in Art. 34, have to distinguish
the terms of an expressionwhich are preceded by the sign
which
from the terms
+
are
preceded by the sign , and
the following
definition is accordingly
adopted. The terms
in an
are
expression which
preceded by the sign + are
called pontive terms, and the terms which are preceded
This definition is
by the sign are called negativeterms.
the
and no meaning
introduced
sake of brevity,
merely for
37.
"
"
^^ to be given to*the ^\*ord3
positiveand
in
is
the
definition.
wrhat
expressed
38.
It will be
seen
that
a
term
negativebeyond
m
may occur
the first term.
an
pression
ex-
Such
with the positiveterms, that is it is
is counted
term
a
treated as if the sign 4- precededit. It \d\\ be found that
in the order of the tenns, as to
if such a change be made
stood first and was
preceded
bringa term which originally
will
be
precededby
by no sign,into any other place,then it
the sign 4For example,
preceded by
no
dgn, namely
.
a-k-'b"c='b
+
a"c=h
"
c +
a;
sion,
here the term
no
a has
sign before it hi the first expresis
it
preceded
but in the other equivalentexpressions
the following
important
have
Hence
we
by the sign +.
addition to the definition in Art. 37 ; if a term he preceded
by no sign,the sign + is to be understood.
39.
at
or
all,
Terms
differ
they
are
said to be like when
they do not difiFer
wise
only in then* numerical coefficients; othersaid to be unlike,, Thus a, 4a, and la are
are
number
2d; 3a; for whatever
a may
6a
a"
present,
rehave
5a left,
if we
subti-act a from 6a we
and then
have
from
subtract
2a
6a
3a left. Similarly
we
the
Thus
8c.
proposed expression
h=2o; and 5c + 3c
form
the
in
be put
simpler
Now
we
be
is equivalent
to
expression
6a-a"
if
Zb
+
may
6"
"
=
=
3a
+
2"
+
8c.
This ia
consider the expressiona
46.
3"
3" from
7". For if Ave have first to subtmct
equal to a
and
46
then to subtract
from the remainder,
a number
a,
the
in
obtain
shall
one
we
operation by
requiredresult
this
the
follows
from
from
tions
nocommon
subtracting76
a;
Thus
of Arithmetia
Again;
"
"
"
a
"
There will be
to such a statement
41.
36-46=a"
no
as
76.
in gimg
now
difficulty
the folloidng,
-36-46=
a
ing
mean-
-76.
We
cannot subtract 36 from nothing and then subtract
46 from the remainder,so that the statement
just given is
not here intelhgible
in itself,
separated from the rest of an
it may
in which
be
sentence
algebraical
occr.r, but it can
of an
easilyexplained thus: if in the course
algebraical
operationwe have to subtract 36 from a number and then
to subtract 46 from the remainder,
we
may subtract 76 at
instead.
once
be led
to conjecture
meaning to
that
give some
the proposed statement
by itself,that is,apart fi'om rmy
this conjecture will bo
and
other algebraicaloperation,
As
found
the student
when
correct,
in the
advances
it is possibleto
a
subjecthe
larger treatise
on
may
Algebra
can
be
15
IV.
EXAMPLES.
consulted with advantage;but the explanationwhich
have given will be sufficient for the present.
we
of expressions
The
hke
8imj)Iifying
by collecting
essential
is the
terms
part of the. processes of Addition and
shall see in the next two
Sabtraction in Algebra,as we
42.
Chapters.
be useful for the
It may
ing
beginnerto notice that accordfo!lo^ving
expressionsare all
defmitions
the
to the single
symbol
equivalent
to
our
:
a}, 1
X
",
a
1
X
a,
4- a
o^
+
a
4-
1, -,
X
X
1,
find the
"=2, c=3, 6?=4, ^=5,
:
following
expressions
valu"s of the
-,
IV.
Examples.
If (2=1,
4-
a"l^
numericaJ
1.
a-2b+^c.
3.
{a+ l){h+ c)-Q)+ c){c-^d)
+ {c-\-d){d-"e).
2.
4.a + ^h
+
"? + (P.
5d + Ae
^c + Zd
^
b^t'j
5.
(a
6.
a*
d
26 4- 3c)'
-
-
AaV)
-
b^-2hc
^
9.
0 +
+
(Z"
-
+
2c 4-
d
+ e
Zd^f+ {c-2d+ Zef.
6"'Z"2 4^2,3+ ^a
_
"^
+
a^
4a^c 4- 6a V
-
4ar'
-
4-
^
7a-2b-Zc-4a
5a'
11.
3.i"-2a-
+
+
5b
Sab-2h'-ab
10.
a^-"2ab
+
+
5a4-a'
b^
a+b
15.
a
^{2e^+7b).
+
+
+
4c + 2a.
9b^- 2ab
-
W.
"4-9a'-4a'-6a.
b*+2bc
+
c*
b+c
c'^+2cd+d*
c+d
16.
!J{21^^
+ d*
a).
-
c*
iG
ADDITION.
V.
Addition.
three cases in Addition,
to make
It is convenient
namely, I. When the terms are all like terms and have the
all like terms
the terms
but
are
sarao
sign; II. When
have not all the same
sign; 111. When the terms are not
in order.
We shall take these thi-ee cases
all like terms.
43.
which
have the same
I. To add like terms
the numerical
prefix tlie common
coefficientSy
the common
letters.
annex
44.
Add
and
For
example,
Ga
+
^a
+
sign.
sign^
1a=lQa,
-2bc-1b"-9bc=-lSbc.
In the first example 6a
See
to + 16a.
II.
is
equivalentto +6a,
and
IGa
Art. 38.
To
add
which
like teims
have not all the
all the positive numencal
same
sign. Add
coefficients
and all the negative numerical
inio one
coefficients
sum,
the
into another; take
difference of these two sums,
the
the
the common
sign of
prefix
greater, and annex
45.
letters.
For
example,
la-^a
Ua
+
2bc-7bc-3bc
in
a-5a-^a=l^a-Wa=^a,
4bc-t-5bc
-
6bc=llbc-l6bc=
^^^'
^^ ^^^ terms which
together the terms which
are
the second case, and put down
A^}'
Add
+
+
preceded by
For
its proper
example;
4a +
5b-lc
+
add
not all like terms.
like terms
by the rule
the otiier terms
each
are
sign.
together
M,
and
-5bc.
3"-"
-a
+
+ 2c +
5rf,9a-26-c-4
Sb + 4c-3d
+ e.
It is convenient to
the terms m columns,so
an^ange
terms
shall stand in the same
colunm; thus we
to take 7 + 3 from 12; the result
if we
firsttake 7 from 12, and then take 3
as
remainder; that is,the result is denoted by
12-7-3.
Thus
Here
because
Art
12-(7
we
we
3)=12-7-3.
+
enclose 7 + 3 in brackets in the firstexpression,
of 7 + 3 from 12; see
to take the whole
are
29.
Sunilarly 20-(5 +
In like manner,
the result is the
then take c from
denoted by a
6
"
6 + c from a;
if we
first take h from a, and
as
same
the remainder; that is,the result is
we
enclose h
are
to take
have
c.
-
a-{h
we
2)=20-5-4-2.
we
suppose
Thus.
Here
because
4 +
c)
+
=
a-h-c.
c'in brackets in the first expression,
to take the whole of " + c from a.
Similarly
a
"
+
{b+
d)=a-b"c
c +
d.
"
If
from 12.
Next suppose we
have to take 7-3
but
have
thus
take 7 from 12 we
obtain 12-7;
we
we
from 12, for we
had to take,not 7, but 7
taken too much
49.
diminished
and thus we
by
Hence
3.
obtain 12
Similarly 12-(7
we
increase the result
must
-
(7 3)=
+
3-2)=12-7-3
-
12
-
7
by
3;
3.
+
+
2.
In like manner,
have to take b"c from a.
suppose we
If we
have thus
take b from a we
obtain a"b; but we
taken too much
from a, for we
had to take, not b, but b
dimuiished by c. Hence we must increase the result by c ;
and thus we obtain a"{b
c)=a b + c.
"
Similarly
50.
a
"
{b+
c
"
"
d)=a
b-c
"
+
d.
Consider the
example
a-{b-^c-~d) a-b-c
=
that is,if b
+
c-d
be
subtracted
+
from
d;
a
the
result is
2"2
20
SUBTRACTION.
a-b~c+d.
Here we
see
that, in the expressionto be
subtracted there is a term
d, and in the result there is
the correspondingterm +d; also in the expression to be
"
in the result there is a
there is a term
4- c, and
to be subtracted there is a
c ; also in the expression
is
and
the
there
in
result
b.
a term
b,
subtracted
term
"
term
"
in the
followingrule for
consideringthis example,and
From
the
others
preceding Articles we obtain the
Subtraction : change the signs of all the terms
and then collect the
to be subtracted,
two
in
the
terras
pression
ex-
as
in
Addition.
subtract Zx
y + z.
the terms to be subtracted ; thus
then collect as in addition;thus
For
example; from
Change the signsof all
we
obtain
"3x
4a;
From
+
y"z;
3y
"
4x
+ 2z"3x
Sa!*+ 5a^-
+ y-z
6x^ -7^
Change the signs of
and proceed as in addition
3y
"
+
2z
"
a;~2y
=
5 take 2a^
+
2a?
-
all the
3;zr*
+ 5^-
to
terms
have
; thus we
Qa?-1x-^
5
bx^*+ %x^"
7
-2s!^+2!Ji?-
ar*+ 7^-lla;2-
a?
+
z.
+
be
fiX^ -6^-7.
subtracted
+ 12
beginnerwill find it prudent at firstto go through
here ; but he may
have done
the operation as fullyas we
the result
himself
to putting do\Mi
graduallyaccustom
without
actuallychanging all the signs,but merely -supposing
The
it done.
We
51.
Thus
to be
have
seen
that
correspondingto
subtracted
in the expression
Hence
it is
result.
an
example as the following
the result
subtract
c ; and
the
term
in the
"
have +c
find
such
to
uncommon
proposed for exercise : from a
The
required is a-vc.
beginner may
we
not
of Art. 41, by considering it as
not in itself,
but in connexion with some
manner
algebraical
operation.
";
"
explainthis in the
having a meaiiing,
other parts of an
EXAMPLES.
It is usual however
serve
and
c
"
"
a
on
a +
may say that a
there remains a-\-c.
c
=
may say that +
of each other ; tluis
( c) will denote the
Or
so
we
remarks
which will
attention of the beginner^
some
the
time to suggest
to
Tlius
from
offer
to
impress results
perhaps at the same
21
VL
and
we
c
"
c,^o that if we
"
subtract
operationsthe
denote
denotes the
reverse
of + c, and
of the
reverse
of
"
reverse
"
for them.
reasons
+
verse
re-
c, that
to + c.
is, ( c) is equivalent
But, as we have implied in
"
"
be content
to defer until
Art. 41, trie beginnermust
later period the completeexplanation
a
of the meaning of operationsperformed on negative
quantities,that is,on quantitiesdenoted by letters with
the
sign
-
prefixed.
It should be observed
addition
and
that the words
subtraction arc not used in quitethe same
in Algebra
sense
In Arithmetic addition always produces
as in Arithmetic.
increase and subtraction decrease; but in Algebra we may
speak of adding 3 to 5, and obtainingthe Algebraical
sum
2; or we
speak of subtracting"3 from 6, and
may
"
It will be seen
in these examples that,to prevent cob
fusion between
brackets
various pairsof brackets,we use
23
BRACKETS.
usingbrackets
of different shapes; we might distinguish
by
different
sizes.
of the same
shape but of
\'inculum
A
equivalentto
is
bracket;
a
see
Art. 30.
Thus,for example,
a-\p-{c-{d-J^]-]=a-\b-{c-{d-e^f)]-]
=a-\b-{c-d
=a-h
+
The
55.
brackets
e-f}'\=a-[b-c
+
c-d+e-f.
beginner is
in
d-e+f]
+
the
order
recommended
always to remove
preceding Article;
nermost
pair,next the in-
shewn
in the
innermost
namely, by
ever
We
pair of all which remain, and so on.
may howif we
remove
a pair of brackets
vary the order; but
includinganother bracketed
expression"vithin it,we must
make
no
chaiKje in tke signs of the included expression.
In fact such an included expression
counts as a smgle term.
followingrule for the multiplication
compound expressionby a simpleexpression;multiply
term
pression,
of the compound ex2')ression
by the simple exand
put the sign of the term
beforethe result)
collect these results to form the comj^leteproduct.
correspondingto the +a which occurs
and the + c which occurs
in the niulti*
multiplicand
plierthere is a term + ac in the product ; correspondingto
d there is a term
ad in the product;
the terms
+ a and
correspondingto the terms "h and +c there is a term
"c in the product ; and correspondingto the terms
6
and "c? there is a term -\-'bdin the product.
that
see
"
"
"
"
Similar
be made
observations may
obseiTations
three results ; and these
respectingthe other
are
brieflycollected
: like signs
multiplication
followingimportant rule in
produce + and unlike signs
shall often
Rule of Signs, and we
in the
This rule is called the
refer to it by this n.T'^ie.
~
.
give the generalrule for multiplying
plicand
term
of the multialgebraical
expressions
; multiply each
bfjeach term, of the multiplier; if the terms have
the same
sign prefix the sign + to the product^ if they
collect these
have differentsigns prefix the sign
; then
residts to form the completeproduct.
We
62.
now
can
"
For
example ; multiply2a
(2a + 36
-
4c)(3a
46)
=
6a2
=
6a' + 9a6
9a6
+
This
is the result
the result and
simplify
6a'
1 2ac
-
8a6
-
which
the
reduce
it to
+
46.
(2a +
1 66c)
-
Here
36
-
4c)
1 66c.
will
rule
"
46
-
1 26' +
-
Sa
4c)
-
1262
a6-12ac-
+
by
(8a6 + 1262
12ac-
-
4c
"
36
(2a+
3a
=
-
36
+
^VQ', we
may
166c.
the
might illustrate the rule by using it to multiply
that on working by
3 + 2 by 7 + 3
4; it will be found
the terms, the result is 30, that is
rule,and cpllecting
5
6, as it should
We
6
"
"
X
63.
student
will sometimes
4c
by 3a, or multiply
The
4c
"
results which
"
46.
requiredare
are
2a
by
X
-46=
~4cx
-4";x-46^
examples as
by "46, or multiply
find such
followingproposed: multiply2a
the
"
The
be.
-
the
8a6,
3a=-12ac,
166c.
following,
28
The
in the
student may
manner
Thus
to
attach
meaning to these operations
already explained;see Article 41-
have
we
the statement
that
mean,
multiplicandand
thel-e will be
Lie A TION.
TIP
MUL
46
-
a
occur
term
=
166c
may
the
be
stood
under-
terms
of a
among
the terms of a multiplier,
among
166c in the productcorresponding
Ac
-
-46
4cx
"
if
a
occiir
to them.
Particular
cases
of these
2ax-4=-8a,
Since then
examples are
2x-4=-8,
such
2x-l=-2.
be
those
in the preceding Article,it becomes
count
necessary to take acof them
in our niles ; and accordingly
the rules for
multiphcation
may be convenientlypresented thus :
64.
examples may
To
given
as
multiplysimple terms; multiply together the
merical
coefficients,
put the letters after this product
determine the sign by the Rule of Signs.
nw-
and
To
multiplyexpressions;multiply eojch term, in one
tiplying
expression by each term in the other by the ride for mulsimple terms, and collectthese partialproducts to
form the complete product.
We
shall now
give some
arranored in a convenient form.
65.
examples of multiplication
MUL
Consider
TIP
29
Lie A TION,
example. We take the first term in
the multiplier,
namely a^,and multiplyall the terms in the
multiplicandby it,paying attention to the Rule of Signs;
thus
we
term
the last
obtain
3rt^
4M^h
"
+
6a%\
take next the second
2ab,and multiplyall the
it,paying attention to the
We
of the multiplier,
namely
in the multiplicandby
obtain
of Signs; thus we
take the last term of the
we
"
terms
Rule
Then
multijjlyall the
and
paying
attention
the Rule
to
multiplier,
namely 3"',
multiphcand by it,
qf Signs; thus we obtain
in
terms
-Qd^b-\-Sa^l)^"\Qah\
the
We
the terms which
thus obtain, so that
we
arrange
like terms
stand in the same
column; this is a very
may
useful arrangement, because it enables
to collect the
us
in order to obtain the final result.
terms
and safely,
easily
In the present example the final result is
3a*
-
10a'" + 22a'"2_
student should
bringinglike terms of the
66.
22a63 + 156*.
observe that
The
product
with the view of
into the same
colunm
the terms
of the multiphcand and multiplier
are
arranged
We
fix on some
letter which occurs
in a certain order.
in
of the terms and arrange the terms
according to the
many
of that letter. Thus, taking the last example,we
powers
fix on the letter a ; we
put first in the multiplicandthe
which
contains the highest power
of ", namely
term
3a''*,
the second
the
which con"Aab
term
next we
tains
put
power;
of a, namely the first power;
the next power
and last
we
put the term 5h\ which does not contain a at all. The
multiphcand is then said to be arranged according to
descending powers
the
same
of
We
a.
arrange
the
multipUer in
way.
might also have arranged both multiplicandand
multipherin reverse
order,in which case they would be
of a. It is of
arranged according to ascending powers
We
no
consequence
the sam^e
order
67.
We
which
order we
adopt, but
for the multiphcand and the
shall
now
Multiply l+2aT-3^
to doscendinrr
cortling
give some
+
a?*
powers
more
by
of
must
take
multipher.
examples.
a;"" 2a?" 2.
x.
we
Arrange
ao*
30
UL
M
Lie A TION,
TIP
ai^-'^x*+'2x 4-1
a^-2x
-2
+6^-4a;-2
-2^
^'Ja? + 2x'^-Qx-2
^7_5^5
Multiplya? +
"
+
c'^-db'-hc-cah^
a
h
+
+
a
a^
ah
"
b
+
a^b
"
+
^-h^ ~hc
ac
"
a^c + ah-
"
c^
+
"
abc
ac^
+
-ab^'-abc
a^c
+
+b^-b^c
-abc-ac^
b"^ + "^
^"?
example might also be worked
brackets,thus,
+
with
the
aid of
c"
(6+ c)
+
-
"
bc'
^-b^
This
o2-a(5 + c)+ "2_"c
+
+b^c
-2"abc
a^
a?
a.
c
a^b
a
c.
of
Arrange accordingto descending powers
a^
+
6c 4- c2)
a2(6+ c)+ a(6^
-
-fa2(6+ c)-a(6+ c)(6+ c)+ (6+ c)(""-"c+ c8)
Then
we
have
a(6^-bc-v c^) a{h + c){b+ c)
-
=a{W-bc
=
Mid
+
c^-{J}+ c)Q)+ c)}
o.{"2"c + c= ("-+
_
-
2"C
+
c3)}
(6+ c)(62_"c+ c^)="3+ c3.
Thus, as before,the
result is a^
+
b^ + (^-oabc.
TIPLICA
MUL
Multiplytogetherx
X
-a
X
-h
81
TION.
a,x"hyX"c.
"
aP-"ax
hx-vdb
"
x^-{a
X
+
b)x + ah
+
b)x^+
"c
a^-{a
-cx^
a^
b
{a +
"
{a + h)cx
+
c)x^+ {ab+
+
dbc
"
should
The student
abx
+
ac
notice that
bc)x" abc
he
make
two exercises
in whioli the
can
from
multiplication
every example
pressions,
multiplicand and multiplierare different compound exby changing the original multipher into the
and the originalmultipUcandmto
nmitiphcand."
multipher.
in
should be the
t*"t of the correctness of his work.
obtained
Th-? result
which
same,
will
oe
a
VIII.
EXAMPLEa
Multiply
by 4;c'.
by 4a'.
3a*
3.
2a'6
by 2a6'.
1.
2x^
4.
3j^fzhj5xYz\
5.
Ixh/^hy 7y*z*.
6.
4a^-36by3a5.
7.
Sa'-9abhjZa\
a
+ 5z'^hj2x'h/.
3ar'-4!/'
9.
a?V
-
y^^^+
2.
by aef^y'z^.
^*^
10.
2xy^z^+ S.vh/^z 5x^yz^by ixy^z,
11.
2x
12.
2u,'3
+ 4jr^+ S.r 4- 1 6
13.
""
-
"
+
yhy2y
^2
+ j;
-
+
1
jt:.
by
a;
by
-
3a;
1.
-
6,
2
VIU.
EXAMPLES.
14.
1-6^
l+4;"-10:i7''by
15.
a?'-4a;'4-lla;-24
3^;".
+
by ^
+ 4a; +
5.
~T6.
"" + 4;c"+
5;"-24by
17.
a^-Tar' +
Saj + l
18.
;r'+ 6;c'+24^
19.
d;"-2;"2 +
20.
a^-2ix^^^x^-1x^\
21.
x-^Za.
a(P-~Zax\ys
22.
a2 +
23.
252 + 3a"-a2
24.
a2_Q5
25.
dr-ab-v- 2b-
26.
4:X^
27.
x^
28.
2;c'+
29.
^c'+
y^
30.
x*-+
+ 4."V + ^-^y'^
+ 1 62/*
by
2.z-*y
31.
8 la?*+
32.
a? +
33.
a'" 0^+6;?;
34.
d^ + lr +
"^"bc"ca"ahhya-"b^-c.
35.
a2 + 4^^
+
452^j
36.
a^~2ah
+
62 +
+ ll^
by 2;i;'-4a;+l.
6:0bya^-6a;2+l2.c+12.
+
3^
3^_4i3y4^.3+
+
by ^
a2 +
2aa;-:c'by
+
2^.4-l,
2d;3+ aa;"+ 2^
"
2a;c+A''.
by 7a- 5".
2,si)y^2+^5_5a^
+
by a^ +
ab
3x1/ y^ by Sx
-
;c2_4^
-
+
2bl
2i/.
-
xH/ + xy*"y^hyx-k-y.
"
3;ry+ 4^^ by 3^
"
;c2^+
llx^y+
:" +
y
+
1
"
9a;V
Axy A-y'*.
by .?:4-2/-l.
ic
Sj?/+ rr*by
+
2^^.
-
3.c -y.
2y-3^by;"-2y-"-33:.
+
"^ by
+
a
by a^
-
6+
:c.
46.c + 4/.="u.-".
a^ 4- 2ab
c^.by
^I/-
d"^-
the following
together
expressions
Multiply
37.
x"ay
ZS.
x
39.
x'-axA-a^
40.
a~2a,
+
a,
x' + a*.
x-ha,
+
x
b,
x-a,
x
x'^+
+ c.
ax+a'^,
x +
a,
x*
-
ar;i^"{-a*.
x4-2a.
I.
34
DIVISION.
It may happen that all the factors which
Thus
in this manner.
divisor may be removed
example,that 24a"a? is to be divided by Scix :
3"
lAabx
8a^_^
X
_
~
~
Sax
Sax
70. The rule with respect to the
examination
an
may be obtained from
in
occur
For
in the
suppose, for
occur
sign of
of the
the
quotient
cases
which
Multiplication.
have
example,we
Aab'x.2c=\2dbc;
I2abc
12abc
.,
,
.
"
therefore
=Bc,
,
4ab
,
4:abx
-"
3c
-3c=
12abc;
"
-llabc
.,
-\2abG
.
=
.
3c,
-
,
therefcff"
"
3c,
=
-r-r-
^
12a^C',
"
=
-
^aby. "3c=l2abc;
\2abc
\2abc
_
therefore
"
4ab.
6e
"
=
"-
-4ab
Thus
Division
=
"
"
"4ao
-
-"
4abx3c=.
"
,
-
3c,
'
^
---
=
-
,
4.ab.
-3c
it will be seen
that the Ride of
as well as in Multiplication.
Hence
,
4ab.
"
therefor"
.,
^
=4ao.
"
Signs
holds
in
have the following
rule for dividingone
another:
Write
the dividend
over
simple expressionby
with a line between
the dimsor
them; if the expressions
have common
the common
factors; prefix
factors,rcTnove
the sign + if the expressionshave the same
sign and the
if they liave differentsigns.
sign
71.
"
we
4
is divided by arwtJter
One power
of any number
of the same
number, by subtracting the index qf
power
the latterpower from the index of the former.
72.
r
35
DIVISION.
For
By
example,suppose
Art. 16,
have to divide a"
we
by a\
a"=axax"xaxa,
a?
therefore
".
,c'
.,
"jxcxc
=
_.
c^ = c' *.
cxcxcxc
the rule may
In'like manner
other case.
may
_
=
=
*'c*
we
a' = a" 3.
=
cxcxcxcxcxcxc
Similarly
-i
Or
" x a
=
=
-5
be shewn to be true in any
shew the truth of the rule thus
c*'x(?
by Art. 59,
=
therefore
:
c^^
(P
(P
I4="^^
^="^"
in the dividend
If any power of a number occurs
in
number
the divisor,
and a higherpower of the same
the
Arts.
and
be
72.
can
71,
quotient
by
simplified
Suppose,
for example,that 405^ is to be divided by 3c6^;then the
73.
quotientis denoted by
"-rg
dividend and
divisor;this
tient denoted
by
^
The factor IP'occm's
.
may
be
in both
removed, and the
quo-
|^'^,
; thus
=
.
II.
The rule for dividing
a compound expression
tion
by a simpleexpressionwill be obtained from an examinain Multiplication.
of the corresponding
case
74.
For
example,wc
have
{fl-h)c
therefore
ac-'hc\
=
=
a
"
6.
c
(a-6)x
therefor"
-c=
-(3WJ +
"C;
=^a-6.
"
"
c
3"2
36
DIVISION.
Hence
qf the
collect the
dividinga
pound
com-
simpleexpression: divide each term
the divisor,by the rule in the first
results to form the completequotient.
expressionby
dividend
by
case, and
For
have the followingrule for
we
a
example,
4a2
=
"
^bc +
ac.
To
divide
III.
75.
one
compound expressionby
another we must proceed as in the operationcalled Long
The followingrule may be given.
Division in Arithmetic.
dividend
both
Arrange
of
powers
some
and
common
of some
descending powers
Urst term
of the dividend by
and
ing
according to ascendboth according to
or
letter,
divisor
common
letter.
titefirstterm
put tJie result for the firstterm
of
Divide
the
the
divisor,
oftJiequotient; m,ul-
tiply the whole
product from
divisor
the
subtract
and
by this term
t/te dividend.
To the remainder
join as
be
of the dividend, taken in order, as may
terms
many
required, and
repeat the whole
until all the
process
taken
dowru
terms
of
the
Continue
dividend
have been
operation.
the
that for the
The reason
tor this rule is the same
as
rule of Long Division in Arithmetic,namely, that we
may
how
often
and
find
the
break the dividend
into
parts
up
in each part, and then the aggi-egate
divisor is contained
is
the
of these results
completequotient.
76.
shall
We
arranged in
a +
a
now
convenient
give
examples
of Division
form.
+ b
b)a^A-2ab-^W{a
a^ + ab
some
a+bja^-b\a-b
a^ + ab
37
DIVISION,
a^- 2"6 + 3 "2j 3^4- 1 Oa^?,+ ^^cC-y^-22a"+
-4a^b
1 56"
{Sa^-4ab
+
56="
lSa^b--22ab^
+
Consider the last example. The dividend and divisor
both an-anged according to descending powers
of a.
are
The first term in the dividend
is 3a* and the first term in
the divisor is o-; dividingthe former
by the latter we
obtain 3a^ for the first term
of the quotient. We
then
multiplythe
whole
that each term
contains
"
the
divisor
below
comes
same
^a^b+lSa^b^;
power
and
by 3a^, and
the term
of a;
we
bring down
we
dividend,namely, "22ab^.
4"^",by the first terai in
We
the
"
Aab for the next
the whole diWsor
under
those terms
place the
result so
of the dividend which
subtract,and
obtain
the next
term of the
divide
the
first term,
divisor,a^; thus
we
obtain
in the quotient.We then multiply
4ab and place the result in order
by
of the dividend with which we are now
subtract,and obtain 5a'^b^ lOab^ ; and we
"
occupied; we
bring doAvu the
We
term
"
"
next
divide
term
5a^2"'by a^,and
in the quotient. We
by 55^,and
place the
of the
term
tlius
then
terms
dividend, namely, 155*.
obtain
55" for the next
multiplythe whole divisor
we
as
before; we
subtract,and
there is no remainder.
all the terms
in the diridend
As
have been broughtdown, the operationis completed; and
the quotientis 3a^
4a5 + 55^.
-
It is of great importance to arrange
and divisor according to the same
order
letter;and
to
attend
to this order
both
of
some
in every
dividend
common
part of the
operation.
happen, as in Arithmetic, that the division
be exactly performed. Thus, for example, if we
cannot
divide d^ + 2aJ" 4- 26- by " -!-5, we shall obtain,as in the first
example of the preceding Article,a + 6 in the quotient,
Tl.
It may
and there will then be
a
remainder
b'\ This result is
ex-
38
DIVISION.
pressedin
we
may
ways
similar to those
used in Arithmetic ; thus
say that
b
+
=a
7
+
=:;
a+D
a+o
that
is,there
In
is i^guotient
a+",and
B
general,let y^Vp-nd
denote
is divided
t
.
and
expressions,
is
the quotient g, and
is expressed algebraically
student Vill observe \hat each
an
convenient
an
"
two
B
A
by
suppose that when
then
this
result
the remainder
R;
in the followingways,
The
fractional part
a
feimpleor
expression,
'
for .distindtnessand
expressionby
a
letter hero may represent
it
often
is
compound;
'^"
^^
"
*
'"
"
*
singleletter.
"We shall however
consider
in subsequent
at present shall confine ourselves to
Chapters,and
examples of Division in
fractions
algebraical
which
performed.
the
operationcan
be
exactly
V
^
78.
We
givesome
Divide x^-5a^
Arrange both
scendingpowers
+
"
examples :
more
lx^
+
2x'-6x-
2
by
l + 2x-Bx'
+
divisor accordingto
divideij^""^d
of ^.--
a^.
de
""^'^
a*-3x'+2a+ljx''~5x^
2aT^
-
2x^
"2a^
+'7x^+2x^-ex-2{x^-2x-2
+
Qx^
+
2x^-
Qx
+Qa^-4:a?-2x
-2x*
^-6x'-4x-2
'-2x^
+6a^-Ax-2
39
DIVISION.
Divide
a'
"3
+
c^
+
as
select
we
contains
that
seen
descending
terms, sucli
a,
h
+
e.
a.
It will be
to
+
a
according to descending powers
Arrange the dividend
of
by
3a6c
-
a
powers
a*6 and
we
arrange
of
then
a;
these terms
according
when
two
there are
of
a^c,which involve the same
power
letter,as b, and put the term which
wliich does not; and again, of
the term
abc, we put the former first as involving
new
b before
the terms o^ and
of b.
the higher power
example might also be woi-kod, with
brackets,thus:
This
a-hb
+
-3abc
cjd^
a^
+
a\b
-a\b
-
a\b
+
c)
+
c)-3abc
4
c)
-
b^ +
+
+
a(62+
a(b^-
the
c^{a*-a(b+ c)+b^-bc
b^ + c^
2bc
+
c^)
bc
+
c^ + b^ + 1^
of
aid
+
c^
40
EXAMPLES.
x^
Divide
-{a + h-"c)x*-^{ab+ac+hc)x-abchj
x-c
+ {ab+ a"-i'bc)x-dbc{x'-{a-"b)a+a
c)ar^
+ i +
it-cja^~{a
{a+ b)a^+ {ab+
"
IX.
a"
be)x-ab"
+
dbx
"abe
abx
abe
"
in which
the mn?.t"
Every example of Multiplication,
plierand the multiplicandare different expressions,Wiil
if the produ'jt
furnish two exercises in Division;because
be di\'ided by either factor the quotientshould be the other
factor. Thus from tht" examples given in the section on
derive exercises in Divlsio^j,
the student can
Multiplication
And
from any example
and test the accuracy of his work.
of Division,in which
the quotient and the divisor are
be obtiiJned
different expressions,a second
exercise may
tliAt
so
by making the quotienta divisor of tlie dividend,
ttie new
divisor.
quotientought to be the original
which
examples in Multiplication
operationsthat they deserve
algebraical
some
The
three examples are of great importance.
following
The
first example
Is,of (a + 6)*; thus
we
of (a + 6)(a + 6),that
value
gives the
have
Thus
the square of the sum
is equal to
of two numbers
the sum
increased bp
of ths squares
of the two numbers
twice their product.
Again, the second example gives
(a-bf
Thus
=
d^-2ab
+
lr'.
tJie square
of the difference
of two numbers
the
the
sum
equal to
of
of the two numbers
squares
their
ticice
diminished
product.
by
The
is
last example gives
("+ 6)(a-6)
Thus
numbers
=
a2_6^
and
the product of tJie sum
differencecf two
is equal to the differenceof their squares.
results of the preceding Article furnish a
simple example of one of the uses of Algebra; we may
80.
Tlie
Algebra enables us to prove general theorems
respecting numbers, and also to express tfiose theorems
briefly.
say
that
IN
43
MULTIPLICATION.
For example, the veBvM{a + h){a"h)=a'^-l^ is
to be ti-ue,
and is expressed thus by symbols more
than by words.
A
proved
generalresult thus expressed by symbols is
called
B.
pactly
com-
often
formula.
81. "We may here indicate the meaning of the sign "
which is made
and which
by combining the signs+ and
,
is called the double sign.
^
"
Smce
we
(a+ 5)2=a2+ 2a5
+
6", and
these results in
mfty express
one
{a-hf=a^-2ab
thus
formula
+
l^,
:
where =t indicates that we
take either the sign + or
may
the sign , keeping throughout the upper
sign or the
lower sign. a"6
is read thus, ^' a plus or minus
6."
"
"We
shall devote
Articles to explainingthe
some
"We shall
of the formuloe of Art. 79.
that can be made
use
them for tJie sake (f
repeat these fomiul^e,and number
easy and distinrt 'p/erenceto them.
82.
=a"
{a + bf
+
2"6
"'
+
(1)
=a''-2ab-hP
(a-bf
{a + b){a-b) a''-b^
(2)
(3)
=
metical
be of use in ArithThe formulae will sometimes
calculations. For example; requiredthe difference
of the squares of 127 and 123.
By the fomr.ila (3)
83.
(127)'-(123)*(127+ 123)(127- 123)
Thus
it would
the
be
=
=
thus the square
29 by
multiplying
Or suppose
53
X
47
we
formula
=
4
=
1000.
the first.
Again,by the formula (2)
(29)2 (30-1)2 900
By the
x
is obtained more
easilythan
requirednumber
by squaiing 127 and 123, and subtractingthe
second result from
qnd
250
=
=
(50 +
-60
is found
of 29
1=841
+
more
;
easilythan by
directly.
have to multiply53 by 47-
29
(3)
3)(50 3) (50)^ 3'
-
=
-
=
2500
-
9
=
2491.
44
GENERAL
We
that we
obtain
Suppose
84.
of
can
course
multiphing 3^
4-
RESULTS
2y by
require the square of 3^ + 2y.
ordinaryway, that is by
it in the
2?/. But
'ix +
we
also obtain
can
it
another
namely, by employing the formula (1).
way,
number
The formula is true whatever
a
may be, and whatin
.
h may be;
obtain
we
number
ever
for b.
Thus
(3^ + 2yf
{^xj"+
=
2
so
we
put
may
for a, and
^x
(3:c2y)+ {2yf dx'-" I2xy
4y*.
+
=
ly
beginnerwill probablythink that in such a case ho
of the formula,for
does not gain any thing by the use
he will believe that he could have obtained the required
result at least as easilyand as safelyby common
work
of the formula.
This notion may be correct
as by the use
in this case, but it will be found
that in more
complex
wiU
formula
be
of
service.
the
cases
great
The
85.
x
+
Then
x
+ y +
"
(a + by
x'^+
=
Then
by the
Thus
{p
+
y'^
"pt
"
+
+
h^
of
use
(2)we
2pq
+
q^+ 2{pr"ps
g2^
^Q _^ ^2 ^
"
have
z^ -{-^xy+
^yz
+ 2xz.
"
g'=
of p"q-',r
+ r
Denote
s.
"
+
s=a
"
(p
2^r
+
qr
"
2qs
"
+
"
qs)+
2pq
"
require the product
/"-^
a
and
+ r-*=a
z'i
.
b.
+
{r-s)\
qf and (r- *)'.
r'^ 2rs + s'
2ps
"
"
2qi
"
of p"q-\-r"s
q~r-"s.
Let jo
note
De-
z.
sf
r
we
+
'lyz+ z"^.
+
express
+
Suppose
y
(1)we
(j)-q)^+ 2{p-q){r-s)
=
q
"
+
have
(1)we
a^ + 2ab
=
=p^
p
zf
of
use
of
use
+ 2xz
9.xy-t-.y2
X'^-
Suppose we requirethe square
by 6; then p"q
qhj a and r-s
By the
"
+
x
+ '2.i,x
+ y)z +
a'-"^az-{-z'^'-{x-"yf
=
=
Th\Ji"{x + y
of
square
by the
and
z^a+z;
{a+zf
p
require the
Suppose wc
yhy a.
r-*
+
=
"; then
", and jo-g-"
r-t-*=a"
6.
2rs.
aad
IN
Th^i
the
by
45
MULTIPLICATION.
of
use
have
(3)we
"
{a + b){a-h) a*-"={p-qf-{r-9fi
=
and
the
by
{p-q
use
safe,and
+
should
he
3)=p'*-'2pq+ q^-{r^-'2r8-^-s?)
=p'^+ q^-t^-s^-2pq + 2rs.
in
exhibited
method
The
have
(2)we
r-s){p-q-r
+
86.
ts
of
the
preceding Article
be adopted by the beginner;
therefore
famihar
with the subject he may
of the work.
Thus in the last example,
becomes
more
dispensewith some
he will be able to omit that part relating
to
simplyput down the followingprocess ;
as
{p-q
r-s){p-q-r
+
+
=
j?' 2pq
=
or
r-s)(jp-q-r
+
+
through the
The
formulae.
Find the
c, b
+
Take
b+
c"-a"
+
q^" r^ +
2r8
^;
"
c
"
+
2rs
"
s^.
probablyfind it prudent to
the precedingArticle.
will
all the thi-ee
employ
the four factors
a +
h + c,
a +
b-Cy
a.
c){a+ b-c)={a
the last two
obtain
factors;by (3)and (1)we
+
by-c'=a'- +
2ab + "-c'.
obtain
factors;by (3)and (2)we
{a-b+ c)(b+ c-a)
We
in
fullyas
product of
the firsttwo
{a+
Take
+
work
will
followingexample
87.
a-b
q^-{r^-2rs + s^)
s)={p-q)^-{r-s)*
=p^"2pq + q^"f^ +
But at firstthe student
go
"
+
still,
briefly
more
{p-q
h, and
s) fj"-q + {r-s)]{p-q-{j'-8)}
ip-qf-(r-sy-=p^-2pq
=
and
a
=
{c+ (a-b)}{c-{a-b)}
=
c'-{a-b)*
=
c'-a''
have now to multiplytogethera^ + 2ab
2a6 -ft^. y^^ obtain
+
+
2ab-b*.
"-c^
and
46
RESULTS
(a'+
2a"
""
+
-
MULTIPLICATION.
IN
(c" a2 + 2a6 62)
c2)
+ 62-c2)}{2a5-(a2
+ 6"-c2)}
+ (a2
={2aft
-
-
+ 62_c2)2
(2a5)2-(a"
=
+ 62)2_2(a2+62)c3
+ c4}
=4a''5'-{(a"
=
40=62-.(a2
+ 2(a2+ ?^)c2_^
+ 52)2
=
^aW
=
2a26a+ 262c2+ 2a2c2_a4-6*-c*.
a^
-
2a262 54 + 2a2c2+ 252^2 c*
-
-
_
whicli are
88. There are other resultsin Multiplication
than the three formulae givenin Art. 82,
of less importance
of attention. We placethem here
V)ut which are deserving
in order that the student may be able to refer to them
when they are wanted; they can
be easilyverified by
actual multiplication.
ffl3
+ 63,
these operationsas far ai
student
can
carry on
in the truth of
he pleases,
and he will thus gain confidence
shall now
which
the statements
we
make, and which are
in the higher parts of largerworks
demonstrated
strictly
the statements :
are
on
Algebra. The following
The
af'"y^ia divisible by
y if n
ic"
be a7iy whole
a;""
2/"is divisible hj
x
+
y if nhe
of* +
y" is divisible hy
x
+
y if n
any
even
number
number;
whole
be any odd whole
;
number.
might also put into words a statement of the forms
will
of the quotient in the three cases; but the student
most
readily learn these forms by looking at the above
examples and, if necessary, carryingthe operations still
"We
farther.
We
x"y,
may
when
add
n
is
that
an
;r**
+
even
y"
whole
is
never
divisible by
x
+
y
or
number.
The student will be assisted in remembering the
results of the x"receding
Article by noticingthe simplest
93.
51
FACTORS.
in each of the four results,
and referring
other
wish
consider
For
to
whether
to it.
example,suppose we
the
index
3^ "y' is divisible by x
7 is an
y or by x-vy,
of this kind is
odd whole number, and the simplestcase
which is divisible hj x
so
we
y, but not by a"+y;
x"y,
and
infer that x^ -y"^ is divisible by x
not
by x-\-y.
y
Again, take a?"y^\ the index 8 is an even whole number,
of this kind is oc^"y"^,
and the simplest case
which is
and x + y; so we
divisible both hy x"y
infer that a^"y^
"
ed
Suppose that {a^ 5jty+ 6y*){x Ay) is to be dividWe
by a^"lxy+\2y*.
might multiplya^"6xy-"rQy*
and
divide
the result by x^-7xy + 12y^.
then
x
by
4y,
of the questionsuggests to us to try if
But the form
x-4y is not a factor of x^-lxy + Vly'';and we shall find
that x^-7xy + l2y^^{x- 3y){x Ay). Then
95.
"
"
"
-
{a^"5xy + 6y^ (x Ay)
{x"3y){x-Ay)
a^
-
and
by
division we
_
~
-
5xy + 6y^
*
x"3y
find that
a^-5xy
*-3y
+
6y^
=^-^^4"2
52
EXAMPLES.
XL
The student with a littlepracticewill be able to
resolve certain trinomials into two binomial factors.
Sf6.
Algebra an expressionwhich
exactlyis said to be a m"asure
it; an expression which divides two
measure
exactlyis said to be a com,7non
In
divides
of
measure
said to be a
whole
number
which
divides two
exactlyis said to be a comtnon
In
98.
which
number
exactlyis
it; a
measure
Measure.
Common
a
55
MEASURE.
the
Arithmetic
whole
numbers
divides
of
another
it,or
pression
ex-
measure
expressions
more
or
to
of them.
greatest
of
me"isure
comm/)n
is the greatestwhole number
which will measure
them
all. The term
greatest common
is also used in Algebra, but here it is not very
measure
because the terms
dom
appropriate,
greater and less arc selapplicableto those algebraicalexpressionsin which
definite numerical
values have not been
assigned to the
whicli
various letters
It would be better to speak
occur.
of the highestcommon
onea^ure, or of the highestco?nmon
in
but
dicisor;
conformity with established usage we
two
or
more
shall retain the term
greatest
The letters g.c.m.
instead of this term.
We
in
have
to
now
will
comm,on
be
often
explainin
measure.
what
used
sense
for shortness
the term
is used
Algebra.
99.
measure
It is usual
of two
or
by the greatest common
simple expressionsis meant
to say, that
more
greatest expression which
common
of
measure
The
them, all; but
finlyunderstood until we have
rule for findingthe greatest
will
this definition will not be
given and exempUfied the
the
measure
simpleexpressions.
followingis the
simple expressions.Find
for findingthe g.c.m.
of
the g.c.m.
o/
hy Arithmetic
this
number
after
put every
to all the expressions,and
give
Rule
the numerical
letter which
is commmi
to each
letter respectively
the least index
coefficyfnts;
in the expressions.
which
it has
5e
COMMON
GREATEST
the numerical
Here
20a^b^d.
their
and
example; requiredthe
For
100.
MEASURE.
G.aM.
expressionsare
the least index
g.c.m.
of IGa^^e
coefficients are
and
and
16
20,
to both the
The letters common
is 4.
b ; the least index of a is 3, and
a and
of b is 2. Thus we obtain 4a^b^ as the required
G.C.M.
Again; requiredthe
of Sa^c^ar^yz^,
12a^bcxh/^i
Here
the numerical
coefficients are
and IQa^c^x^.
8,
is
The
letters common
4.
to
12, and 16; and their g.c.m.
and
and
their
indices
least
allthe expressionsare a, c, x^
y ;
and
Thus
obtain ^d^cx^ as
1.
we
are
2, 1, 2,
respectively
the requiredg.c.m.
g.c.m.
The following
statement
gives the best practical
is meant
of what
the
term
by
greatest common
in
it
the
of the word
shews
sense
Algebra, as
measure,
expressionsare divided
greatesthere. \Vhen two or more
101.
notion
hy
their
greatestcommon
measure,
the
quotientshave
no
m,easu7^e,
common
^.
the
Take
and
these
first example of Art.
100, and divide the
by their g.c.m.; the quotientsare
have no common
measure.
quotients
4ac
and
pressions
ex-
bbd,
second example of Art. 100, and
di\ide the expressions
by their g.c.m.; the quotientsare
and ^a"i^, and these quotientshave no
iWca^z^, Zcv^hy"^^
Again,
common
take
the
measure.
The
notion which is suppliedby the preceding
Article,with the aid of the Chapter on Factors,will enable
the student to determine in many
the g.c.m. of comcases
pound expressions. For example; required the g.c.m. of
4a" ("+ ")'and 6ab{a^-b%
of the
Here
2a is the g.c.m.
factors 4a^ and Gab; and a + b is a factor of (a+ b)^and
of a^ b\ and is the only common
factor. The product
is
then
the g.c.m. of the given expressions.
2a{a + b)
102.
"
But
this method
cannot
be applied to complex examples,
because the general theory of the resolution of
expressionsinto factors is beyond the present stage of
the student's knowledge; it is therefore necessary to adopt
58
GREATEST
The
106.
the
MEASURE.
COMMON
rule which
is
given in
principles.
two
following
Art. 104
depends on
If P
mA.
For let
measure
A, it will measure
denote the quotient when
is divided by P; then
A
a
therefore mA
therefore P
A=aP',
measures
maP;
quotient,and D the remainder.
Divide C by Z", and
suppose
that there is no remainder,and let r
denote the quotient
Thus
have the
we
A
=pB
+
(p
B
Kjj
qC
D)
C Cr
rZ)
results :
following
a
B=
qC+ A
C=
rD.
"We
shall first shew that Z" is a common
of
measure
A and B.
therefore D
Because
C=rD,
measures
(7;
also
and
therefore,by Art. 106, D measures
g(7,
qC+D;
that is,D measures
B and C,
B. Again, since Z" measures
it measures
Thus
D
A.
pB + C; that is, D measures
measures
"We
A
and
common
A
have
B;
and
B.
thus shewn
shall now
we
of
that Z" is a common
measure
shew that it is their gr cutest
measure.
of A and B meaArt. 106 every common
measure
sures
of
A -pBj that is (7j thus every common
measure
of B and C,
A and 5 is a common
measure
Similarly,
C
is
and
B
common
measure
of
a
measure
evei'y common
By
Therefore
of C and D.
^
is
of D.
measure
a
than
D
divide D.
of A and B.
can
measure
common
every
But no
It is obvious
108.
of
measure
of those
two
or
of A
and
sions
expressionof higher dimenD is the greatest
Therefore
common
measure
that, every measure
expressionsis a
more
59
MEASURE.
COMMON
GREATEST
of
a
common
measure
common
expressions.
is
in
107 that every
common
of A and B measures
D; that is,every common
their greatest common
of two expressionsmeasures
It
109.
measure
measure
shewn
Art.
measure;.
state and exemplifya rule which
is adopted in order to avoid fractions in the quotient; by
refer to the
the use of the rule the work is simplified. We
in the larger
Measure
Chapter on the Greatest Common
We
110.
Algebra,for
shall
now
of the rule.
the demonstration
Before
placing a fresh term in any
the dividend,
divide the divisor,or
is common
icliich has no factor which
is
whose
m,easure
greatest common
dividend
at such
a
w,ay multiply the
has no factor that occurs
which
111.
and
but
For
Sx^-7x
example; requiredthe
quotient,wer may
hy any expression
to the expressions
required; or, we
pression
exstage by any
in the divisor.
of 2^2-7^
g.c.m.
+ 5
Here
take 2;^^ 7^ + 5 as divisor;
we
if we divide Zx^ by 2^^ tlie quotient is a fraction ; to
avoid this we multiplythe dividend by 2, and then divide.
+ 4.
"
2x'^-1x
+
5) 6x'^-Ux-h
8
{3
6;i;2_2iic+i5
IxIf
we
now
the
di\'idend,
make
7^-7
a
7
divisor and
2^2
"
7^ + 5 the
firstterm of the quotientwill be fractional ;
factor 7 occurs
in every
of the proposed
term
and we remove
divisor,
this,and then divide.
but the
COMMON
GREATEST
60
MEASURlS.
2a^-7a;
x-\)
i,2x-6
+ 5
2x^~-2x
"
Thus
we
obtain
it will be
Here
1
a?"
6^ + 5
the
as
that
seen
required.
g.c.m.
the second part of
used
we
the rule of Art. 110, at the beginningof the process, and
the first part of the rule later. The firstpart of the rule
should be used if possible
part. We
; and if not, the second
have used the word expression in statingthe rule,but in
the student will have
the examples which
to solve,the
removed
wiU be almost
factors introduced
or
always nur
merical factors,as they are in the precedingexample.
We
will now
give another example ; requiredthe
of 2a?*- 7a;*- 4;c'4- a; -4
and
g.c.m.
Zxi^-lia^-lx^-Ax-lQ.
Multiplythe latter expressionby
2
and
take it for
then
dividend.
^^-la^-^-{-x-4:)
Qie^-21x^6a;*
-
-
_
i2ar2+ 3a?
-
(3
12
a;*+ 8a;'-lla?-20
multiplyevery term of this remainder by -1
divisor;that is,we may change
using it as a new
signof every term.
We
before
the
21a:3
4a?2-8a;-32
"^-8a"
may
+
ll"+20j
2a?*-
Ta;"-
4a?"+ a?-4
+ 9
(,2a?
9a;3-26a;2_
^^^_
9a^"-72a;2+
99a; +180
46a;2_ 138a;-
Here 46 is a factor of every term
reoioye
it before using the remainder
4
184
of the remainder;
as
a
new
divisor.
wo
61
EXAMPLES.
XJI.
is the
required.
Thus a^-3:"-4
g.c.m.
Suppose the originalexpressions to contain a
is obvious
factor I^, which
on
common
inspection
; let
bF.
ai^ and ^
^
Then, by Art. 109, i^ will be a factor
of a and b, and multiply it
Find the g.c.m.
of the G.C.M.
by F; the product will be the g.c.m. of ^ and B.
112.
-=
=
than two
proceed to the g.c.m. of more
compound expressions.Suppose we require the g.c.m. of
Find the g.c.m.
of any two of
tht'ee expressionsA, B, C.
them, say of ^ and B; let Z" denote this g.c.m.; then the
of D and C will be the requiredg.c.m. of A, B, and C.
G.C.M.
113.
a
"We
now
For, by Art. 108, every common
of ^, B, and
common
mea.surc
of
Therefore
measure
common
D and C.
of .4,^, and
C;
and
C is a
the g.c.m. of D
^, B,
of D
measure
and
by
common
and
Art.
and
109
measure
C is the
C is
every
of
g.c.m.
G.
In
similar
find the g.c.m. of
may
/our expressions. Or we may find the g.c.m. of two of
the ^ven expressions,
and also the g.c.m.
of the other two;
then the g.c.m.
of the two
results thus obtained
will be
the G.CM.
of the four given expressions.
114.
a
manner
Examples.
Find
the
greatest common
we
XIL
measure
in the
following
examples :
20aHA
2.
I6a^,
Z6j^y^z^,48a^y'z^.
4.
35a'I^x'y\49a^b*a!*y^.
4{x + l)\ 6(;c"-l).
6.
6(^ + 1)3, 9(a;^-l).
1.
I5x*,
3.
C.
ISxK
C2
XII.
EXAMPLES.
7.
12(a''-"-6")",
8(a*-n
!X^-y\ -P*-2A
8.
11.
a?2+ 2;2?-120, ic2_2^_80.
12.
a;*- 15^
13.
:p3+
14.
a;3_ 9-^2+ 23.?;-12,
15.
aj3_29a; +
16.
!X?-A\x-Z%
17.
a^ + 7^+l7^
18.
0^-10.?;^ + 26^ -8,
19.
+
4(a;2-a?-hl),3(a;^
20.
+ l), 4(a?6-l).
5(a;2-aj
21.
6ii?^
+
22.
;i;8-4^
23.
a^-^x^-^,
24.
a;^-2a;2 + 3a?-6,
25.
ar*-l, 3ir'+
26.
a^-9;zj''-30a;-25, af'^-a^-'lx^-v^x.
27.
35a^
28.
a^-'^m^
+
a;"-9^-
36,
g^s^.i3^+12,
a;
+
36.
a^
7a!2+16;2?+16.
+
^-10a?2
42, ;c3+ -^2
35-^
_
^
V5,
+
+ 2a? +
47^2
+
49
+
ii^-11^2^25;c
+
+
25.
8.^^+19^
a?-^3s'
;c2+
12.
+
1Zx-\%
+
,-.
l).
+ 48a;2+
2, 9;i?3
-
+ 28a;-15.
52.r
3, 2;c4_9^3
+
1 6.
+
i2a;2-7.
Q!^-'ix'^-"f%
+
x^
-
a?
x^ -'Ix.
-
2;c*+ 4a?3+ 2;c2+
i3"p +
i^ 42a^
+
^\
41;ir3_9^2_9"p_l,
^x'^-lx^-^^x^-'^x+Y,
x^-o(?^1x'^-^^1x'--x^V
29.
2:c^-6:c3 4.3;2;2-3:c+l,x^ -^x^^x^-Ax^'-^Vlx-^,
30.
a?^-l, a7i"+a^ +
31.
^-3a;-70,
32.
a^-xy-\ 1y-,
33.
2^2 + 3^^
34.
ic3_3^2^_2a3^
36.
S^c' 3;c-y+ xy^
"
4.
a?8+ 2d7^+ 2^*
+
2^
+
^2
+ a? +
ofi-^^x^-IQ), a;8-48."+7.
x"^+ 5.t:y
+
^2^ 3^2
^ 2aa:
62/'.
a\
-
a:3-rt;z;2-4al
-
y\
4:X'^ybxy^ + y^
-
l.
COMMON
LEAST
XIIL
Least
In Arithmetic
another whole number
115.
hj
Common
a
116.
In Arithmetic
more
whole
Multiple.
whole number
is said to be
is measured
number
which
whole
numbers
is said to be
or
a
63
MULTIPLR
common
is measured
which
a
multiple of
by two or more
inultipleof them.
it ;
a
whole
the least connmon
multiple of two
is the least whole number
which
numbers
them all.
The term least common
is measured
by
multiple
is also used in Algebra,but here it is not very appropriate;
The
letters l.c.m.
will ofteybe used for
Art. ^^.
see
shortness
We
in
instead
have
now
of this term.
to
explainin
what
sense
the term
is used
Algebra.
It is usual to say, that by the least common
117.
tiple
mulof two or more
is meant
the leasi
simpleexpressions,
nition
expressionwhich is measured
by them all; but this defiwill not be fuUy understood
until we have given and
exemphfied the rule for findingthe
of simple expressions.
least
multiple
common
followingis the Rule for findingthe l.c.m. of
the l.c.m.
simple expressions. Find by Arithmetic
of the
numerical
coefficients
; after this jiumber
put every letter
ichich occurs
in the expressions,and give to each letter
it hat in the eX'
respectivelythe greatest index which
The
pressions.
118.
For
Here
ZOaWd.
and their L.O.M.
example; required the
the
numerical
L.C.M.
of 16a*bc
coefficients are
16
and
and
20,
in the exThe letters which occur
is 80.
pressions
are
a, h,c, and d; and their greatest indices are
respectively
4, 3, 1, and 1. Thus v/e obtain SOa^b^cd as the
required L.C.M.
of Sd-'b^c^x-^yz^,
12a*bcar*y^,
of the numerical
coefficients
and
l.c.m.
in the expressionsai-e
is 48.
The
letters which
occur
their greatestindi(X's are
a, b, c, x, y, and z ; and
rei"pecas
lively4, 3, 3, 5, 4, and 3. Thus we obtain ^bal^h'^i^a^y^z^
Again; required the
Here the
liia^c^xh/*.
the .jquircdL.C.M.
L.C.M.
LEAST
i)4
notion
MULTIPLE.
gives the best practical
followingstatement
is meant
by the term least common
multiple
The
1 1 9.
COMMON
of what
of the word least here.
in Algebm, as it shc"v-p the sense
WTien
the least common
?nuldple of two or more
sions
expresis divided by those expressions the quotientshave no
measure.
common
Take
the first example of Art. 118, and divide the l,c.m..
expressions; the quotientsare ob^d and 4ac, and
by the
these quotientshave
no
common
measure.
Again; take the second example of Art. 118, and divide
the L.C.M.
by the expressions; the quotients are Sa^cy^,
and these quotientshave no common
^(?a?yz^^ and ZaJfia^z^^
measure.
Tlie notion
120.
which
suppliedby the preceding
Chapter on Factors,will enable
is
the aid of the
in many
the l.c.m. of comto determine
the student
pound
cases
expressions. For example, required the l.c.m. of
and Q"ab{a^-V). The
of 4a* and 6a" is
l.c.m.
4a"(a + ")='
factor
Also (a + hf and a-"h^
have the common
12a-6.
that (rt
+ ^)("
+ 6)(a-") is a multiple of {a + hf
a + h, so
and of a'^ V^; and on dividingthis by (a+ 6)'and a* li^we
Article,with
"
"
obtain
the
quotientsa
Thus
measure.
we
h and
"
a-\-h,which
have
no
V2a'^'b{a
+ hf{a"h)
obtain
common
the
as
quired
re-
L.C.M.
L.C.M.
or
more
The
be
following
may
given as the definitionof the
of two or more
compound e.rpressions. Let two
of some
compound expressions contain powers
121.
letter;then the expressionof lowest dimensions
in that letter which
is measured
by eicU of these expressions
is called their least common
multiple.
common
We
sh'illnow
how
to find the l.c.m. of two
however
will
compound expressions.The demonstration
not be fullyunderstood
at the present stage of the student's
122.
shew
knowledge.
Let
and
B
denote
freatest
hen from the
measure.
A
common
nature
the two
of the
and
expressions,
Suppose
greatest
A
="
aD.
common
and
D
B
their
=
measure,
bD.
a
66
Hence
LEAST
COMMON
the
is
l.c.m.
MULTIPLE.
-x+l)
{x^-Zx- 4){'Ix''
124.
It is obvious
multiple of tico or more
of those expressions.
(Sors 2ar + 4).
-
that,every multi2)le
of
is
expressions a common
a
common
multiple
Every common
multipleof two expressionsis
multipleof their least comm.on
multiple.
125.
a
A
aiid B
denote the two
their
expressions,M
multiple.SupL.C.M.; and let N denote any other common
pose,
if possible,
that when N is divided by M
there is a
remainder i2 ; let g denote the quotient.Thus R
N"
qM.
Now
M
and iV,and therefore they meaA and B measure
sure
R (Art.106). But by the nature of division R is of
than M ; and thus there is a common
lower dimensions
than
multipleof A and B which is of lower dimensions
there
is
absurd.
This
Therefore
be no
their l.c.m.
can
iV"is
of
M.
that
remainder R;
a multiple
is,
Let
=
that we
require the l.c.m. of thre^e
Suppose now
compound expressions.A, B, C. Find the l.c.m. of any
of them, say of A and B, let M
denote this l.c.m. ;
two
then the l.c.si. of M
and C will be the requiredL.C.M. of
A, B, and C.
For every common
multipleof M and C is a common
multipleoi A, B, and C, by Art. 124. And every common
multipleof A and ^ is a multipleof M, by Art. 125; hence
multipleof M aiid (7 is a common
multiple
every common
Therefore
M
of A, B, and C.
the l.c.m. of
and G is the
l.c.m. of a, B, and C.
126.
1 27.
In
a
similar
manner
we
may
find the
L.C.M.
of four
expressions.
12S,
of the
and
The theories of the greatestcommon
measure
least common
multiple are not necessary for the
subsequent Chapters of
the
present work,
and
any
be
difii-
find in them
culties which
the student may
postmay
of
Boned until he has read the Theory
Equations. The
however
attached
to
the
exitn^ples
precedingChapter and
to the nresent
count
Chapter should be oaiefully
worked, on acof the exercise which
mental
fundathey afford in all the
processes
of
Algebra.
EXAMPLES.
XIII.
Examples.
Find
the least
07
XIII.
multiple in
common
the
amples
followingex-
:
1.
4rt'", 6aZ/l
3.
8aVy,
o.
4"i;a4-6), 66(a3+ 63).
7.
^-3j7-4,
9.
l2a;^ + 5j7-3,
126Vy'.
2.
Y'laW-c, 18a6V.
4.
{a-Vf,
6.
a^-ft^, "'-"".
a"-"".
ar"-d7-12.
6^^ + ^-d7.
10.
a;5-6i;'+ll^-6,
11.
A-^-7^-6,
12.
ar*+ z3+2ar'
13.
dr*-2a;3-t3.r2 + 8a;--4, a7^-5^
14.
;r*+ aV
15.
4a^"2^, 6053^2, I8a-6c3^
16.
8(a"-6"), 12(a4-6)", 20 (a-6)*.
17.
4(a+6),
IS.
15(a'6-aZ"'), 21(a3-a^"'),35 (06*+ 6^
19.
ar2-l, ;"3+l,
a?3- I.
20.
a?'-l, a^+\,
ar^+l,
ar^-l.
21.
a;"-l, a?+\,
x^-\,
a?"+ l.
22.
a;'+ 3a; + 2,
x^-^Ax
23.
a^2+ 2;c-3,
;r'+ 3^-a;-3,
24.
a:"+ 5a?+10,
+
^
a^
8^+17ir
+
a^
a^
-
+
-
24.
10.
a^^-l.
d;+l,
+
9 j;^+ 260?
-
aa^
-
a^x
+
""
20"-16,
a^.
6(a2-"2)^ 8(a3+ 5').
+
^,
a^
;c'-19a;-30,
+
5x + Q.
a^
+
^'^-k-x-Q,
;c3-15;c-5a
FRACTIONS.
XIV.
Fractions.
this Chapter and the following
four Chapters
shall treat of Fractions ; and the student will find that
we
the rules and demonstrations
closelyresemble those with
which he is alreadyfamiliar in Arithmetic.
-
129.
In
130.
By the expressiont
equalparts,and
to be divided into b
are
to be taken.
indicate that
we
Here
is called
r
that
a
of such
fraction;
a
a
a
unit is
parts
is called
b is called the denominator.
Thus
into
how
the denominator
indicates
equal parts the
many
and
the
numerator
unit is to be divided,
indicates how
be
to
taken.
of
those
parts are
many
the numerator,
as
and
expressionmay be considered
Every integer or integral
fraction with unity for its denominator; that is,for
a
h+c
a
,
,
example,
"
=
t
b +
"
^
c
"
"
"
.
Algebra, as in Arithmetic, it is usual to give
the followangRule for expressinga fraction as a mixed
by the denominator, as
quantity:Divide the numerator
to the
and
annex
quotient a fraction
Jar as pr-)ssible,
remainder
the
for numerator, and tliedivisor for
having
131.
In
denominator.
Examph
^
FRACTIONS.
is recommended
student
The
it is
last
the
to
step;
brackets,namely,
for
Rule
132.
Either
Let
will "
by
denote
^
xc=
^
parts
taken
are
For
in each
This demonstrates
let
be
in
J,
;
^
^
the
is
tf
hence
For
numerator
Let
in
times
tmies
as
="
many
.
b
each
of
c
the
integer;
any
fractions
c
^
parts is taken,but each part
of
each
large as
as
times
into
c
part in
times
as
r-
because
in
parts as
in
,
many
-
.
,
denote
b
r-
b
be
will i-r"j=^.
therefore
c
the
-r-
of the Rula
the
se
be
For
be
is
r
b
is -th of
c
This demonstrates
integer. Either
integer,or divide the
by
and
fraction,
any
"
c
of the Rule.
jo:kI form
for dividing
a fraction
the denominator
by that
by that integer.
T
\"
-r-
fraction,and
Rule
multiply
c
and
jb
be
This demonstrates
133.
divide
integer; then
any
b
any
number
times
is
integer.
0
same
i-
c
hence
;
^
the first form
the unit is divided
Ob
fraction by an
or
by that integer,
of the fractions
b
will xrx^=""
then
and
in
denote
t-
of
-
equal parts, and
as
use
(j? 2).
-
fraction,and
-v-
of the
integer.
b
Again;
=
numerator
into b
m
2)
j? +
-
that
any
.
b
b
unit is divided
reallyan example
multiplyinga
multiply the
the denominator
(
+
tion
particularatten-
to pay
times
c
r-,
be'
an
any
by
integer;
Art.
"^
.
9
the firstform of the Rule.
then
132; and
70
FRACTIONS.
QC
let
A^n;
then will
-r-
denote
^o^
y^"'=i'
and therefore
is -th of
~h
-^
,
he
^
times
^
c
any
by
r,
integer-
Art. 132;
.
h
This demonstrates
and
fraction,
any
the second
and
If the numerator
he multipliedhy t?ie same
fractionis not altered.
134.
form
of the Rule.
denominator
of any fraction
value
the
integer^
of the
For
if the numerator
of a fraction be multiplied
by any
integer,the fraction will be multiplied by that integer;
and the result will be divided
nominat
by that integer if its debe multiplied
by that integer.But if we multiply
by an integer,and then divide the result by
any number
the same
the number
is not altered.
integer,
The result may
also be stated thus: if the numerator
and denominator
of any fraction be divided
by the same
the value of the fraction is not altered.
integer,
Both
these verbal statements
statement
v-
=
h
ir
be
included
are
in the
braical
alge-
"
of the
This result is of very great importance; many
operationsin Fractions depend on it,as we shall see in the
next
two
135.
Chapters.
The
demonstrations
only when
satisfactory
given
in this
Chapter
are
letter denotes some
positive
whole
but the results are assumed
to be true
number;
whatever
the letters denote.
For the grounds of tliis
assumption the student may hereafter consult tlie larger
Algebra. The result contained in Art. 134 is the most
forth
important ; the student will therefore observe that hencethat it is always true in Algebra th^
assume
we
^^^^^^^^
h~
hr*
For
^'
every
^' ^^d
example,if we
put
^
-
"^y
denote.
1 for
c
we
hare
=
,
"
v
.
71
XrV.
EXAMPLES.
IBoalso
In like maimer,
to
obtain
we
-^
a
that
by assaming
such
results
the
as
36gg
7
as
8a'
4g
+
always
equal
2g
XIV.
following fractions
25a:
is
-2a_
Examples.
the
c
following:
"
__,_3i^__"
Express
x
v
mixed
quantities:
3ft
+
Aa
9
12.ir2-.'"y
Qx
^-
*^-
"
"
^
a^
+
+
^-3
3
asi^-
Sa*je-Sa^
jp^-2st'
'
'
x-2a
x*
+
ai^-x+l'
x^-l
l
,^
^
9.
-,
10.
"
"
.
Multiply
11.
g^by3".
12.
^-^-^by3("-").
Divide
16.
^'by2ic.
3y
16.
^^'"fby
a
+
o
3a~26.
72
OF
REDUCTION
FRACTTONSl
Reduction
XV.
of Fractions,
result contained
The
in
Art. 134 will now
be
the reduction of a
appliedto two important oper^ons,
lowest
anJIihe
fraction to its
reduction of fractions
terms,
to a common
denominator.
136.
Rule for reducinga fraction to its lowest terms.
the numerator
and
denominator
of t?i4fra"tion
137.
Divide
by
their greatestcommon
For
example; reduce
The
4a^h^ ;
we
measure.
,
g, g
to its lowest terms.
of the mimerator
i"
and the denominator
and denominator
dividingboth numerator
by 4a'^b\
G.c.M.
obtain for the required
residt
^
That
-r-,.
50d
is,
-,
I6a'^"c
.
,
^
equal to
J.
^^^
a^a^?
.
.^
^*
*^
.
,
.
expressed m
more
a
i*
"
5bd
I
snnplo
form ; and it is said to be in the lowest terms, because
cannot be further simplified
by the aid of Art. 134.
rg2 4X
+
"
Again
; reduce
,_
3
to its lowest terms.
g_
""
of the numerator
3; dividingboth numerator
we
obtain for the
The
it
G.C.M.
is
and the denominator
and denominator
by x"2
a?
requiredresult
"
1
"
"
.
g
_
In
and
some
examples we
denominator
liave a
rule for findingthe g.c.m.
may
common
perceivethat
the numerator
factor,without using tiia
Thus,for example,
a-b
+ e
{a-b + c){a-b-c)
cy~ {a+ b + c){a-b-c]~a + bTc^
{a-by-c'
_
a^-(p +
_
74
EXAMPLES.
XV.
Examples.
Reduce
the
XV.
li-actions to
following
their lowest tenus:
lOa'^x
5.
5a^x"15ay^'
a^-hSx
+ 2
^
^'
a?2+ 6;c + 5'
2y^4-a;-15
^ ^-{a-^b}x+
x^-{a+b)x + ab
x^+{c-a)x-ac'
(x+af-{b
"
-^
3T
'
+
4x'
6a^-ll;r^.5
3^3_2;^J5_i-
21a:+18'
"
rT-3
"
T^
2a^^-llx^+\Qx+\Q'
23.
^^:^\
X*
+
2.r3+ 2;r-r
+ Q
16.
.
a^-2ax
"
12a;3_5^4.5^_6-
te^
SSx-27'
a,-3
+ ^+io*
20a^'-4-;g-12
21.
+
x^-i-5x
a?2+ ;g-42
;c3-10^'^
ab
3x*^2Zx-36
c)*
cf
+
{x-^hf-ia +
15.
'
x^^-2x-l5
~^.
;c3_2^
22.
'
2x^
+
+
+
a\
2rt2av-a8
aa^-^a^x~^
^^^'
24.
2.r*-9^3_i4y
+
3
EXAMPLES.
Reduce
the
denominator
33
XV,
following fractions
75
their
to
lowest
:
^
^
-^
^
^
34.
a
x
"
a*
X
36.
a*
a"x*
a^"a**
b
a
1
3
(^-1)"'
a
x"a*
a?+l'
(^+1)2'
a^+ax
a**
+
1
a^
"
d?'
1
+
a^^
ip"-l'
ax
39.
tfi-ax
4
a+sf
38.
a^'
"
y
J?
:f-1'
a*
ab
36.
37.
ax
x^
+
ax
g'
+
.r*TaV
a^^
1
1
40.
fl?'-(a+2")^+ a"'
+
3c^-{a
1
+
c)x-\-ae*
fl?'
common
76
OR
ADDITION
Addition
XVI.
SUBTRACTION
Subtraction
or
Rule for the addition
the fractions to a
Reduce
then add or subtract the numerators
denominator.
inon
140.
Add
Examples.
to
-j"
"j-
or
of Fractions.
subtraction
a-c
T'^
+
_a
and
retain
From
take
36
denominator,
b
'
.
c
4a-
com-
_2a
a"c
l
b
the
.
c +
~
tions.
frac-
denominatn\
common
Here the fractionshave alreadya common
and therefore do not requirereducing;
a-\-c
of
c
3a -46
4a -36
-(3a -46)
_
_
~
c
c
c
4a-3b-Sa
+
4b _a
+
'
c
c
The
full,as
b
~
~
student is recommended
have done in this
we
to
put dovm
the work
example, in order
to
at
ensure
accuracy.
Add
-^
a +
-^.
to
a"b
a+o
will be the
denominator
Here the common
b and a" 6,that is d^"VK
_c{a-b)
c
^+6"
mrefore
a-b~
a^-fta'
b
+
b)
a^-b^
'
c^"^Jfl^^l"^p")
a^-b^
c
a +
_c{a
c
product of
a-b
_ca"cb
+
ca
+
cb
2ca
_
a^~6"
~a2-62*
"
From
"2 +
a"b
6
,
,
take
i
;
+
a
a"o
The
77
FRACTIONS.
OF
.
o
denominator
common
{a+ hf
h
a +
is a'-Wi
a-b
{a-h)*
^
_
a-b~a'-b^*
_
b~a'-b^'
a +
J^"Vf_-^=K
"i"\_^
b
b
"r"b^
Therefore
a
+
a
"
a'^+2db
b''-{a^-1ab + b'') 4db
+
_
From
By
-s
"
"
"
take
z-
:
Art. 123 the
of the denominators
l.c.m.
{x- 1)(^-3) {^x-+
is
6);
3^-
(:e4-l}(4.c^4-3.r-6)
+ 3ar-6)'
3~(a;-l)(^-3)(4u,^
a?+l
_
a^-4a;
4^-3a-
+
+
+ 2)(^-1)
(4.i;2-3^
2
Therefor"
;B*-4a?
+
Aj^-dx^-lbx+lB,
3
+ 33;-6)-(4^-3a;
(j;+l)(4:e"
+ 3a?
(a; 1)(a; 3)(4;c'
-
4:g"+
-
+
-
2)(a?-l)
6)
+ 5^-2)
7aT'-3:g-g-(4a.-^-7a:'
+ 3^6)
(a? 1)(a; 3) (4^;''^
-
-
-
lAa?-^x-A
+ 3u;
{ps-\){x-Z){Ax"-
-
6)
78
ADDITION
141.
to
OR
SUBTRACTION
We
have sometimes
to reduce a mixed
quantity
fraction; this is a simple case of addition or subtraction
of fractions.
a
h
+
a4--=-
^
c
c
a^ + Sab
2^6
_
a-\-b~
\
x-9.
x
b
+
a
3
+
+
a
b~
b
a +
x-2
3'
+
a
.
c
(a-f ^)
a
h
+
ac
=
-
_
b"
+
+
"
c
2ab
a
_
h
ac
=
-
I
c
lab
a
h
a
,
-,
Examples.
+ 3)(:g'-3^-f-4)
(a?
x-2
^
X"-Sx
a^-Zx-V4:
^
+
a^-5x+\'Z-{x-2)
_ar^-bx-"\2-x
*"
+
^a^-Qx+\4
2
~~
~
a^-Sx
x^-3x
+ 4^
"*-3jr
+ 4
addition
Expressions may occur
involvingboth
and subtraotion.
Thus, for example,simplify
142.
6"*"^^^^
a +
The
L.G.M.
is
that
a*-b\
of
^^+"'
+
a^
b^
a^
-
{a^-b^{a^-i-b^
is
denominators
the
_a{a-b)ia-
_a
a^
ah
a
+
a~b^
-
~
a^-b'
_
+
a^'b+
a^
_
a^-b*-
d'
'
a*-b*
abja'+ b*)
ab_
a^
ab*
_
a+b~
a*-b'
'
a^-b^
^a^a^-y^^a^-a%^
tf"'
Therefore
a*-a^b
+
"
'
a*-b*
a*-b*
r
+
a+b^a'^-b^
aW-(a^
+
a^b
a^ + 62
+
ai)'-{a*-aW)
a'^-b*
a*-"^b
+
aW
-
ab^ ^a^^-^ab^-
a*
+
d^b"^
2i^ff
"
_
a'-"
'
+ 4
a'^l?'
Simplify{a-b){a-c)'^
ib-c){b~a)^
{c-a){c-l"y
The
He
example.
is very
for the
denominators
render
pay
the factor b-a
contains
fraction
second
this factor differs from
of the
in the denominator
the
and
which
occurs
in the
and
denominator,
common
thus
to
operationsextremely laborious.
the
The
particular attention to this
the product of the
liable to take
should
befriimer
sign
of each
and
term;
Art.
by
in its denominato
factor
first
a-",
fraction,
only
135,
b
b
^
{b-c){a-b)'
{b-c){b-a)
the
Also
in
form
a
which
Rule
the
is
of Signs
the
for
convenient
more
be put
object; for by
our
can
have
we
{c-a){c-b)
Hence
third fraction
of the
denominator
{a-c){Jt""c).
=
be put in the form
proposed expression may
b
a^
c
{(^b){a-c) {b-c){a-b)'^
{a-c)(b-c)'
and
in this form
is
we
at
see
{a"b){a
c)(b
"
that the
once
(b"c)"b
nominato
de-
c).
"
By reducing the fractions to the lowest
tlie proposed expressionbecomes
a
of the
L.C.M.
{a"c)
+
c
minator
deno-
common
{a"b)
{a-b){a-c){b-")
ab
it,
ac
"
"
'"'"'""^
X
"
ab
bG-\-ac"bc
+
.,
"""*
(a-")(a-c)(6-c)
In
143.
two
or
hand
two
or
may,
if
we
"
we
into
a
have
""
shewn
*
how
to combine
the other
singlefraction; on
up a singlefraction into
example,
please,break
For
fractions.
more
Zhc
Chapter
fractions
more
we
this
.
,
'
Aac^^ab
^bc
Aac
5ab
3
4
5
a
h
o'
_
_
abc
abc
abc
abc
80
EXAMPLES.
Examples.
Find
the value of
XVL
XVL
82
EXAMPLES,
XVL
30.
ae^+
y"^
31.
_^
'^_
_
y^
xy +
xy
x^-2x
+
Z
a^+xy*
x-2
1
32.
af^+1
^-ar
l^^+l
+
1
2
33.
+
(^-3)(a;-4)
(a;-2)(^-4)
1
2^-3
1
34.
x{x
x{x+l)
+
i){x+
2)'^
x{x +
x+l
\-2x
35.
^_J_
x-y
3"
2(^
x'^"xy + y"^
1
x
+
xy
.
x^
y
y^*
+
xy-lofi
a/^-y^
ai*+xy+ y^
x"y
6(a;+ l)'
1)
+
x-y
37.
a-l
x+1
2
38.
a^-x+1
x^ + x+1
a
+
h
+
by
x*
a-h
by
ax"
x'^+l'
+
2{a^x+'b'^y)
a^x^ + b'y^
39.
ax
2y
1
"^
3(y--j:+l)
(^-2)(a?-3)'
2x
40.
x^"x'^+l
ar^+ x+l
x^"x+1
3
12
41.
a^-4x
x^'-lcc+U
1
+
x
a
+
JL
a-b
44.
2a
4a
42.
43.
x'-5x
+ 3
ofi c^
x"a
"
a
+
a^-vb'^
b
1
1_
x"oa
2b
_2_
x
+
x'
+
4.b^
a^-"."4"
3__
_3
Za
x-^a
a^'
x"a
+
4:'
1
G
4
83
XVI
EXAMPLES.
1
4
_
45.
^
a
a"b
2b
"
L +
r+
a
a
c
46.
b
+
'Zb
a +
c
{a!-a){a-b)'^
ia!-b){b-ay
^
47''
48.
,
{x-b){b-ay
{x-a){a-b)
fr" TT
,-r"
{x-a){a-b)
+
{x-b){b-a)'
1
1
49.
{a b){a-c)
"
(b a){b"c)'
"
b
a
50.
"^
{a-b){a-c)
51.
,_U_,.
52.
L^
a{a-b){a-c)
'
{b~a){b-c)
1
.1"_,^
{a-b)(a-c)'^
{b-a)ib-c)
{c-a){c-hy
1
L
b{b-a){b-c)
abc'
+
a*
c"
ft'
53.
{a-b){a-c)
{c-a){c-h)'
{b-a){b-c)
1
54
\
+
si?"{a+b)x+ab
x^-{a
+
c)x +
ac
1
x^"{b + c)x + bc'
x+b
x+c
55.
u^"{a + b)v +
ab
a?
"
{a+ c)x +
x
ac
+ a
x^"(J}+ c)x-{-bc'
1
1
56.
{fL-b)(fl-c){x-a) {b-a){b-c){x-b)
1
{c-a){c-b){x-c)'
6"2
84
MULTIPLICATION
FRACTIONS,
OF
Mtdtiplicationof Fractions.
XVII.
of fractions. Multifor the multiplication
ply
for a new
numerator^ and
togetherthe numerators
denominator.
the denominators
for a new
Rule
144.
followingis the
The
145.
Let
Rule.
and
r
be
-,
0
usual
demofistration
fractions
two
which
of the
are
to be
a
"1
C
therefore
together; put^=a?,
multiplied
and^=?/;
and
a=bXf
therefore
ac
c
=
bdxy
=
di/;
;
CM
divide
by bd,thus
hd~^'
But
*2^=6^^5
a
-
.,
therefore
=
d
146.
and
product of the numerators,
denominators; this demonstrates
Similarlythe
two
bd
is the
ac
productof the
than
tj"
j
b
And
ac
c
""
r
Rule
fractions
are
We
now
shall
togetherthe
be
may
demonstrated
bd
the
the Rule.
when
more
multipliedtogether.
pxamples.
give some
factors of the
new
Before multiplying
numerator
and
the
factors of the new
denominator, it is advisable to examine
in both the numerator
if any factor occurs
and denomibe
stinick
and
the
it
of
result will
out
as
both,
luitor,
may
thus be simplified;
Art. 137.
see
Multiply
a
by
-
.
c
a
a
11
Hence
a
-
C
4-
=
and
are
"
b
ab
c
c
equivalent
;
G
-;and^(2ar-3)
-^.
=
so, for
example,
Multiply
by
-
85
FRACTIONS.
OF
MULTIPLICATION
-
.
y
y
s^
xxx
X
X
y^y~yyy~^'
thus
\y)
~y''
Mulhply-jbyZa
8c
^
46
_
~
46
9a
6')
4(a"
-
^
X
_
~
36
9a
(a + 6)
4a(a-6)x3a
'
4a
_
~
"
(a -h)
I
+
-
+
by
1
I
--1.
+
{a^ +
a^+b'^-ab
F- +
(a^+b^^-a^b^
^O'
a^b^
+
b^
+
~
a^b"-
d'b'^
may
ab){a''+ b^-ab)
_
~
ab
ab
6(a46)
6(a+6)x3a(a+6)
^
we
36
12a
x
_
3a*
a'4-y-fq6
Or
2c
12a
X
'
(a + 6)"
Multiply
2g
8c
X
~
^^--by^^^
Multiply
3a'_
3a
_
proceed
thus:
a^b''
86
MULTIPIJCA
TJON
FRACTIONS.
OF
therefore
fa
\b
a
.
_,.
The
\fa
J\b
b
two
^^
\
b
b-
J
a
^?
5a
a^
h-
a^
b'^
a-
,.
results
for
agree,
^
'
_
Multiplytogether ^--^,
-
"
-f- 1
and
^"
add
the
mixed
quantityb +
have
^
^
?"+-_-.
b and
convenient
singlefraction.
already done
,
factors,aud
aud
by
to
,
reduce
the
Thus
in former Chapters,we
results which
the student
here give some
must
must
sume
asof
and
which
he mi^^t use
to be capable
explanation,
which
be
jis rules in working examples
proposed See
may
and
135.
Arts. 63
147.
As
^ZT
more
,-
first t\vo
multiplythe product separatelyby
it is
"
a-b^
then
results ; but
=
a^
b-
"
,
a*-k-b* + arll^
,
+
might multiplytogether the
We
ft^
"
we
XV
EXAMPLES.
Examples.
6ftc
2a
5a2
36
a^
b^
x^y
a^
__,
+
^
2ax
+
c"
he
ac
eS)
a^-{a
h)x
+
11
^+^^/
"
7
X
r
a^+'j^
"
+
2
x'"l
x-1
+
^
y
^^
a^"y^'
"
x
y
^".
^
\
+
yj'
("_"-"_?).
(,-_^Y
a*
h
V
\6(j
cj
oc
or
\r
a
a/
a
\a
J
X
+
+
xj
c?-?+f-"\.('?_?-|+*y
yj
yj
\a
0
X
a^-2a?
+
l
\a
a;'-4a7
+
X
4
0
^-6"+9
'
(x + 2)*
x''
+
-J
a?
\x
x
a?-(?
ah
+
l
x"l
a*-2ax
a^
+
^''+ ?/'
ar"-y^
^^
a?* + 2;?;V+ 2/^ x^-xy-^y'^
-5
14.
6*
x
'
z-x'
y-z
10.
12.
a2
"
,
g
XVII.
following:
the value of the
Find
87
IL
88
DIVISION
FRACTIONS.
OF
XVIIL
Division
Fractions.
of
Rule for dividingone fraction by another.
the divisor and proceed as in Multiplication.
148.
Rule.
and
followingis
The
149.
Suppose
have
we
therefore
demonstration
divide
to
hx, and
=
ad
c
r
by
j;
hdx
ad
dy;
=
Idx, and
=
therefor"
he
hdy ;
-
X
_
_
~
hdy~ y'
be
But
-y=''^y"b^d'
a
,
therefore
ad
a
r-=%.'^
he
DC
c
a-^ j=
d
o
160.
usual
therefore
3=y;
a
^,
the
We
shall now
Divide
by
"
a
~
'
examples.
-
.
h
a
Divide
d
givesome
a
c
ac
Tb^^scZa
9a
"*"
46
_
~
8c
3"5"^
8c
4"
9a
'
{a+hf
ab-b*
"a
.
(a+ft)3 ^np
x
12a
2c
_
_
~
"
36
X
12a
36
a^-1^'
_""-62
a"-"8
^
-
"
2c
(o + 6)2
52
(a-")g
"(a-6)(a+6)(a-")^
b\a
+
b)^
Invert
bia + b)'
"
put
of the
=
x
^"
90
DIVISION
Find
OF
(-^^
the value of
FRACTIONS.
?)
"
^" ^
-
\2.v-bJ
2a6
_2ab"a{a
a
when
-^,
a?=
h-x
b)
+
+
rt
6
ah-a-
_
_
'b'*
~
a
\~
b
+
2ab
b
~~a
b
+
a
2ab-b{a
j^_
b)
+
+
ab-b^
_
a+b
a+b
1
f^
2a?-6
Therefore
a+b
^" 6
=
?^'
"^"b
b
^
+
a
db
+
a
a^
"
""
x
=
a{b
a)_
"
i^
ab-b^
+
a
a
~ab-b''~b{a-b)~~b'
gz|J=(_g^"\
therefore
.
ab
a
.
Again,
^
'
a-x=-
=
j-
1
6
+
a
a'
a(a+b)"ah
=^
^-^
_b{a
ab
_b
"
;
=
,
b
a
+
+
b)-ab^
a
+
b'
b*
~
"~1
'
-
-
b-iv
(^r
Therefore
1
r
-
b'
""'
x
b
+
a
+
_^
-
=
"
b-x
\2w-b/
a
-
b
+
a
b
+
a
'''
^^^^
Therefore
b~
+
a
-"
Ir
a+b
Ts
~
s
5'
6'
b^
=0.
be
results given in Art. 147 must
again here in connexion vrith Division of Fractions.
The
152.
Since
ac
c
a
t-..
rx
-
b
a
d
ac
and
have
"
5-j^
"
c
a
:"=
i-
bd
a
...
Also
d
since
04^
c
""ft'^""^
d"
o"
c
j-j-s-;"="
bd
d
"
b^ ,
=
M"^
ac
,
and
(zc
"T^j=~r:j;
b
d
bd^
,
we
c
^
j=-j-j,
,
^"
y
"*^"
:
bd
a
r,
b
T
given
EXAMPLES.
XVII
91
L
XVIII.
Examples.
Divide
1
a*
a'
Ax*
"
2""
a*" lax
^
Aax
+
6(^;"-6^)
J_
,
3.
S.t^
'
Aa^
+
ax
Ax'
,
"^
x^-j^
x*
+
xy
+
o^ + 3a^ar+ 3a^
+
^
y^
a^
^-
h' + 2nb-c*
+
"f-a*-b^
ar^-6ar
+
9
+
^
2ab
^
(a-"-j;)*
,
^
a
+
b+
b
+
c^a'
a^-2x
+
e
V
12.
fl+^Vi-'^hv-^.-
13.
J.
5^^-^bya:
+
a*
ar*
15.
a?
a
a
X
,
_^
-1
a*
-
-i
X*
by
.
14.
a'-]ibya-^.
92
EXAMPLES.
16.
8a +
^^08^
17.
18.
-i
_
"
ar
1,^1
-J +
+
-
-
.
_by--l+-.
i+
+
,
1
by
-
-
"'
\a-\-xj
20.
by
^
XVIIL
\a-\-xJ
^ g-3(^-2-V5?by?
+
+
?.
+
the following
Simplify
expressions:
3a?
23.
x-\
^^-^
-^
X-"r\
25i
6
.
X
1
2
2
24.
.
{x-h)(x-c)
ar+a
L.
1+i
1
"^
26.
1+
i+^-.-H"L
1-0?
"
^Y"-
27.
28.
^
1
1 +
1 +
1 +
-
-^
0? +
l-x
X
29"'
(^
\x-y
( ^^
30
-
y
\^+y
\.
(
^'
y' \
,
yj \x^+ y'^x^-yy'
'
x
+
y
\
y^
^
\(
x^-7f)
^-y
^
\
'
\x
+ y
\
x^"y'^y
I
EXAMPLES.
08
Xrill.
"+i
31
y{xyz
1
+
x
.33.
when
T
o
r.
+
a
0
x-b
x-a
a^
.
"
when
T
0
.
r
a
o-a
"
-
-
a; +
+
when
a
2/
x
2a
x=
=
and5
-
=
^^
yL_when2,=^.
"
4
Aab
x"2a
a6
,
rr
/x-ay
t)
I
x-y
ZT
-V9
+
x-2a
+
-
"
;c=
9
4"-x^
x
=-
a
b
a
+
+
b
,
when:r="
"
^,
x
+
-"
.
a-2b
2
ab+a
a+1
x+y-l
"
when
"
2b
-X
\x-bj
40.
-.
3
x^-y^
;c-2/
h
2b
39.
,
3
_-
38.
.
,
2/
;c +
+
when
6
-^-+_iL_
37.
j-
a^(b-a)
,'
qf
6(6 + a)
"
r
a
36.
=
a"b
X
+
-
x
a
X
.^
35.
followingexpreasionB:
x=
x
"
_.
34.
zy
+
x
of the
the values
Find
+
"
,
r
+ 1
when
,
x=
,
ab+V
.
.
,
ajiay=-j"-:,
^
ab+1
b
94j
equations.
simple
XIX.
Simple Eqiiationa.
When
two algebraical
expressionsare connected
by the sign of equalitythe whole is called an equation.
called sides of the
The
are
expressions thus connected
of the equation. The expression to
equation or members
is called the first side,and
the left of the sign of equality
the expression
to the rightis called the second side.
163.
identical equation is one
the
numbers
sides are equal whatever
identical
for example,the following
are
the two
in which
letters represent;
An
154.
(a;+ a){x-a)
=
equations,
x'^"a^,
statements
true \vhatever
that is,these algebraical
are
and a may
numbers
x
represent. The student will see
that up to the present pointhe has been almost exclusively
occupied with results of this kind, that is,with identical
equations.
identical equation is called briefly
an
An
ideiitity.
An equation o/ condition is one which is not true
the letters represent, but only when
whatever
numbers
the lettei*srepresent some
numbers.
or
particularnumber
I'or example, a; + l
be true unless x=6.
An
7 cannot
equation of condition is called brieflyan equation.
155.
=
156.
A
letter to which
a
value
particular
or
values
givenin order that the statement contained in an
quantity. Such
equationmay be tnie,is called an laiknown
value of the unknown
quantityis said to satisfy
particular
the equation^ and is called a root of the equation. To
solve an equation is to find the root or roots.
must
be
An
quantityis
equation involvingouo unknown
said to be of as
dimensions
the index of the
as
many
miknowu
of
the
quantity. Thus, if x denote
highestpowo"
157.
SIMPLE
95
EQUATIONS,
the unknown
quantity,the equation is said to be of
dimension
when
x
occurs
only in the firstpower ; such
equation is also called
the
the
one
an
simple equatimt,or an equation of
no
higher power of x,
a
first degree. If a^ occurs, and
equation is said to be of two
di:nensions
; such
an
quadratic equa'ion,or an equation
degree. If a^ occurs, and no higher power
equationis ;ilsocalled
a
of the second
of X, the e( Jnation is said to be of thr-ee dimensions
; such
cubic
equation,or an equation
an
equation is also called a
And
of the third degree.
so
on.
both
It must be observed that these definitions suppose
members
of the equationto be integralexpressio7is
so Jar
as
relates to
x.
simple ecjuations.
which
shall shew
how
to solve
first to indicate
some
tions
opera-
presentChapter we
In the
158.
may
We
be
have
performed
destrojingthe equalitywhich
If
159.
term
every
multi2}lied
hy
the
an
equation without
it expresses.
each
on
number
same
on
side oj an
the results are
equation
equal.
truth
of this statement
follows from the
be
if
that
equals
multipliedby the same
principle,
the results are equal ; and the use of this statement
The
seen
be
obvious
number
will bo
immediately.
Likewise
be divided
if every
by the same
term,
on
number
each side
tiieresults
of
are
equation
equal.
an
of Ai*t 159 is to clear an equaa
of fra^ctions; this is eflfected by multiplying evei^
of the fractions,
lerm
by the product of all the denominators
least
the
common
by
please,
multipleof those
or, if we
for
denominators.
example, that
Suppose,
160.
use
principal
The
?
+
?
Multiply every
term
4x6xa?
+
that
divide
e\'cn'
term
by
3x6xa;
3
+
4
3
x
4
+
?=9.
6
x
6 ; thus
3x4xa;^3x4x6x9,
is, 24a?+18.r+12.i--^648;
by
6 ; thus
4a; + 3:i;+2.i'=108.
96
EQUATIONS.
SIMPLE
Instead
multiplyingevery
term
by 12, which
of
multiplyevery
3, 4, and
6 ;
is,
that
divide both sides
4
x
6, we
maj
minators
deno-
9^=108;
12 is the root
=
12.
=
"
of the
rerify this by putting
proposed equation. We may
for x in the originalequation-
12
side becomes
fii*st
The
12
"
1^
which
12
+-"
3
+
0
with
agrees
Any
equation to
that is 4
"
4
161.
an
x
9 ; therefore
by
;r
Thus
3
is the l.c.m. of the
then obtain at once
should
we
by
term
,
the second
term
he
may
the other side
Suppose,for example,
Add
to each
a
b from
we
side.
transposedfrom ons side
by changing its sign.
that
x
x"a
see
that
+
=
b
y +
"
"a
has
=
suppose,
from
for
Art.
a-y.
as
one
r";m("ved from one
side
the other side ; and
+//
^n
and
side
"6 on
as
appears
J^ the sign of every
changed the equality still holds.
follows
a.
been
l(}2.
Thus
h-y.
=
a"y"b
of the equation,and appears
"*-bhas been removed
from
the other side.
This
q/
each side ; thus
x"b=b
Here
9 ;
side ; then
that is
Subtract
2, that is
3 +
+
of
term
an
equation
161, by transposingevery
example, that
x"a
=
h"y.
he
term.
08
if
SIMPLE
put 2
we
16-10-6,
shall obtain
for x in the original
equationwe
that is 0, as it should be.
Solve
166.
EQUATIONS.
a?-
2
-(2^ -3)
^^^.
=
the brackets; thus
Kemove
l-a?=
that IS,
multiplyby 2,
2-2x
2
transpose,
-
=
1
=
therefor"
g"
;
;
;
Zx + 1,
3a?;
2;c +
1 ="r,
that is,
"
6x="\
or
V
"=-.
o
Solve
167.
icr=S|
-2
2"-
28
6|
=
-"
the
;
L.O.M.
of the
is 10;
denominators
multiply
by 10;
+ 4)-(7a;+6) 28x2-5(aj-l);
6(5;c
thus
=
25:" + 20-7a?-6
that is,
=
66-5^
25a?" 7a?+ 5ii;=56
transpose,
23a?
that is,
+
5;
+ 6-20
+
6j
=46;
46
therefore
a?
=--=3
2.
to put down all the work
beginneris recommended
in this example, in order to ensure
at full,
as
accuracy.
ing
in clearMistakes with respect to the signsare often made
the
above equation the
an
equation of fr"^tions. In
The
*Ix 4- 5
fraction
"
to
"
has
to be
multipliedby 10,
put the result first in the form
ftftenK'ardsin the form
to the signs.
and
it is advisable
"(7a?+ 5),and
"7a; -6) in order to
secure
tion
atten-
EXAMPLES,
37-4",
3)-^(16-5ar)
^(5a;
Solve
lea
+
=
7
o
By
Art
this is the
146
thus
Multiplyby 21;
that is,
aa
same
7(5;"+3)-3(16-5;p) 21(37-4a?),
=
35;c + 21-48
transpose, 35:c +
that
99
XIX,
15a;
+
777-21
15a? + 84a; =
is,
134;c
therefore
a?
777-84:c;
=
+
48;
804;
=
=
)r^
6,
=
-
"
134
c
1
Solve
,/,"
169.
by
Multiply
6a?+15
8^-10
-.^3^
^ __=___.
the
f
4:r-7
productof 11, 7, and
5 ; thus
+ 15)-65(8:c-10) 77(4;z?-7),
35(6a;
=
that is,
210^
+ 525 -440a;
transpose, 210a?
-
440:z?
that is,
308a;
-
change the signs, 440a;
+ 550
=
+ 308a;
308^
=
539
-
210a?
-
-
=
-539;
525
-
550 ;
539 + 525 + 550,
638a?=1614;
1614
r
therefore
*u
a
-
"^
Examples.
=
,
3.
XIX.
7a?+ 70.
2.
16a?-ll
4.
3a? + 23 ==78 -2a?.
7(a?-18) 3(a:-14).
6.
16a? =
7.
7(a?-3) 9(a;+l)-38.
8.
6 (a;
-7) + 63
9.
69(a?-7) 61(9-a;)-2.
1.
5a; +50=
3.
24a? -49
6.
IL
4a? + 66.
19a; -14.
=
=
=
=
28^
+
jg=27C46-'").
10.
\%
=
(4-a?).
38 -3
72 (a;
-5)
=
=
63
9a?.
(5-ar).
*+| |="^^+
7-a
100
15.
BXJJ\IPLES.
f 1=^-7.
+
7x
2x
19.
56-?2=48-f
.
2:c
21.
4a?
+ 6.
12=-'^
+
^
23.
V^^-?f-8.
25.
^
27.
4(a?-3)-7(;r-4)
MZ:^=29.
+
=
3
3
X
^^
30.
4'^4~6
2x
33.
f-8 74-^^.
26.
1 ^=^-2.
=
+
6-a:.
5
6"''6*
^,
2.-HrH=?^.
^
32.
2.
3x
,
+
1
3a?- 1
^,
31.
XIX,
^
6-='^-2-
x+?^"=4-5^^.
""-+-?_
?^7^io;r-l*
EXAMPLES.
37.
7
=
^3.^-4
+
101
XIX,
;r--^.
3a;-l
6a;-5
38.
16
8
39.
^^_ 5^,25
0.
=
^-3
a;-l
^^-5
40
"
4
9'
6
7a;+ 5
"a; + 6
_
8--5a?
~
6
12
2a?+7
"-l
^^
*^-
"
4
:"+2
^
-2-^-1
x-\
=
9ar-3
x-2
'-
2
,^
2.X-5
47.
6fl?+ 3
=^."
+
^"
6
Rj + 8
5a?
T.
a;
5a?-17|.
=
"
2:c-9
48.
6
3
49.
3^ -Hi^^10-^^=0.
60.
J(3aJ-4)
^(5a;
3)
7
+
+
=
43-5:r.
3
"
'""
a?
a;
+
2
3-4
a?
ar
^
_-
^6='^-
^^
''"
^--2
2-^-
a?-"-3
4
9
-r
102
EXAMPLES.
5-3iO
,"
^3-
55.
6x
XIX.
3 -5a?
3
,
-4--*-3"
=
3--
2
5^-[8af-3{16-6;?T-(4-5:c)}]
=
l-2;r
4-5;c
13
^^
57.
^
?"i_5rl
+
5a;-
+
3
=
-^
9.r-5
1
^a; +
i2^?"zl,
=
b
-^4-21+
^"
"""
4.,
4
o
58.
e.
-TT-
^-.24.
9ar-7
=
^-
x-2
Zx-b
3
12
1
60.
2
61.
^(8-a:)+a;-l|
2(a:6)-|.
+
=
3a;-
1
7a;
13-a;
^^
62.
"
2x-l
"
_
6x-4
=
_-
7X+12
11,
".
-(^+3).
Simple Equations, contintied.
XX.
170.
We
l^Z
EQUATIONS,
SIMPLE
shall
give some
now
examples
of tlie solution
little more
difficult than
are
a
simple equations,which
those
in the preceding Chapter. The student
will see that
it is sometimes
advantageous to clear of fractions partially,
and
then
to effect some
reductions, before we
re
the remaining fractions.
move
of
"1.
Here
Solve
we
may
+
_
-j^^
convenientlymultiply by 12; thus,
^:j^-4(2;"-18)
3(2a;
+
+
that is,
3)
+
By transpositionand
11 ; thus
72 + 6a; + 9
reduction
12
12;c + 72
by transposition,
72
that
is,
+
1 43
x
=
dSx-
=
+
3^
+
4.
3.z;+ 4.
(5a?- 13),
11
=
12 +
obtain
=
43:z? =
therefore
64
=
we
(;"+ 6)
that is,
^x
=
^^^|^-"c
Multiply by
5i+-^.
=
__
55;c
143;
-
1 2;r,
215;
"^
=
6.
43
6a?-13*
,"^
Here
a^
we
16.r-15
"
may
^.
convenientlymultiply by 24;
20|-8^
thus
104
EQUATIONS,
SIMPLE
that is,
144:u-320
4-
48.C -4- 1 Gx
15
-
165
154-
=
+
64ft
16 -2x
and
By transposition
reduction
144a; -320
~
'
15-2;c
multiplyby
15
-2a;; thus
144;i;
320
-
therefore
that
(15
4
1 44;i; + Sx
is,
152a;
2a;) 60
=
-
320 +
=
-
8jf ;
60,
380;
=
*=f|^2^2=2|.
therefore
173.
=
=
Solve
=
.
x-1
x+d
+ 9);
Multiplyby {x 7)(a;
-
thu"
+ 3",
(a;+9)(a;-6)=(a?-7)(;i;
that is,
x^
+
4:X"45
x^-4x"2l;
=
subtract x^ from each side of the
4a;-
transpose,
that
4"i;+ 4a;
is,
will be
8a;
seen
-4a;-
45=
=
45
=
24;
that in this
we
"
thua
equation,
21;
21,
3c^ is found on hoth
cleared of fractious;
example
have
by subtraction,ami
equal ion.
so
the
106
SIMPLE
EQUATIONS.
1
Change the signs; thus
that
{x-5){x-Q)
therefore
-
llo; + 5;"=
is,
-
6
30 ;
-
6a; =24;
therefore
a?
1
Solve
176.
as
=
-Say-'e
"
:9"
it is advisable
fractions ; thus
common
"^
10
1*2
.
-Sar +
accuracy
ensure
decimals
=4.
-75
-45^?o
,"^
\100
^
+
10
2
100/
3\
/9a;
^5
If
Simplifying,
"
to express
ol
"-"'"
^
~
6
"
o-
{x- 2)(a? 3);
=-24;
-6a?
therefore
To
=
a^-Ux-^-ZQ^x^-Zx-^Q'f
is,
that
-(^_2)(^_3)(^-5) (^-6)*
of fractions ; thus
Clear
1
"
9
(x
VlO
10/*
2\
-3J
^(^--j 6-(^^
;
=
-
a;3a;5^a;2
....
thatis,
2-^4-4="-3-'3-
Multiplyby 12,
transpose,
6a; + 9a;
19a;
=
15
-
72
=
72
8 + 15
+
4a; + 8 ;
-
=
95 ;
95
therefore
x
=
"
=
6.
be proposed in vrhich letters are
may
used to represent known
shall continue to
quantities
; we
represent the unknown
quantityby a;, and any other letter
will be supposed to represent a known
will
quantity. We
177.
Equations
M"lve three such
equations.
SIMPLE
Solve
17"
-
f=a
b
+
a
Multiplyhy ab;
that
thus ba;+aa!
is,
{a+ b)x
divideby 0+":
107
EQUATIONS.
=
(ibc;
abc ;
=
x=^^,
thus
a +
179.
Solve
Here
ab-^ax-^bx
{a+ x){b+x)=a{b
b
+
c)-^^-^(A
a^^db+ac+^+a^'f
+
therefore
ax+bx=ac
that
{a+b)a}=ac(l+^
b
is,
divide
by
180.
a +
+
~]
^="'
6 ; thus
_=L_^.
Solve
Clear of fractions ; thus
{x-a){2x- bf ={x-b) (2x of;
-
that
is, {x-a){^^-4xb
Multiplyingout
we
l^={x-b)iAa?-4Mi
+
+
c^
obtain
Aa^-4LX^{a + b)-^x{4ab+ b'^)-(ai^
=^^-Ax'^{a-\-b)-"rx{4"iib-"a^-a^
therefore
therefore
,,
xW-ai^
xa?-
a^b ;
x{([i'^~")a^b-dti^=ab{a-b)}
=
ab
ab{a-b)
-
thereforo
=
a?=
=
"
"
,
"
7.
108
EXAMPLES.
Although
181.
belong
no
to
difficulty
will
serve
as
resembles
we
the
in
a
following
model
only
single
a
Ja;
transposition,
square
both
for
already
Solve
By
following equation
give
Chapter
we
present
those
obtain
the
sides
+
XX
the
examples.
solved,
in
value
therefore
transpose,
1 6
therefore
;^^
a?
Zx-\
tjx
2x-6
x-Z
it
equation
that
quantity.
=
(8
=
64
-16
=
64
+
;
sjxf
"
=
64
^/;i?;
1 6
=
=25.
X
_
.
The
be
and
circumstance
Jx=6\
therefore
will
there
solution,
unknown
8"
=
-16
as
strictly
rot
S.
=
16
ic-
the
the
of
^(x-ie)
thus
the
similar
^(a;"16)
;
of
steps
doe^
it
,
80
;
-
16
V^
+
^
;
"
^^-
"
,
^^^-^^^
*^
2-
=
x-^'
^2
7^326-
2^7
+^I^-
2
2:g-5
2a;- 6
14.
=
^^'7
-
x-l
7^-21
a;-l
A'2+ 3
*^-
109
XX.
EXAMPLES.
_
3a?-8~3;c-7*
,
^(a;-3)+
8.
15.
"-3-(3-;c)(a;+l)
16.
^-x-2{x-\){x
17.
^^-l ^=7:c.
19.
15-i(57-;pX
|(2;r-10)-^(3a;-40)
=
2)={ps-2^){b-2x).
+
18.
+
+ 3)".
(^+7)(:?;+l)=(;z7
=
6;g +
90
8_2a;
2;c + l
38_
+
a;+12~'
"*"
~
4"+l7
8*
2
40
32
4
3^-10_
23.
(a;-l)".
^+^(a?-2)
24.
^g^ (a?-l)(a?-2) a^'-2d?-4
=
+
=
^'
=
6
(3;c-2)(2^-3)
2,"
--^-g(3.-4)..(^-^)i^-3)
i5"
x+lO
2C.
"
35
3
,,
5
v"~
-
"
*/
g
3^-1
4;p-2
1
24;-l
3;i;-2
6'
a"-3
x-2
27.
"
.
34? +
2*
2
110
x-4:
a?-5
a?-7
a?~8
x-6
x-Q^
x-S
x"d*
29,
4a;2-l
3-23?
2"r-5
1-20?
2x-l~
^^'
3
+
T^^ie^T^*
^_?+"_l"^^l
'
34.
l-x
2-x
S-x
33-
-7--^-3-
=
^^
"
-2
6
{x + \){x+ 2){x+ 2k)
+ 3(4a;-2)(a?4-l).
(:p-l)(;r-2)(a;-3)
=
35.
a?-8
a;-9_a;-t-l
X
32
XX.
EXAMPLES.
{x-d){x-l){x-5){x-\)
(a;-2)(a;-4)(:r-6)(ar-10)
=
36.
(8a;-3)2(;c-l)(4;c-l)2(4:c-5).
=
iP-1
*25a?+-2:c-l.
38.
"5:c-2
39.
"5a;+ *6a?--8
40.
-15^; 4-
41.
a"
T
*^
l
a; +
=
'75aj+ -25.
=
"1350?- -225
-36
"6
-2
"
D95?--18
-9
42.
=x.
o"
i"
+"
=
",
~
6a;
6
a;-
6
44.
+ ar(a?-6) 2(a;-a)(a?-6X
a;(a;-a)
46.
+ 2"
(a;-a)(a"-")(a?
=
+
26)
=(aj+2a) (a?+26){x-a-h\
46.
{x-a){x-'b)^{x-a-'b'f,
b
a
47.
a"b
x"b~x"o'
x~a
^
^^+^
x+b
x+c'
^
4a
,
x
49.
111
XX,
EXAMPLES.
+
a
_1
a-b
1_
^
x"a
x
x^"ab'
b
"
1
1
1^1
50.
x"b
x-'b"c
x"a+e
x"a
mx"a"b
mx"a"c
51
nx"c-d
62,
54.
nx
b-d'
"
{a-b){x-c)-(b-c){x-a)-(e-a){X'-b)='QL
x"a
x
+
a"b
a+b~
lax
a_
a^"b^'
{a"x)(b-x)
(p + x)(q + x).
=
x"a-i
x
a
"
x"b"\
x"b
^x"a-\_
x"a
2~x
"
"
56.
{x + a){2x +
b
57.
{x + 2a) {X
a)2 {x + 2b) {x
68.
{x-af{x
69.
Ji4x)+^{4x-1)
60.
^{x
61.
+ ll)+ ^/(a;-9)=10.
^/(:c
62.
V(9^
63.
V(^
6^
V("-a)+"/Ca-2a="/("-J).
U)
+
+
+
cY
+
=
{x + b){2x +
=
-
a-2b)
+
=
=
J{x-U)
+
-
7.
=
l4,
=
2a
-
^/aJ.
a
cf.
+
6)".
(x-bf{x-2a
4)+V(9a;-l)=3.
4a")
x-b"*!*
b"l
+
b\
112
PROBLEMS.
XXI.
PrMems,
"We shall now
in the
applythe methods explained
precedingtwo Chapters to the solution of some
problems,
the
student
exhibit
to
of
the
of
and thus
specimens
use
certain
In
these
problems
quantitiesare given
Algebra.
and another, which has some
assignedrelations to these,
has to be found is
has to be found; the quantity which
called the unknoicn
quantity. The relations are usually
expressedin ordinary language in the enunciation of the
problem,and the method of solvingthe problem may be
thus described in generalterms : denote the unknown
in algebraical
quantity hy the letter x, and
express
tchich
hold
between the unknown
langvLagetlie relations
the
given quantities;an equation will thus
quantity and
be obtained from which the value of the unknown
quantity
be found,
may
182.
is 85, and
of two numbers
The sum
183.
is 27 : find the numbers.
denote the less number
of the numbers
is 27, the
denoted by ;z?+ 27 ; and since the
have
we
Let
X
"
that is,
therefore
2;c
=
+
their difference
the difference
will be
greater number
of the numbers
sum
is 85
;
then,since
a;4-27=85;
2a; + 27
=
85;
27
=
58 ;
=
29.
85
-
68
therefore
the less number
27, that is 56.
Thus
29 +
a?
=
-^
is 29 ; and the greater number
Divide
184.
"2. 10*. among
have
than A^ and G
5*. more
may
and B together.
A, B, and (7,so
may
have
as
much
that
as
is
B
A
Let X denote
the number
in ^'s share,
of shillings
in jB's share,
of shillings
then a; + 5 will denote the number
in C's sliare.
and 2;ir+ 5 will denote the number
of shiUings
11 4-
PROBLEMS.
7
is
And
-(100-a?) shillings.
100
shillings
; therefore
100-a?lb8.
the
whole
value
5
is to be
X
^
^xl002:z?+^^(100-^);
=
multiplyby 2, thus
600
7x
therefore
-
tliat is,
=
4a;
=
3a? =
tijcrcfore
x="r-
=
4tx + 700
700
500
-
7a?;
-
;
200;
66f
.
o
Thus
there must
be
3 2 lbs. of the second sort.
66|lbs. of
the
first sort, and
187. A line is 2 feet 4 inches long; it is requiredto
divide it into two parts,such that onepart may be three-
fourths of the other part.
Let
denote
X
the number
of inches
in the
largerpart
;
3a?
then
will denote
"
The number
the number
of inches in the other part.
of inches in the whole
line is 28 ; therefor"
a?+^=28;
therefore
4a; + 3a?
therefor"
inches
a?
one
1 12 ;
7a?=112
t'.iatis,
Thus
=
part is 16
J
=16.
inches
long,and
the other part 12
long.
had "1000, part of which
lie lent at
person
aiuujal
4 per cent.,and the rest at 5 per cent. ; the wnole
lent at 4 per
inteiest received was
^44 : how much
was
?
ctnL.
188.
A
115
PROBLEMS.
denote the number
of pounds lent at 4 per cent. ;
of pounds lent at
the number
;c will denote
tlicu 1000
The annual interest obtained from the former
5 per cent.
Let
X
"
IS
-
and
"
,
from the latter
therefore
^^
-^--Tqq \
"
^ ^^Too"""^^
"^
=
(1000 x) ;
therefore
4400
=
4^ + 5
that is,
4400
=
4;c + 5000
therefore
Thus
x
"600
was
-
=
5000
-
-
-
4400
5x ;
=
600.
lent at 4 per cent.
will find that the only difficulty
In
in
statements
pressed
extranslating
solvinga problem consists
in ordinary language into Algebraicallanguage;
if he is sometimes
and he should not be discouraged,
a
him
little perplexed,since nothing but practicecan
give
in this process.
readiness and certainty
One remark
may
what
is
be made, which is very important for beginners
;
called the imknown
quantity is reallyan unknown
number^
noticed in forming the equaand this should be distinctly
tion.
in
the
second
Tlius,for example,
problem which ^e
189.
The
student
beginby saying,let x denote the number
^'s share; beginners often say, let x
of
A's
which
is not definite,
because
A'% money
be
money,
may
or
as
expressedin various ways, in pounds,or in shillings,
fraction of the w^hole sum.
a
Again, in the fifth problem
have solved,we
which we
begin by saying,let x denote
of inches in the longerpart ; beginnersoften
the number
the longerpart,or, let ;p=
to these
a part,and
say, let x=
the
same
objectionappliesas to that already
phrases
solved,we
in
shillings
liave
=
noticed.
in translating
a
Beginners often find a difficulty
problem from ordinarylanguage into Algebraical
language,
t)jcausethey do not understand
is meant
what
by the
ordinary language. If no consistent meaning can be iisto translate
to the words, it is of course
si-;ned
impossible
them; but it often happens that the words are not ab190.
116
XXL
EXAMPLES.
but appear to be susceptible
of more
solutely
unintelligible,
than one
meaning. The student should then select one
and
meanhig,express that meaning in Algebraicals}Tnbol8,
deduce
from it the result to which it will lead.
If the
result be inadmissible,
or
absurd, the student should try
another
But if the result is satisfactory
meaning of the words.
he may infer that he has probablyunderstood
the
words correctly
it
still
be
interestingto try
; though
may
the other possiblemeanings,in order to see if the enunciation
is
of
than
one
reaUy
susceptible more
meaning.
A student in solvingthe problems which
191.
are
for
find
which
he
can
some
exercise,
readilysolve
^ven
may
process of guess and trial ; and he
be thus inclined to undervalue
the i)o\ver of Algebra,
may
and look on
its aid as unnecessary.
13iit we may remark
that by Algebra the student is eiuiblcd to solve all these
he ^^^ll
problems,without any uncertainty; and moreover,
find as he proceeds,that by Algebra ho can
solve problems
which
would
be extremely difficult or altogether
if he relied on Arithmetic
alone.
impracticable,
by Arithmetic,or
by
a
Examples.
Find the number
1.
XXI.
which
exceeds
its fifthpart by 24.
father is 30 years old, and his son
is 2 years old :
many
years will the father be eight times as old t^a
\
A
in how
the son
.3. The
is 33
:
4.
than
B
5.
is 48
:
diflFerenco of two
find the numbers.
numbers
of ^155
The sum
B contributed "15
raised
:
how
muck
was
more
did each
The diflfercnceof two
find the numbers.
than
is 7, and
their
sua
by ^4,^, and C together;
and
C
"20
more
A,
contribute ?
numbers
is 14, and
twice as old as B^ and seven
A\^
imited ages amounted
to as many
years
the age of ^ t find the ages of A and B.
6.
years
as
now
their
ago
sum
their
represent
117
XXL
EXAMPLES.
to a certain
7. If 66 be added
treble that number : find the number.
number, the
result is
child is bom
in November, and on the tenth
he is as many
of December
days old as the month was
he bom
?
the day of his birth : when
was
A
8.
Find
9.
that number
24 exceeds
80
much
as
as
the double
the number
day
on
increased by
itself is below 100.
of which
is 9
of which
is a certain fish,the head
inches long; the tail is as long as the head and half the
back ; and the back is as long as the head and tail together
is the length of the back and of the tail?
: what
There
10.
Divide
three times one
the
11.
12.
as
number
part may
be
and
A
as
B
parts such
two
equalto
four times
raised
by A^ B,
of "1Q was
The sum
B contributed as much
much
into
84
as
together:
that
the other.
and
C ther;
togeand C
A and ^10 more,
how
did each conmuch
tribute
?
Divide the number
of one part may be
13.
seventh
60
parts such that
into two
equal to
a
eighth of the other
an
part.
gallonshad been dra^vn out of one of
equal casks, and 80 gallonsout of the other, there
After
14.
two
remained
other
:
34
just three
times as much
in
did each cask contain when
what
cask
full ?
one
as
in the
Divide the numbei*
75 into two
parts such that
3 times the greater may
exceed 7 times the less by 15.
15.
distributes 20 shilUngs among
20 persons,
each
and
to
sixteen
each
givingsixpence
some,
pence
received
to the rest: bow many
sixpence each ?
persons
tbe
16.
A
17.
Di\'ide the number
of three times one
sum
person
20
into
part,and
parts such
that
five times the otlier
two
be 84.
part,may
The priceof a work
which
out in parts is
comes
"2. 16.9. 8i/.; but if the price of each part were
13 {lunco
than it is,the priceof the work would be "Z. Is. iUi.:
more
18.
how
many
19.
by
parts were
Divide
2 sbaP
be
45
there ?
into two
parts such that the first divided
by
equalto the second multiplied
2.
118
EXAMPLES.
XXL
times as
four times
years ago the father was
is th age of each ?
v/as
: what
20.
old
father is three
A
his son;
old as his son
as
four
as
then
'
21.
Divide
into two
that the
parts such
the
eighth of the
meeting a company
exceed
part may
one
188
other
by
fourth of
14.
of beggars gave four
person
to each, and had sixteen pence left ; he found that
to enable him to
more
e should have
requireda shilling
give the beggars sixpence each : how many beggars were
22.
A
Eence
there ?
Divide 100 into two
part be subtracted from
p;irtssuch that if
23.
one
a
fourth
a
of the other
tliirdof
the
mainder
re-
be
11.
may
and B, engage
Two
at play; A
has
24.
persons, A
they begin,and after a ccilain
"T2 and B has ."52 when
have been won
and lost between
of games
number
them,
B
how
much
did
A has three times as much
as
:
A
money
win?
25.
Divide -60 into two
parts such that the difference
be
the greater and (54 may
between
the less and 38.
difference between
equal
to twice
the
of ."276 was
raised hy A, B. and C together;
and iJ 12 more,
B contributed twice as much
{xb A
B and\"12 more:
how much
as
and (7 three times as much
did each contribute ?
26.
The
sum
Find a number
such that
27.
exceed
the sum
shall
its seventh
tAvclfth by 113.
28.
An
3000
in
domg
men,
force
gin::l
29.
; there
so
of its fifth and
sum
its
eighth
and
its
defeat loses one-sixth of its number
w^ounded, and 4000 prisoners; it is reinforced
but retreats, losing one- fourth of its number
army
in killed and
by
in
the
of
a
remain
18000
men
:
what
was
the
ori-
1
Find
a
number
its seventh shall exceed
seventh by 99.
One-half
that the sum
of its fifth and
the difference of its fourth and its
of persons received
certain number
eighteen-penceeach,one-third received two sliillings
each,
each ; the whole sum
and the rest received half a crown
how
there ?
distributed was
"2. 4". :
many
persons were
30.
of
such
a
EXAMPLES.
A
31.
ef 4 per
received
much
119
XXI.
had
^"900 ; part of it he lent at the rate
cent.,and part at the rate of 5 per cent.,and he
person
equal sums
interest from
as
the two
parts : how
did he lend at 4 per cent. ?
father has six sons, each of whom
is four years
his next
brother; and the eldest is
younger
three times as old as the youngest: find their respective
A
older than
32.
ages.
Divide the number
92
the firstmay exceed the second
the fourth by 24.
33.
A
34.
gentleman
A, B, C, D;
servants
much
as
much
as
^, (7
C and
much
as
B
left "550
of whom
as
together :
A
how
into
by
four such parts that
10, the third by 18, and
to be divided among
four
B
to
have
twice as
was
and B together,and D as
much
had each 'I
numbers
such that the half
Find two consecutive
and the fifth of the first taken together shall be equal to
the third and the fourth of the second taken together.
35.
three
is to be distributed among
of money
A, B, and C; the shares of A and B together
persons
amount
to "G0 ; those of A and G to "80 ; and those of B
and C to "92 : find the share of each person.
A
36.
sum
together; A
persons A and B are travelling
has "100, ana
B has "48; they are met
by robbers who
take twice as much
from A as from
B, and leave to A
taken from
much
three times as much
to B-. how
as
was
Two
37.
each?
of "500 was
four persons,
divided among
The sum
that the first and second
80
together received "280, the
third together "260, and the first and fourth
first and
together"220 : find the share of each.
38.
39.
money
"40
:
After A has received "10 from B
B and "6 more
as
; and between
had each at first ?
what money
A
40.
worth
2
wine
has two sorts of wines,one
sort
and
the
other
worth
Ad.
3s.
a
quart,
of
mixture
to make
100
he wants
a
merchant
a
shillings
quart; from
quarts worth
he has as much
them they have
these
2^. 46;?.a
take from each sort ?
quart: how
many
quarts
must
he
120
EXAMPLES.
In
XXI.
mixture
of wine and water the wine composed
than half of the mixture,and the water
25 gallonsmore
less
than
5 gallons
a third of the mixture
: how
lons
galmany
there
each
of
1
were
41.
In
of
a
of 10000 tickets,
half the
lotteryconsisting
number
of blanks
prizesadded to one-third the number
3500 : how many
was
prizeswere there in the lottery?
In a certain weight of gunpowder the saltpetre
43.
than a half of the weight,the sulphur
composed 6 lbs. more
the charcoal 3 lbs. less than a
5 lbs. less than a third,and
42.
fourth
:
a
how
many
lbs.
were
of each
there
?
ingredients
A general,after having lost
44.
had
left fit for action 3600
a
of the
found
battle,
three
that
than half of his
than one-eighthof his army
more
were
uiiiiy ; 600 men
the remainder, forming one-fifth of the
wounded
; and
were
or
slain,taken prisoners,
missing: what was
army,
of
?
the army
the number
he
How
men
more
sheep must a person buy at "7 each
for folding
them at
that after paying one
a score
shilling
them at "8 each ?
night he may gain "79. 16s. by selling
45.
many
of money
shared among
five
A^ B, 6',Z",and E; B received "10 less than A ;
posons
D received "5 less than C;
than B\
C received "16 more
than Z"; and it was
and jE^received "15 more
found that
and B together: how much
E received as much
did
as A
46.
each
A
certain
sum
was
receive ?
of money
starts with a certain sum
A tradesman
;
end of the first year he had doubled his original
stock,all but "100 ; also at the end of the second year he
had doubled the stock at the beginning of the second year,
all but "100;
also in like manner
at the end of the third
47.
at the
year ; and at the end of the third year he
rich as at first : find his original
stock.
as
was
three
times
with a certain sum
of
borrowed
much
he had about him,
as
as
money;
and spent a shilling
out of the whole : witli the remaindei
much
he went to a second tavern, where he burrowed
as
as
he had left,
and also spent a shilling
tlien
went
to
and
he
;
third tavern, borrowing and
a
spending as hefore,after
which he had nothing left: how much
had lie at firet?
48.
A person
there he
went
to
a
tavern
122
PROBLEMS.
perform a piece of work
and B alone can perform it in 12 days : in what
they perform it if they work together%
Let
A
alone
A
1 94.
denote
X
perform
can
perform
z^th of
requirednumber
the
-
the
of
days.
th of the work ; therefore in
ths of the
-
in 9
can
work;
work.
In
therefore
day
one
in
days
x
x
days,
time
will
one
day
In
days he
can
B
can
perform
he
can
perfoim
X
~
of the
ths
work.
perform
of the work
A
195.
must
the
be
And
whole
since in
work, the
sum
days
A
and
B
gether
to-
of the fractions
equal to unity ; that is,
cistern could
be filled with
alone
pipe
in 6 hours,and
alone in 8 hours ; and it could
one
x
by
by
water
of
means
of another
means
pi{)e
emptied by a tap in 1 2
time will the
: in what
the tap are all open ?
be
closed
Iiours if the two pipes were
cistern be filled if the pipesand
Let
hour
hours
denote
x
the
first
it fills
-
the
of hours.
required number
pipe fills
-
th of the cistern
ths of the cistern.
In
one
In
therefore
;
hour
the
oue
in
x
second
o
pipe
fills
-
th of the cistern ; therefore in
x
hours
it fills
o
1
X
q'ths
of the cistern.
In
one
hour
the tap
empties
-
th
123
PROBLEMS.
of the cistern ; therefore in
we
filled,
since in
And
cistern.
the
hours
co
it empties
the whole
-
ths of
cistern
is
have
6
"^8
It is sometimes
196.
hours
x
the
unknown
other
some
this will be
12""
convenient
to denote
by ^, not
required, l"ut
quantity which is explicitly
quantityfrom which that can be easilydeduced ;
illustrated in the next two problems.
A colonel on attemptingto draw up his regiment
in the form of a solid square finds that he has 31 men
in his regimore
ment
over, and that he would require24 men
in order to increase the side of the square by one
197.
man
:
Let
how
many
men
there in the regiment ?
were
denote the number
of men
in the side of the first
of men
in the square is x^ and
square ; then the number
the number
of men
in the regiment is iC^+ 31.
If there
in a side of the square, the number
x+\
of men
were
men
X
be {x + Vf ; thus
in the square would
in the regiment is (:"+1)2-24.
Therefore
that
From
occur3
is,
{x +\f-'2A
ar24- 2."+
^^
=
+
2.^+1-24
2^7
=
+ 24
31-1
men
31.
these two equalexpressions
we
in both ; thus
therefore
of
x^ + 31,
=
1-24
the number
=
31;
=
54;
=
27.
can
remove
x^ which
54
therefore
x
=
"
of men
in the regiment is
the number
Hence
that is,729 + 31, that is,760.
+ 31j
(27)'
124
PROBLEMS.
A starts from a certain place,and travels at the
rate of 7 miles in 5 hours ; B starts from the same
place
direction at the
8 hours after A, and travels in the same
rate of 5 miles in 3 hours : how far will A travel before he
198.
is overtaken
Let
by
B
1
before he is overtaken;
in
5
which
travels
B
therefore
travels 7 miles
since A
Now
of hours
represent the number
X
x
A
"
travels
S hours.
hours, he travels
of
-
a
5
*Ix
mile in
miles.
; and
hour
one
hours
x
he travels
"
5
B travels
Similarly
of
-
he
hours
therefore in x"S
when
therefore in
travels
A they have
B overtakes
Theiefore
of miles.
multiplyby 15; thus
25
mile
a
-8)
(a;
-
=
21a;,
=
21;c;
25;c-200
therefore
25;r-21a:=200,
4;r
=
hour, and
one
{x"S)
travelled
that is,
that is,
in
miles.
the
same
And
ber
num-
200;
x^"^-.50.
therefore
4
Ix
Therefore
7
x50
=
"
=
_
before he
199.
was
70 ;
so
that A
travelled
70 miles
o
o
overtaken.
ProWems
student to have
sometimes
given which
obtained from Arithmetic a
are
suppose
the
knowledge of
1 25
PROBLEMS.
pr()})ortion
; this -willbe ilkistrateJ in tl"o
shall conclude
we
next
tvvo problems. After them
the
of
difficultcharacter
a more
Chapter \\v\a\ three problems
tlian those hitherto given.
the
meaning
of
It is
200.
the number
56 into two
be to the other as 3 to 4.
requiredto divide
parts such that
one
may
the first part; then the other
Let the number
x denote
part must be 56 "x., and since ;z; is to be to 56 ;r as 3 to 4
"
we
have
3
X
66-^~4*
Clear of fractions;
thus
4.c
3(56-x);
=
that is,
42?
=
168-3^;
therefore
7^
=
1 68 ;
therefore
ar
1 fi"^
the
Thus
that
IS
=
-"
first part is 24
=
and
24.
the other
32.
of solution
preceding method
however
beginner; the following
The
for
a
Let
second
soLond
to 56
;
part is 56"24,
the
part
as
number
must
3 to 4.
denote
is the
is much
natural
shorter.
most
first part ; then the
be \x^ because
the first part is to the
Then
of the two parts is equal
the sum
2x
the
thus
Thus the first part is 3
pa:t is 4 X S^ that is 32.
x
8, that is 24; and
the second
126
PROBLEMS,
A
contains
cask, A,
gallonsof wine and 18
gallonsof water; and another cask, B, contains 9 gallons
of wine and 3 gallonsof water: how many
gallonsmust be
from each cask so as to produce by their mixture
drawn
201.
7
gallonsof
wine
and
7
12
gallonsof
water
?
Let X denote the number
A; then since the mixture
will denote the number
\A-x
the number
Now
B,
of gallonsto be drawn from
is to consist of 14 gallons,
of gallonsto be drawn from
(^gallonsin A is 30, of which 12 are
12
is,the wine is
wine ; that
of the whole.
"
Therefore
the
tJO
X
gallonsdrawn
the
Similarly
14
-;r
wine.
gallonsof
A
from
contain
-^^^gallonsof
gallonsdraAvn from
the mixture
And
i5 contaiii
wina
-
is to contain 7
-^o
"
-
gallons
of wine; therefore
12^ ^ 9(14-^)_
30
3(14-^)
2a;
.,
.
,
that is,
one
.
"5""'""4
"
Thus
10
202.
At
hand
~^'
12
gallonsmust
what
of
a
be drawn
time between
watch exactlyover
^
^^*
from
A^ and
2 o'clock and
4 from
B,
3 o'clock is
the other?
of minutes
after
the required number
will
hand
the long
In X minutes
over
move
2 o'clock.
and
hand
the
fiioe;
as
moves
long
X divisions of the watch
will
the
short
short
hand
twelve times as fast as the
hand,
Let
move
X
over
denote
"
\.2i
divisions in
x
minutes.
At
2
o'clock the
127
PROBLEMS.
of the longhand; so
ghort hand is 10 divisions in advance
the long hand
must
10
that iu the x minutes
pass over
divisions than the short hand; therefore
more
A hare
takes four
leaps to a greyhonnd'sthree,
but two
greyhound'sleapsare equivalentto three of
the hare's ; the hare has a start of fifty
leaps: how many
leapsmust the greyhound take to catch the hare ?
203.
of the
of leaps taken by
Suppose that Zx denote the number
the number
the greyhound; then 4;2J will denote
of leaps
taken by the hare in the same
time. Let a denote the number
of inches in one
of
the
hare
then
3a
denotes
the
leap
;
number
of inches in three leaps of the hare,and therefore
also the number
of inches in two leaps of the greyhound;
therefore
the
denotes
"
the number
greyhound. Then
tain 3a; X
contain (50+
leapsof
3^
Inches.
"
of inches in
And
50
+
the greyhound
4x
leapof
one
will
leaps of the hare
con-
will
4;p)ainches; therefore
-=(60
"
+
4a;)a.
9x
Divide
by
a; thus
therefore
9x
therefore
Thus
The
a;
the
that
formed.
we
=
50 +
=
100
will see that
a, to enable us
can
remove
4a;;
+
8a; ;
=100.
greyhound must
student
symbol
and
"
take 300
leaps.
have
to form
the
we
introduced
it by division when
an
liary
auxi-
equation easily
;
the
equationia
128
EXAMPLES.
XX
IL
Four gamesters, A, B, C, D, each with a different
204.
stock of money,
sit do^Mi to play; A wins half of Ba
first
stock,B wins a third part of C's, C wins a fourth part of
and D \v\ns a fifth part of ^'s ; and then each of the
/"'",
gamesters has ;"23.
Let
denote
Find
the stock of each
the number
of
at first.
which D won
from
the number
A\
in A'" first stock.
Thus
with
what
A won
from B, make
Ax, together
up 23 ;
therefore 23
the number
40? denotes
of pounds which A
B.
from
since
A
half
of ^'s stock,23
won
won
And,
4^?
also denotes what was
left with B after his loss to A.
X
then
poimds
will denote
5x
"
"
B
from
won
4.?;,together with what
C,
denotes
make
therefore
tiie
number
4^x
of
23
;
pounds
up
from C.
which B won
third of C's
And, since B won
a
Cs
first stock, \2.x denotes
first stock; and therefore 8:j;
Again,
denotes
23
what
"
left with
was
C after his loss to B.
Again, 8x, togetherwith
23 ; therefore 23
from D.
C won
stock,4 (23
3
"
And,
Sx)
(23 8x) denotes
"
C
since
D's
denotes
what
won
first stock
left with
was
make
up
first
therefore
after his loss to C.
D
won
from
23 ; thus
23=3(23-8x)
therefore
23x=46
therefore
x=%
Thus
D
of D's
; and
x, which
-
J
of
X",make up
pounds which
fourth
a
Finally,3 (23 8x),togetherwith
A
from
won
the number
Sx denotes
"
C
what
+
x;
;
the stocks at firstwere
Examples.
10, 30, 24, 28.
XXII.
privateerrunning at the rate of 10 miles an hour
18 miles off,running at the rate of 8 miles
discovers a sliip
the sliiprun
before it is
miles can
hour : how many
an
1. A
overtaken?
Divide the
of
three-fourths
other part the sum
2.
50
number
one
part
may
be
be 40.
into two parts such that if
to five-sixtlisof the
added
130
EXAMPLES.
13.
Divide
the number
the first increased
third
be
multipliedby
equal.
14.
as
3
for
If 20 men,
women
week's work
a
15.
Divide
of their squares
16.
88
into four
There
40
and
women,
50
1
into two parts such that the difference
may be 1000.
100
places154
two
are
apart, from which
with a design to meet ;
in two hours, and the
miles
time
start at the same
travels at the rate of 3 miles
one
other at the rate of 5 miles in four hours
meet?
two
persons
17.
Divide
by
parts such that
by 2, the second diminished by 3, the
4, and the fourth divided by 5, may all
children receive "50
for a week's work, and 2 men
receive -as much
5 children,
what does each woman
receive
or
them
among
XXII.
5
:
when
will
they
into two parts such that the greater increased
be to the less increased
by 7, as 4 is
44
may
to 3.
18.
as
A
much
can
as
in 24
the work ?
work
do half
(7,and
days:
as
much
work
together they can
in what
time
could
as
B,
B
can
do
half
complete a piece of
each alone complete
Divide the number
90 into four parts such that if
the first be increased by 5, the second diminished
by 4, the
third multipliedby 3, and the fourth divided by 2, the
19.
results shall all be
20.
Three
equal.
persons
can
together complete a piece of
in 60 days ; and it is found that the first does threefourths of what the second does,and the second four-fifths
of what the third does : in what time could each one alone
work
complete
the work
?
Divide the number
36 into two
part may be five-sevenths of the other.
21.
partssuch
that
one
draw up his army
in
the form of a solid square finds that he has 60 men
over,
and that he would
in
41
his
in
men
more
require
army
order to increase the side of the square by one
man
: how
there in the army ?
were
many men
22.
A
general on
attemptingto
I
131
XXII.
EXAMPLES.
Divide the number 90 into two partssuch that
part may be two-thirds of the other.
23.
one
of eggs, half of
person bought a certain number
them at 2 a penny, and half of them at 3 a penny ; he sold
them again at the rate of 5 for two pence, and lost a penny
by the bargain: what was the number of eggs ?
24.
A
A and B are at present of the same
age; if A'"
age be increased by 36 years, and ^'s by 52 years, their
ages will be as 3 to 4 : what is the present age of each ?
25.
For 1 lb. of tea and 9 lbs. of sugar the
8*. 6rf.; for 1 lb. of tea and 15 lbs. of sugar the
12*. Qd. : what is the priceof 1 lb. of sugar ?
26.
so
was
charge is
cliargeis
divided between A and B,
27. A prizeof ;"2000 was
that their shares were
of 7 to 9 : what
in the proportion
the share of each ?
hired for 40 days at 3*. Ad. per
day,for every day he worked ; but with this condition that
for every day he did not work he was to forfeit 1*. Ad. ; and
the whole he had ."3.3^. Ad. to receive : how many days
on
out of the 40 did he work ?
28.
A
workman
29.
A
at
much money
and "6 more,
was
play first won
as B ; but B,
on
had five times
had each at first?
money
B, and had
winningback his own
"5
as
from
much
money
as
A
then
as
money
: what
Divide 100 into two parts,such that the square of
their difierence may
exceed the square of twice the less
30.
part by 2000.
A cistern has two supplypipes,
31.
which will singly
fillit in 4^ hours and 6 hours
and it has also
respectively;
a leak by which
it would
be emptied in 5 hours: in how
many
hours will it be filledwhen
all are
workingtogether?
A farmer would
mix wheat at As. a bushel
so that the whole
mixture may
rye at 2". 6d. a bushel,
of 90 bushels,and be worth
3". '2d. a bushel:
many bushels must be taken of each ?
32.
with
9"2
sist
con-
how
132
XXII
EXAMPLES.
bill of "3. Is. 6d. was
paid in
and the whole number
of coins was
florins,
there of each kind ?
coins were
33.
A
34.
mix
A
sort
coarser
a
28
56 lbs. of fine tea at 5*.
3*. 6d. a
at
lb.,so
lb. : what
4*. 6d. a
togetherat
as
how
:
lb. would
a
to sell the
quantityof
many
whole
the latter sort
he take 1
must
35.
the
on
with
grocer
and
half-crowns,
A
person
agreement
hired
labourer
a
for every
that
day
do
certain work
he worked
he should
to
a
receive 2*.,but that for every day he was
absent he should
lose 9d. ; he worked
twice as many
days as he was absent,
and on the whole received "1. Ids.: find how many
days
he worked.
drawn
up in a solid square ; when
time after if was
some
again dra\vn up in a sohd square
fewer in a side ; in the
it was foimd that there were
5 men
had been
from the field: what
interval 286 men
removed
of men
in the regiment ]
the original
number
was
36.
regiment was
A
of money
divided between
A sum
A
and "\
was
to that of ^ as 5 to 3 ; also the
was
so that the share of A
share of A exceeded
five-ninths of the whole sum
by ^"50 ;
what was
the share of each person ?
37.
38.
left his whole
gentleman
A
estate
share
of the eldest was
"800
the estate; the share of the second was
one-fourth of the estate; the third had
The
sons.
his four
among
less than half of
than
"120 more
half as much
as
the eldest ; and the youngest had two-thirds of what the
second had.
How
much ("d each son receive ?
39.
A
B
and
began
to
playtogetherwith equal sums
of money ; A first won
"20, but afterwards lost half of all
half as much
he then had, and then his money
that
was
as
had each at first?
of B : what money
A
lady gave
consistingof
had 12d.,each woman
40.
of poor,
of
the
women
was
number
muuber
of
two
of
women.
a
guinea
in
charityamong
a
number
and children ; each man
women,
men,
6d.,and each child Sd. The number
less than twice the number
children
How
four
many
less than
person;^
of men
; and
three times
the
were
there
lieved?
re-
EXAMPLES.
133
XXII.
piece of
cloth at Zs. 2d. per
of
it
at
sold
one-third
45. per yard, one-fourth of
yard. He
and
the
Sd.
remainder
at 3*. Ad. per
it at 3".
per yard,
the
whole
his
14*.
and
2du
How
was
gain on
yard;
many
contain
the
?
did
piece
yards
41.
draper bought
A
a
"ZZ. *ls:6d. in buying sheep of
the
first
different sorts.
sort, which formed one-third
For the second
of the whole, he paid 9*. 6d. each.
sort,
one-fourth of the whole, he paid 11*. each.
which formed
42.
grazierspent
A
For
For the rest he
did he buy ?
paid 12^.
6d. each.
What
mmiber
of
sheep
woman
bought a certain number of e^s,
of
for
at the rate
5
at
twopence; she sold half of them
and
half
of
them
at 3 a penny,
and gained 4"f.
2 ft penny,
number
what
the
of
was
by 80 doing:
eggs 1
43.
A
market
44.
A
pudding consists
and
raisins,
and
the
suet
4
8d.
of 2 parts of flour,3 parts of
parts of
suet ; flour costs 3d. a lb.,
raisins,
6d^
cost
of
the
the
several ingredientsof
Find
pudding,when
the whole
cost is 25. 'id.
employed together
persons, A and B, were
for 50 days,at 55. per day each.
During this time A, by
spending 6d. per day less than B, saved twice as much as
How
much
did
B, besides the expenses of two days over.
45.
A
Two
spend per day?
income.
A
persons, A and B, have the same
laysby one-fifth of his ; but B by spending "60 per annum
than A, at the end of three years finds himself ."100
more
of each ?
is the income
What
in debt.
46.
Two
A puts 7
47. A and B shoot by turns at a target
the
bull's
bullets out of 12 into
eye, and B puts in 9 out of
in
bullets. How
32
them
they put
12; between
many
fire?
did
each
shots
the
two
Two
are
casks,A
and
mixtures
of wine
of
is
to the quantityof
wine
and water; in A the quantity
the
like
i?
4 to 3 ; in
water
as
proportionis that of 2 to 3.
what must B contain,so that when
If A contain 84 gallons,
48.
B, contain
put together,the
wiuc and half water ?
new
mixture
may
be half
134?
The
49.
one-
XXII.
EXAMPLES.
hundredth
squireof a parishbequeaths a sum
equal to
part of his estate towards the restoration
church;
of the
of the
the
towards
less than
^200
school; and
of the
"
value
"200
After
County Hospital
39
cies,
deductingthese lega-
to the heir.
remain
estate
this towards
the endowment
less than this latter sum
What
was
the
of the estate?
How
50.
many
of an
three-quarters
does
ago it
hour
was
it want
twice
to 4
as
many
o'clock,if
minutes
o'clock?
past two
casks,A
Two
51.
minutes
and B,
filledwith
are
two
kinds
of
sherry,mixed in the cask A in the proportionof 2 to 7,
and
in the cask B in the proportionof 2 to 5 : what
quantity
mixture
which
each
be taken from
to form
must
a
shall consist of 2 gallonsof the first kind and 6 of the
kind
second
hollow
regiment
the hollow
square
is 1296.
form
the
men
square.
person buys a pieceof land at "30 an acre, and
fold,
finds the value increased threeit in allotments
selling
self:
for himthat he clears "150, and retains 25 acres
so
A
53.
by
of his regiment into
of men
in the
12
deep. The number
in the front of
Find the number
of men
oflScer can
An
52.
a
?
how
54.
many
The
acres
there?
were
country was increased by
war.
During a long peace which
and at the end of that
paid ofi",
national debt
of
in a time of
one-fourth
followed
was
"25000000
time the rate of interest
It was
then found
cent.
a
was
that
reduced
the
before the
of the debt before the war
?
was
amount
55.
A
the
same
and B
as
play at
a
game,
winner
one
from
amomit
war.
4^
to 4
per
terest
of annual inthe
What
was
loser
shillingless than half
agreeing that
the
to the
with equal quanthe money
the loser has ; they commence
tities
and
and after B has lost the first game
of money,
than A : how
the second, he has two
more
won
shillings
?
much had each at the commencement
shall
always pay
i
135
XXIL
EXAMPLES.
centre ;
tho same
A clock has two hands tummg
on
twelve hours, and the
a revolution every
the swifter makes
time -willthe swifter
slower every sixteen hours : in what
66.
gainjust one
complete revolution
on
the slower?
At what time between
3 o'clock and 4 o'clock is
67.
hand of a watch exactlyin the direction of the other
one
hand produced %
The
58-
hands
other at 3 o'clock:
certain
amounted
to X297.
months it amounted
A
59.
60.
hours
of a watch are at
when are they next
sum
of money
12*. in
rightangles to each
at rightangles?
lent at
eightmonths;
to "306
:
what
was
and
the
simple interest
in
sum
seven
more
%
1799
watch gains as much
as a clock loses; and
hours
clock
the
to
1801
by the
are
by
equivalent
much
the watch gainsand the clock loses
: find how
A
watch
per hour.
It is between
and it is observed
11 and 12 o'clock,
of minute
number
between
the hands
is
spaces
it was
two-tliirds of what
ten minutes
previously: find the
time.
6L
that the
62.
A
and
made
a
joint stock of jCSOO by which
they gained X 160, of which A had for his share "32 more
than B : what did each contribute to the stock ?
63.
A
distiller has 51
him
B
gallonsof French
brandy, which
to buy some
glish
Enwith the French,
a
shillings
gallon; he wishes
a gallonto mix
brandy at 3 shillings
and sell the whole at 9 shillings
a gallon. How
many
of the English must
he take, so that he may
what
30 per cent, on
he gave
for the brandy of
cost
8
gallons
gain
both
kinds?
officer can
form his men
into a hollow square
and
also
into a hollow square 8 deep; the front in
deep,
the latter formation
contains IG men
fewer
than in the
64.
An
4
former
formation
:
find the munber
of
men.
136
SIMULTANEOUS
XXIII.
SIMPLE
Simultaneous
eqtmtionsof the firstdegree with
unknown
two
EQUATIONSi
quantities.
have an equationcontaining
two unkno^vn
and y, for example 3a?
7?/ 8. For
value which
v/e
please to assign to one of the
every
unknown
quantitieswe can determine the coiresponding
find as many
value of the other ; and thus we
can
pairs
of values as we
which
the
satisfy
given equation.
please
if
find
l
for
3a;
we
Thus,
=15, and therefore
example, y
find cU;
2 we
a?
22, and therefore ic 7J; and
5; if y
205.
Suppose we
x
quantities
"
=
=
=
so
=
=
=
on.
Also, suppose that tliere is another equation of the
snnie
kind, as for example 2a? + 5?/ 44 ; then we can also
find as many
pairsof values as we pleasewhich satisfythis
=
equation.
But
suppose
wc
h"th
equations; we
of
X
anfl
by
5 ; thus
one
x
and
y which
satisfy
value
shall find that there is only one
value of y.
For multiplythe first equation
15a? -35?/
and
of
ask for values
multiplythe
second
40;
=
equationby
14a? +
35y
=
7 ; thua
30"
by addition,
Therefore,
15a?-
35y
-)^14.r +
that 18,
2.9a?
=
351/ 40
=
+
308 ;
348;
348
,"
tuerefore
--Thus
Vut
for
if both
x=
-
=12.
to be satisfied x must
in cither of the two given
equationsare
this value of
example
"
in the
a?
second; thus
we
obtain
equal 1%
equations^
138
SIMULTANEOUS
Then
put this value
for
example
SIMPLE
of
x
EQUATIONS,
in either of the
in the second
given equations,
thus
;
Suppose,however, that in solvingthese equationswe
If we multiplythe first
to begin by eliminatingx.
by 12, and the second by 8, we obtain
96a; + 84?/
=
wish
tion
equa-
1200,
96;z?-40y=:704.
Therefore,by svbtraction,
84y
40^=1200-704;
+
that is,
1242/ 496,
therefore
y=4.
render
multiply the first
thus
Or
the process
may
we
"
=
for
the second
simple;
more
equation by 3, and
24a? +
212/=300,
24a;-
102/= 176.
we
may
by 2;
Therefore,by subtraction,
212/+ 10?/ 300 -176;
=
that
is,
31^
therefore
209.
124;
2^=4.
Second
quantitiesin
substitute
=
method.
terms
this value
Express
one
the other from either
in the other equation.
of
Thus, takingthe example given in
have from the first equation
we
8ar=
therefore
of the unknown
100-72/;
x^}-^:^
.
the
equation^cmd
cle,
preceding Arti-
Substitute this value of
obtain
Then
:
from
a?
in the second
x
=
189
and
equation,
of y in either of the
substitute this value
and we shall obtain
Or thus
EQUATIONS.
SIMPLE
SIMULTANEOUS
we
given equations,
9.
have
the first equation we
7z/ 100-8a?;
=
lOO-So;
therefore
we
y
=
=
.
of y in the second
this value
Substitute
obtain
"
equation,and
^^_5 (100-8^)^33.
84;r-5(100-8;c) 616;
therefore
=
that is,
84a;-500
therefore
124;p
therefore
Third
210.
quantity
in
method.
terms
40:r=616;
5004-616
=
a;
+
=
1116;
9.
=
Express
the
same
unknown
of the other from, each equation^and
equate the expressionsthus obtained.
Thus, taking again the
equation
88
+
^-'
12
x=
5y
"
^^^
"^~~^"
same
example, from
^^""^
^^"
second
the first
equation
140
SIMPLE
SIMULTANEOUS
100-7y ^
^"
_
,
Therefore
88 +
=
"
3(100-7y)
300-21y=176+
tliercfore
300-176
Sly
therefore
y
thus:
from
124;
=
4.
=
deduce
can
first
the
10y;
+
a?
=9.
equation y
12^"88
from
the second
thus
lOy;
21y
=
is,
Or
.
2(SS + 5y);
=
tliat is,
Then, as before,we
5y
i^
"
multiplyingby 24;
Clear of fractions,
by
that
EQUATIONS.
equationy
=
"
=
=
"
and
"
;
therefore
a
=9;
,
o
100-8a;
From
before,we
211.
this
deduce
can
Solve
y
19;r-21y
that ig,
therefore
+
then,as
21;c-l9y
=
140.
be solved by the methods
already
however
to shew that these
them
abbreviated.
be sometimes
we
Here, by addition,
and
4.
=
19;c-21y=100,
equationsmay
explained; we shall use
may
shall obtain
equationwe
These
methods
12;c-88
obtain
21^-19^/
40x
-
x
=
100 + 140;
40y
=
240
y
=
6.
"
;
we
by subtraction,
Again, fi'om the originalequations,
obtain
21;c- 19y- 19ar4-21y
=
140-100
that is,
2x-\-2y-A0\
therefore
"+y=20.
;
SIMULTANEOUS
Then
since
SIMPLE
x
"
y
Q and
=
a; +
a?
13, and
=
student will find
obtain
20, we
14;
2/
=
2j;=26, and by subtraction 2y
therefore
141
EQUATIONS.
=
by
tion
addi-
y=7.
he
proceedsthat in all
examples may be treated by
parts of Algebra, particular
methods
which are shorter than the generalrules;but such
abbreviations can
only be suggested by experienceand
and the beginner should not waste his time in
practice,
seekingfor them.
212.
The
213.
Solve
+
-
X
If
cleared these
-
=
8,
as
?^-=
^
y
X
equationsof
3.
y
fractions
they would
involve the product xy of the unknown
quantities;and
thus strictly
they do not belong to the present Chapter,
be solved by the methods
But they may
already^ven,as
shall now
For multiplythe first equation by 3
shew.
we
and
and the second by 2,
add ; thus
wo
142
XXIII.
EXAMPLES.
Solve
214.
a^x +
y^y^c^,
by
+
ax
c.
=
Here a; and y are supposed to denote unknown
while the other letters are supposed to denote
of money
A sum
divided equallyamong
was
a
certain number
of persons ; if there had
six more,
been
each would
have received
less than he did ;
two
shillings
and if there had been three fewer, each would
have received
two shillings
than he did : find the number
of
more
persons, and what each received.
221.
151
PROBLEMS.
Let
denote the number
of persons, and y the number
of shillings
which each received. Then xy is the number
of
in the sum
of money
which is divided ; and, by
shillings
X
supposition,
{x + Q){y-2)=xy
(1),
{x-'3,){y^Vi)=xy
From
obtain
(1)we
therefore
Qy
+
xy
From
(2).
"
2x
\2
"
(3).
6y-2x=\2
obtain
(2)we
xy-^2x-^y"Q
therefore
From
xyy
=
2x-'"y
xy',
=
(4).
Q
=
3y
(3)and (4),by addition,
=
18 ;
therefore y=6.
Substitute the value of y in (4); thus
2;"-18
therefore 2^?
Thus
24 ; therefor"
=
there
12
were
x
=
6;
=
\%
and
persons,
each
received
6
shillings.
of two" digitsis equal to five
certain number
of its digits
if nine be added
to the
times the sum
; and
the digitsare reversed : find the number.
number
222.
A
Let X denote the digitin the tens' place,and y the digit
is H)x-\-y ; and, by
in the units' place. Then the number
number
the
is equal to five times the sum
of
supposition,
its
digits;therefore
\0x
+
If nine bo added
that is,we
y
obtain the number
From
(1)we
(2)we
(1).
to the number
\(ix-+y + ^
From
6{x+y)
=
=
lOy
its
+
digitsare reversed,
x-, therefore
\Oy+x
(2).
5a?=4y
(3).
obtain
obtain 9a? + 9
=
9y
;
therefore
a; +
1
=
i/=
152
PROBLEMS.
Substitute for y in
(3); thus
therefore
Then
ic=4.
from
223.
A
5.
=
is 45.
requirednumber
the
Hence
obtain y
(3)we
an hour is detained
railwaytrain after travelling
after which
it proceeds at six-fifths of its
late. If the detention
former rate,and arrives 15 minutes
had taken place 5 miles further on, the train would
have
than
later
arrived 2 minutes
it did. Find the original
rate
distance
and
the
travelled.
of the train,
minutes,
24
of miles per hour at which
Let Zx denote the number
and let y denote the number
the train originally
travelled,
will
^x
of miles in the whole distance ti-avellcd. Then y
of miles which remain to be travelled
the number
denote
"
after the detention.
distance
would
be
At
the
rate of the
original
travelled in
"
--z
hours:
train this
at
the
in-
bx
creased
rate
it will be
travelled in
^-r
"
hours.
Since
^x
the train is detained
24 minutes, and yet is only 15 minutes
late at its arrival,it follows that the remainder
of the
journey is performed in 9 minutes less than it would have
been
is
-
if the rate
-
of
an
had not been
increased.
And
9 minutes
hour ; therefore
oO
V~^x
y-^x
__9^
^
~
ex
5x
If the detention had taken
there would have been y"5x"5
Thus
we
60
^ ^*
place 5 miles further
shall find that
dx
bx
on,
miles left to be travelled
60
^^'
154
EXAMPLES.
such that
numbers
sum
XXV.
is two-lbirds
one
of the
other,and
is 100.
We
number,
proceed
may
and
nmnber;
y=3-,
Or
we
number^ then
100"
;c
x
then
denote
the
have
we
greater^
ic+y=ioo.
Let x denote the
will denote the less number;
proceed
may
Let
thus.
y the less
thus.
greater"
fore
there-
1x
Or
we
proceed thus.
may
number, then
will denote
2x
Let ^x denote
the less number;
the greater
therefore
2.r + 3;"=100.
the
By completingany
numbers
I'cquired
The
student
may
of these processes
60 and 40.
are
accordinglyfind
we
that
shall find tliat
he
can
solve
of the examples at the end of the present Chapter,
with the aid of only one letter to denote an unknown
"iuanof the examples at the
tity;and, on the other hand, some
e!id of Chapter xxii.
most
to him
naturally
may
appear
solved with the aid of two letters. As a general rule it
of
may be stated that the employment of a largernumber
unknov/n
quantitiesrenders the work longer,but at the
time allows the successive steps to be more
same
readily
followed ; and thus is more
suitable for beginners.
some
The
beginner will find it a good exercise to solve the
example given in Art. 204 with the aid of four letters to
which are required.
represent the four unknown
quantities
Examples.
XXV.
If A'" money
three times as
increased by 36 shillings
he would
were
have
much
as
wei^B
B; and if jB's money
diminished by 5 shillings
he would have half as much
ao
A : find the sum
each.
possessedby
1.
2.
second
Find
numbers
two
make
may
of the firstmay
20, and
make
20.
such that the firstwith half the
also that the second with a third
If B
3.
of money
would
be
mis
B
:h
were
to
if
A
;
give "25
double
were
155
XXV.
EXAMPLES.
to
they would have equal
give "11 to"B the money
to A
that of A
:
find the
money
which
has.
actually
such that half the first with a
Find two numbers
Lthird of the second
32, and that a fourth of the
may make
firstwith a fifth of the second may make
18.
4.
buys 8 lbs. of tea and 3 lbs. of si^r for
person
"1. 2s.; and at another time he buys 5 lbs. of tea and 4 lbs.
of sugar for 155. 2.d.: find the priceof tea and sugar per lb.
5.
A
Seven years ago A was
three times as old* as B
seven
was
; and
years hence A will be twice as old as B
will be : find their present ages.
6.
the fraction which becomes
7. Find
equal to ^ when
is increased
the numerator
by 1, and equal to | when the
denominator
is increased by 1.
A certain fishingrod consists of two
parts; the
is
of
the lower as
to the length
lengthof the upper part
5 to 7 ; and 9 times the upper
part togetherwith 13 times
the lower part exceed
11 times the whole rod by 36 inches:
8.
find the
lengthsof
the two
parts.
spends half-a-crown in applesand pears,
person
buying the apples at 4 a penny, and the pears at 5 a
he sells half his apples and one-third of his pears
penny;
for 13 pence, which Avas the priceat which he bought them:
find how many
applesand how many pears he bought.
9.
A
has two
wine merchant
sorts of wine, abetter
and a worse;
if he mixes
them
in the proportion of two
the
quarts of the better sort with three of the worse,
mixture will be worth Is. 9d. a quart ; but if he mixes them
in the proportionof seven
quarts of the better sort with
eight of the worse, the mixture will be worth 1*. lOd. a
quart : find the priceof a quart of each sort.
10.
A
farmer sold to one
person 30 bushels of wheat,
and 40 bushels of bai'ley
for "13. 10s.; to another
person
and 30 bushels
he sold 60 bushels of wheat
of barley
for ."17 : find the priceof wheat aud barleyper bushel.
11.
A
156
XXV.
EXAMPLES.
n
farmer has 28 bushels of barley at 2,?. Ad. 9
bushel: "vith these he wishes to mix lye at 3*. a bushel,
and wheat at 4s. a bushel,so that the mixture
consiet
may
^
and be worth
of 100 bushels,
3*. 4c?. a bushel: find hoiHl
bushels of rye and wheat he must take.
^|
many
12.
A
if A loses
of 10 shillings;
A and B lay a wager
he will have as much
B will then have ; if B loses he
as
will have half of what
A will then have : find the mone:
13.
i
of each.
If the numerator
of
by 1, and the denominator
will be 1 ; if tlie numerator
and the denominator
the value will be 4 : find the
14.
15.
in
a
-
A
certain fraction be increased
be diminished
by 1, the value
be increased by the denominator,
diminished by the numerator,
a
fraction.
of posts are placed at equal distances
If to twice the number
of them we add
number
straightline.
two
the distance between
consecutive posts,expressed in
If from four times the distance beis 68.
tween
feet,the sum
subtract
two consecutive posts,expressedin feet,
we
is 68.
of posts,the remainder
half the number
Find the
distance between the extreme
posts.
16.
gentleman distributing
money
A
found
men
able to
to each
that
give
5
man
shillingsleft:
shillings.
4
he
wanted
10
to each man
shillings
shiUingsonly, and
find
the
number
some
poor
order to be
; therefore he gives
finds that he has 6
among
in
shillings,
of
poor
men
and
of
in a tavern found,when they
17. A certain company
if there had been three more
their
that
to
came
bill,
pay
bill,
they would have paid one
persons to pay the same
each less than they did ; and if there had been
shilling
fewer persons they would have paid one
shillingeach
did
find
number
the
of
than
and the
more
:
they
persons
of shillings
each paid.
number
two
certain rectangularfloor,such that
if it had been two feet broader,and three feet longer,
it
would
have been sixty-foursquare feet larger; but if it
it would
had been three feet broader,and two feet longer,
have been sixty-eight
find
the
feet
length
larger:
square
and breadth of the floor.
18.
There
10.
A
is
a
certain number
of two
digitsis equal
to four
EXAMPLES.
157
XXV.
times the sum
of its digits;and if 18 be added
number
the digits
reversed : find the number.
are
Two
digitswhich form
the addition of 9 ; and the sum
on
33 : find the digits.
20.
a
tlio
to
number
change places
of the two numbers
is
of two digits
When
is doubled,
a certain number
and increased by 36,the result is the same
as if the number
had been reversed,and doubled, and then diminished by
of
itself exceeds four times the sum
36 ; also the number
its digits
by 3 : find the number
21.
have together 5 cwt of luggage,
passengers
above the weight allowed
and are charged for the excess
55. 2d. and
ds. 10c?. respectively
; if the luggage had all
belonged to one of them he would have been charged
Two
22.
19*. Id.
without
find how
:
luggageeach
much
passenger
is allowed
charge.
A
23.
B
and
ran
which
race
a
lasted 5 minutes; B
yards; but A ran 3 yai*dswhile B
running 2, and won
by 30 yards: find the length of
and the speed of each.
course
had
start of 20
a
A
24.
A
gives to
and
back
again to
many
B
as
left;each
many
have
B
^
as
A
has
each
many
as
and
left,
of
has
them
each had at first.
A
and B
B
A
as
as
many
B
a
certain number
has
has
returns
now
was
the
of counters
;
already,and B retm-ns
left ; A gives to ^ as
to ^
sixteen
as
many
counters:
A has
find how
as
togetherperform a certain work in
30 days; at the end of 18 days however
B is called off
and
A
finishes it alone in 20 more
days : find the time
in which each could perform the work alone.
26.
A, B, and C can drink a cask of beer in 15 days ;
A and B togetherdrink four-thirds of what C does ; and
25.
can
C drinks twice as much
as A : find the time
alone could drink the cask of beer.
each
holding 1200 gallonsis filled by three
pipesA,B,C togetherin 24 minutes. The pipe A requires
than 6' to fill the cistern;and
30 minutes more
10 gallons
than
less run
through C per minute
through A and B
together. Find the time in wliich each pipe alone wouid
27.
A
cistern
in which
fillthe cistern.
158
EXAMPLES.
XXV.
A and B run a mile. At the firstheat A gives B
28.
him by 30 seconds.
At
the
a start of 20 yards,and beats
of
him
second heat A gives B a start
32 seconds, and beats
hour
Find
the
rate
at which A runa
by 9^ yards.
per
situated 24 miles apart,
and B are two towns
of
river.
A
bank
A to B
man
a
on
same
goes from
in 7 hours, by rowing the first half of the distance,and
walkingthe second half. In returning he walks the first
half at three-fourths of his former
rate, but the stream
A
29.
the
behig with him he rows at double his
the whole
he accomplishes
distance in
rates of walkmg and rowing.
A
30.
railwaytrain
rate in
going ;
6 hours.
Find
and
his
after travelling
hour is detained
an
it proceeds at three-fourths of its
minutes, after which
former rate,and arrives 24 minutes
late. If the detention
miles
taken
further
had
place 5
on, the train would have
15
arrived 3 minutes
tlian it did.
Find
sooner
rate of the train and the distance travelled.
The
the
I
original
which
train takes to travel
an
express
is
of
miles
taken
that
120
to
a journey
by an ordinarytrain
The
14.
time in
as 9 is to
ordinary train loses as much
stoppages as it would take to travel 20 miles without stopping.
31.
time
tinio in
express train only loses half as much
stoppages as the ordinarytrain,and it itlsotravels 15 miles
hour quicker. Find the rate of each train.
an
The
trains,92 feet long and 84 feet long respectively,
with
uniform
velocities
on
moving
are
parallelrails ;
in oppositedirections they are observed
when
they move
in one second and a half;but when
other
to pass each
they
Two
32.
direction the faster train is observed
to
pass the other in six seconds : find the rate at which each
train moves.
move
in the
same
A
from A
to G.
A railroad runs
33.
goods' train
train at 1
and a passenger
B.t 12 o'clock,
starts from A
the
two-thirds
of
the
distance
After
o'clock.
goods*
going
three-fourths
of
at
travel
train breaks down, and can
only
its former rate.
At 40 minutes past 2 o'clock a collision
train
The rate of the passenger
occurs, 10 miles from C.
the diminished
rate of the goods'train. Find
distance from A to C, and the rates of the trains.
is double
the
A
certain
J5, and
C,
that
shares
of B
34.
eighths
in
6
A
and
A
shares
the
four-sevenths
share
of A
and
and
A^
of the
exceeded
by "30;
three-
(7's share
B
by "30.
person.
working together can
G together can
earn
G together can
earn
each
man
can
number
and
number
of each
54
alone
earn
shillings
shillingsin 9
SO
shillingsin 15
per day.
40
earn
days;
find the
time
can
is
times
to 48
tmcted
perforaia piece of
G in 30 days; and B and
which each could p^form
in
There
38.
equal
B
and
A
from
of
coin,
and
A
37.
the
certain
G
between
and sixpences
sovereigns,shillings,
The amount
of the shillings
amount
to "8. 6". Qd.
guinea less than that of the sovereigns,and a guinea
half more
the
than
that of the sixpences. Find
a
a
48
and
B
find what
36.
is
B
and
A
of
^'s
and
of A
of each
and
days;
days;
days :
shares
share
; also
by "30
two-ninths
the
35.
^'s
G
and
of the
exceeded
Find
so
divided
of money
was
share
exceeded
sum
15^
XXV.
EXAMPLES.
the
certain
a
the
sum
number
sum
of the
extreme
digit:
find the
number.
work
together in
G in 26| days:
the
of three
number
work
alone.
digitswhich
is
digits,and if 198 be subthe digits will be reversed
; also
digitsis equal to twice the middle
of its
bought 10 bullocks, 120 sheep, and 46
Tlie price of 3 sheep is equal to tliat of 5 lambs.
lambs.
of
A bullock, a sheep, and a lamb
togetliercost a number
number
of animals
shillingsgi'eaterby 300 than the whole
Find
the
whole
the
"468. Qs.
sum
bought; and
spent was
of
and
a lamb
respectively.
a bullock,a sheep,
price
A
39.
farmer
A
40.
of
man
sold at
and
horses,oxen,
market
a
7*. per head ; while a
sold for "22, "12. 105.,and
one-fourth
than
he
same.
did, the
Find
which
the
number
amount
the number
were
sold.
the
and
of stock
whole
sisting
con-
realised
sheep were
he
"1. UKv. respectively. Had
and
of oxen,
25 more
sheep
horse,
received
of
head
that
sheep,so
"2.
sold
100
an
ox,
would
horses,oxen,
have
and
a
been
stillthe
sheep,respectively
160
QUADRATIC
XXVI.
Quadratic Equations.
quadraticequationis
A
226.
EQUATIONS,
of the unknown
the square
equationwhich contains
but
no
quantity,
higher
an
power.
227. A pure
only the square
quadraticequationis one which contains
of the unknown
quantity. An adfected
quach'atic
equationis one which contains the first power
of the unknown
Thus, for
quantityas well as its square.
is a pure
example, 2:"2=50
quadratic equation; and
^aP" 7^ + 3
0 is an adfectedqujidratic
equation.
=
-
The
followingis the Rule for solvinga pur"
quadraticequation. Find the value of t/iesquare of the
unknown
quantity hy the Rule for solvinga simple equation;
the
the
values
then,by extracting
qfthe
square root,
unknotcn
are
found.
quantity
228.
For
example,solve
Clear of fractions
10
therefore
therefore
"
r
+
"
j-"
by multiplying
by
30 ; thna
=
-
13;i;2^180
a^
=
"
=
+
130 + 15
325;
=
25 ;
"5.
by the Rule for solvinga
is equal to 25 ; therefore x must
number, that if multipliedinto itself the product
example, we
simpleequation,that x^
this
6.
{a^ 13)+ Zia^- 5) 180;
exti-act the square root,thus ^=
In
=
"
find
be such a
That is to say, x must
be a square root of
is 25.
is
the
Arithmetic
6
root of 25; in Algebra
In
25.
square
consider either 5 or "5
as
a
we
square root of 25,
may
5x5.
5x "6
Hence
x
gince,by the RtUe of Signs
of
values
the
have
either
the
5
and
or
equation
6,
may
This
denote
satisfied
willbe
thus, 4;- "5.
we
=
"
"
162
EQUATIONS.
QUADRATIC
followingis the Rule for solvingan adfected
reduction
quadratic equation. By transposition and
the equation so that the terms
which
involve the
arrange
unknown
quantity are alone on one side,and the coefficient
of a^ is + 1 ; add to each side of the equation the square
of x, and then extract the square
of half the coefficient
root of each side.
230.
The
shall now
It will be seen
from the examples which we
solve that the above rule leads us to a point from which
can
we
immediately obtain the values of the unknown
quantity.
231.
Sdve
aj"- 10a; +24=0.
By transposition, a?^10x="2^\
add
(yY,
a^-10d?
extract the square
+
62= -24
root,
a:
6
*
=
a?=5*l=5
transpose,
hence
"
26=1;
+
1j
+ l or
verifythat either of these values
proposed equation; and it will be useful for
thus to verifyhis results.
Solve
232.
3"2-4a?-55=0.
By transposition, 3a:2-4a?=65;
divide
AA
1;
a;=6or4.
It is easy to
the
5"
by 3,
/2V
T^s"'
/2V
4a?
2
65
13
2
extract the square ro"t,
^
=
~
a?=r"
"3
*
11
13
"
00
=*=
o
2
transpose,
169
4
.
^
=5or
"
"
.
*"
satisfies
the student
QUADRATIC
233.
Solve
2a?-k-Zx-Z5
2"*
By transposition,
f^y
0.
=
+ 3a; =
35 ;
^+f f;
divide by 2,
.AA
=
^^^ W
^
35
3
x=
"
Solve
*
"5,
=-or
4
"*-4a?-l
2
=
By transposition,a:^- 4^;=
add2",
i
T
7
17
-""
4
289
17
3
transpose,
9
^
extract the square root,^+t=
234.
163
EQUATIONS.
0.
1;
"*-4x+22=l+4=6;
extract the square
root,
a?
transpose,
-
2
a;=2"
=
,y6;
"
,^5.
the square root of 5 cannot be found exactly;
of it
l?utwe can find oy Arithmetic "i approximatevame
to any assigneddegree of accuracy, and thus obtain the
values of a; to any assigneddegree of accuracy.
Here
In the examples hitherto solved we have found
two difFerent roots of a quadratic
equation; in some cases
however we shall find really
ample,
only one root. Take, for exthe
the equation a;^-14:c + 49
0; by extracting
235.
=
ever
7. It is howx"1
0, therefore x
square root we
found convenientin such a case to say that the quad"
have
ratic equationhas two
=
=
equalroots,
11"3
164
QUADRATIC
Solve
236.
EQUATIONS,
x^-Qx
ic^
By transposition,
6a?
"
add
try to
we
=
0.
=
13 ;
"
a?''-6x + 32=-13
32,
If
U
+
+ 9
extract the square
root
-4
=
hare
we
a?-3="*y-4.
But "4
because any
have
root,exact
approximate,
tiplied
number, whether positiveor negative,if mulgives a positiveresult. In this case the
by itself,
quadraticequationhas no real root ; and this is sometimes
expressed by saying that the roots are imaginary or
impossible,
can
no
square
^
Solve
237.
+
^,
,n
2(a?-l)
Here
"
4
by multiplyingby
nominat
multipleof the de-
2(a?+ l)+ 12=a;2_i^
a?"
By transposition,
24? "
"
a;2-2a?+ 1
1*,
the square root, a;-
therefore
1
Solve
238.
1
15 ;
=
15+
=
=t
1
2;"
~
12ii;+ 70
=
thtt.
\
Multiplyby 570, which
"
,ar.
3(10+a?)
15
16;
-3.
3a?-50
+.
=
4;
"?=l"4=5or
a
000
IS and
7-
first clear of fractions
which is the least common
Thus
extract
=
x^-\
we
A{x^ l\
add
^
or
190
is the least
common
multipleof
190; thus
therefore
190(3^-50)^^^"_
*
10 +
therefore
iB
+ x) ;
190(3" 60) (210 40;r)(10
-
=
-
103
""":."
"
\'"
t^xtractthe square root,ad+
1939
'
19
therefor"
ar=
1
Solve
239.
^
"
"
=10
"
+
2a?-3
^
^-2
a;+2
-29.
or
a?-3
3
^ +
a
t^^c.
39
"
-"
=.^^\
"
=
.
1
a;"
Clear of fractions;thus
+ 2)(x-l)
",{x+Z){x^2){a!-l)4-(x-3){a!
=
that is, ^-7:F
+ 6 +
2a^-2a^-
that is,
therefore
+ 2"(i-2);
(2;c-35(iP
":^-2a?2-5a;+
12x+
:
,
12
=
extract the square root,
;
therefore
a?
2a^
-
2;2;'-3a:'-8ar+12;
=
3a^
-
8^
+
12 ;
a:*" 4a?=0;
a?2-4.?+
add22,
6
2^^4V
2
=
2=fc2
=
^
-
"
2,
.
=
4 or
0.
have ^veii the last thriee lines in order to complete
manner
the solution of the equation in the same
as
the results
in. the. former examples ; ]"iit
may be obtained
simply.For the equationar^ 4.2; 0 may be written
more
obvious
and in this form it is sufficiently
(x~4)x=0;
have
either a; -4
that we
must
0, or x
0, that is,
We
"
=
x
=
4
or
0.
"":''
=
=
,:-"::"
student wili observe that in this example 2x^ is
found on both sides of the equation,after we have cleared
of fractions ; accordingly
it can be removed
by subtraction,
and so the equation
remains a quadratic
equation.
The
166
QUADRATIC
quadralic equation
Every
240.
forvfix* +
px
+
q
numbers^ whole
For
EQUATIONS.
0, where
=
or
p and
be
can
put
q represent
some
in ths
knotcn
fractional,positiveor negative.
contains no
quadraticequation,by definition,
quantity higher than the second
power
be
all
terms
the
Let
brought to one side,and, if necessary,
change the signs of all the terms so that the coeflScient of
of the unknown
the square
quantitymay be a positive
and
number; then divide every term by this coefficient,
the equationtakes the assignedform.
a
of
For
the unknown
example,suppose
therefore
4^^
therefore
a^
"
_
7 -^ + 5
-?+
-
4
Thus
Aa^
*lx
in this example we
7
^
=
0
Here
6.
=
we
.
0.
4
have
"=
and
-~
0'=-.
4
241.
Solve
4
a^-\-px-\-q=Q.
By transposition,
extract
have
the square root,
2
a^+px=
a? +
=^
=
"q\
"
"
"
^
;
2
2
have thus obtained b, generatformula fiSr
the roots of the quadraticequation:^-hpic
+ q=0, namely,
that a; must be equal to
242.
We
,
-y"+V(p"-4g)^ -jp-^(j"2-4g)
*
2
2
this general formula
which wiU hold for any
rery important inferences,
equation,by Art. 240.
"We
shall
now
deduce
from
some
ratic
quad-
two
quadratic equation
A
243.
167
EQUATIONS,
QUADRATIC
have
cannot
than
more
roots.
For
have
we
other of two
that the root must
seen
he
one
or
the
assignedexpressions.
are
quadratic equation where the terms
all on one
the
and
the coefficient
of
of the
side^
square
unknoicn
quantity is unity the sum, of the roots is equal
to the coefficient
of the second term, with its sign changed,
and th^eproduct of the roots is equal to the last term.
244.
In
a
^
For let the
the
sum
equationbe
+
q=0;
of the roots is
-p*^U^-4,)
-"-^/(P'-^),
that is -p;
^
2,
.
the
sfi+px
it
,
product of the
roots is
2^2*
p^-iP^-4q)^
that is g.
that is
The p^receding
Article deserves specialattention,
for it furnishes a very good example both of the nature of
the general results of Algebra, and of the methods
by
which these general results are
The
obtained.
student
should verifythese results in the case
of the quadratic
Take, for example, that in
equationsalready solved.
Art 232; the equationmay be put in the form
245.
0^-3-3=0,
and the roots
are
5 and
"
--
;
thus the
55
4
and the
-
3
sum
,
productof
the roots is
"
"
.
o
of the roots fa
168
QUADRATIC
Solve
246.
EQUATIONS.
"wr*+ 5:" + c"a
By transposition, ax^
Jdivideby
bx=
+
a?2+
a,
"e;
=
"
-
-
;
W-4ac
4a2
extract
the square
.therefo"
root,
,=
'
a; +
"
=
"
'
4a2
-
"
;
Z"i4^rif").
2a
general fonnulse given in Arts. 241 and 248
may be employed in solvingany quadraticequation. Take
for example the equation 80:^"45;"
65
0; divide by 3"
Th"
247.
=
tlius we
have
4a?
55
3
3
-^
,
Take the formula
in Art
241, whidi gives the roots of
4
a^-^px + q=0'y and put
thua obtain the roots of the
(
as
But it ismore
we
thus
avoid
55
andg="
^=--,
;
"
proposed equiUdon.
convenient to use the formufe in Art. 246,
fractiaus. The proposed equation being
Sic^" 4rar 55
4, and
3, 6="
0, we must put a
in the formula which gives the roots of 00?^+ 6.j;
+ c
"
that
shall
we
55,
e="
=
=
=
0,
-^^V^"--4"c).
is,in
Th.swehaye*"^^(H""!0)^
th,tis,*^".
D
D
that IS,
-
,
that is, 5
or
-
"
.
170
XXVI.
EXAMPLES.
35.
5'
2
x-\
x
;p-2
a7-4
+
14
37.
"
15*
x-Z
x"1
x-1
a?-3_ll
x-4:
a?-2~12*
39.
1
1
41.
2(aj2-l) 4(;i?+l)
2:^+1
3:i?-2
11
3;" + 2
2
8
,"
X-1
3^+1
"
3(;"-6)
2;g-7
5^
2;b-8
2
3;2?-2
2a?--6
2a;^5
3a;-2
10
.
47
3
49.
(a?-3)2=2(aj2-9^.
51.
^,
;c +
5
+
2
^-1
2^-1
X'k-2'^x-2'"x-l'
^-^
g-
a;-i-l
:
^-^
S7
I
^
g
56
^-2_2^+13
*
;r+2~
a:+l
a;-f
W-1
2aj--l
_
16
a? +
ar+2
l
TT
"
a?-l
a2-ijj_2a"a?+a*-l
X
-+a
a
X
=
X
^'
X
"
?;
8d?-3
"2-3
"
X
'"
"
;ir+
Ill
^
64.
ir
l
^^x={a^-li*+xy
62.
aA
.
x+\~'
14a?-9
0.
Q'
^-2~
"""
*
+
a?-l
h
i+"
0
=
T~
x
a;-4"2_2a;-H3
5;i?-ll
"
^Jl.
x + 4.
a;+l
^^
H
;
+ 2
*
.
oS.
g
,
'
x
+ Z
^x
2^ x-2~^^^r
+
x
7("-l)*
6
.
oo.
x-2-X-^2
f..
""^
"
65
61.
+ 4
x
a?
^+1
M
"'''
-11.
=
J.
I
"
x
+ o
=
-
a
+
"
-^
a+6
XXTII.
171
QUADRATICS,
LIKE
EQUATIONS
Equations which
he
may
solved
Uke
Quadratics.
There are many
equationswhich are not strictly
but which may be solved by the method of completing
quadratics,
will give two examples.
the square;
we
248.
Solveaj"-7a^=8.
249.
?l;
Addg)V
:".-7^."'=8.?
=
4
4
extract
the square
^
7
aj'=-*-r=8
therefore
extract
the cub6
Sol ve
250*
Subtract
root,thus
4- 3;r +
:5?2
2 from
9
7
root,;""--=="=-;
both
"2 + 3a;
-'.1;
ii?=; 2 or
Add
-
+ 3;"
;"/0"'
namely, J{x'^A-Zx^^)
square of the former J we
2) 6.
-
thus
sides,
oji the left-hand side we
and ar^+ 3^
Thus
1.
"
3 a/{^ + 3;p
2 4- 3
-
or
can
2)
=
-
4.
have two expressions,
2, and the latter is the
"
ihe square.
nowcojiiwip/^e
QY,
a;2+3;"-2
thus
+
3
extract the square
V(^
therefore
^ ^
y(;c2+3a?-2H(|Y=4
+
=
root,thus
+
3;r-2)+
Ay(""+3a?-2)=
--
|=*|;
:
"
-
=
i or
-4.
172
EQ UA TIONS
First suppose
LIKE
J{x^+
3a?
This ia
we
x"
we
+
JiaP'
a?
=
3
or
the whole
on
1
=
.
3a?
2)
=
"
4.
-
^4- 3a?^ 2=
16.
ordinaryquadraticequation;by solvingit
an
shall obtain
Thus
2
-
.
.
suppose
is
1
=
"
""
Square l)oth sides,thus
This
2)
TICS,
ordmary quadraticequationj by,solvingit
an
shall obtain
Next
-
^c^+ 3a?
thus
Square both sides,
UADRA
Q
:
6.
"
-,
have four vahies for x,
we
namely,
3"r-6or-=i4^.
important observation must be made with respect
to these values.
them.
we
Sitjlipose
proceed to verify.
find that a?2+ 3a? 2
If we
16, and thus
put a? 3 we
J{x-+ 3a? 2) " 4. If w", take the value -f 4 the original
4 it
equation will not be satisfied; if we take the value
will be satisfied,' If we
put a?= "6 we arrive at the pame
the result might have
And
result.
been
anticipated,
An
:
=
=
"
=
-
"
because
the
values
a?
3
=
+ 3a?" 2) --4, which
Ay(a?2
put
we
6
"
were
deduced
was
=
equatioij. If;
or
a?.=.-
"'
obtained from
from the original
"
find
w"
that
_
,.
equation will be satisfied
l, anii'the original
a;24-3a?-^2 =
if we
tak"
+ 3;r^2)=
x/(a?2
might have
been
+
1;
the result
and, as Ifefore,
anticipated.
.
;..
arrive at the same
four
fact we shall find that we
of
values of a?,by solving
either
the following{equations,
In
'
'
a^-j.3a?"---3V(a!"+3a?-2)
6,
=
a?2+ 3a? +
but the
values 3
equation,and
or
+ 3a?-2) 6;
3x/(a?2
=
only to
belong strictly
-6
the values
the second equation.
'
-
"
"
~
"
;
the
first
only to
belongstrictly
173
QUADRATICS.
LIKE
EQUATIONS
Equationsmay be proposed which will require
the operationsof transposingand squaring to be performed,
before they are reduced to quador oftener,
ratics
once
will givetwo examples.
; we
251.
252.
9.
2a7-^/(a;"-3a;-3)
Solve
=
Transpose,
2a;
9
Jix^
=
transpose,
3;c
-
4a;2-36;c
square,
divide
-
3);
-
8I=a;'-3;?7-3;
+
Zx"^ Z^x
+
84
=
0 ;
x^-\\x
+
1^
=
0.
-
by 3,
By solving this quadraticwe shall obtain ar=7 or 4.
The value 7 satisfiesthe original
equation; the value 4
to the equation2^ + ^(^2_3^_3^:^9^
belongsstrictly
253.
Solve
Square,
V(a?+ 4)4-V(2:c+ 6) ^/(8^+ 9).
=
a; + 4 +
transpose, 2j{x
Mx
square,
that
+
+
transpose,
+
^Q
The
value
satisfies the
-
1 ;
=
25a;2
=
26x^-\Qx+\;
-
17a;2- 66a; -95
6
+ 6)=8a;+9;
4;,y(2a;
5ar
=
By solvingthis quadraticwe
"
+
4)J{2x + 6)
4)(2^+ 6)
^a^+6Qx
is,
2V(^
2a; + 6 +
=
1 0;c + 1 ;
0.
19
shall obtain
a;
=5
or
originalequation; the
"
-.
value
19
"
"
to
belongsstrictly
the
equation
+ 4) ^/(8a?
+ 9).
V(2a;+ 6) ^/(a?
=
-
254.
that in
an
The student will
in which we
cases
ordinary form, we cannot be
obtained for
trial that the values finally
to the original
quantity belong strictly
equation to
certain without
the unknown
equation.
from the precedingexamples
have to square in order to reduce
see
the
174
LIKE
EQUATIONS
QUADRATICS.
Equations are sometimes
proposed which
are
affd
intended to be solved,partlyby inspection,
partlyby
will
two
give
examples.
ordinarymethods ; we
256.
1
Solve
"
o.^
266.
Bring
=
^-^-^^
the fractions
denominator
common
9 +
/c-4
+ 4
^
;
on
9-0?
a?
g^-^^.
each
side of the equationto
a
thus
IQx
ZQx
is a root
To find the
Here it is obvious that a?=0
other roots we begin by dividingboth sides of the tion
equathus
by 4a?;
9
4
Thus
there
are
three roots of the
proposedequation,
namely,0, 6, "6.
257.
Solve
aj"
-
*Jxa*+ 6a'
=
0.
Here
it is obvious that x"a
is a root.
write the equation x^"c^=7a\x--a);
and
to
other roots we beginby dividingby a? -a.
Thus
We
may
find the
a^+ax+a^=7a\
By solvingthis quadraticwe
Thus
there
are
.lamely,
a, 2a,
three
-So.
roots
shall obtain x=2aor
"So,
of the proposed equation,
175
XXVIL
EXAMPLES,
XXVIL
Examples.
1.
a?*-13a:" + 36
3.
+ 6)
a;4-x/(^
5.
2V(^-2"
6.
a?*-2a;' + a;2=36.
8.
9V(^-9^
9.
2a?2'+6;B=226-V(iB2+3a;-8).
7.
=
4.
a?+
V(^-6:c
7.
28)+
+ 4)
iC*-4j:2_2^(;5*-4a:"
11.
fi?+2V(^+5^
12.
3aj+"/(^+
^{a?-^S) 2l.
=
31.
=
+
7a;+
2)
=
10.
5)=19.
^(4?+9)=2^/a;-3.
a?=7V(2-a?2).
13.
15.
+ 8)- V("+3)=^/".
^/(a7
6V(l-^-^ + 5"=7.
17.
+ ^/(5;B-19)
^/(2a?4-8).
^/(3;^"-3)
18.
V(2^ + l)+ \/(7a?-27)=V(3;"+4).
19.
V(62+"M?)-^/(a2+6;c)=a+5.
20.
+ 2a;2="^-".
2;cV("+ -'P')
=
^ +
21
+ l
N/(12a'-;g)^"
22
a;-v'(12a2-ar) a-l*
a? +
7
x"1
1_
_1
\-\-x
1-a?
3"_
1+^*
x"7
x-\-l
1
1
x-Ji^-x^r^'
x+Ji^-a^)"^
x
+
25.
J{a^-\)
x + a
x"ab+x
x"a
x
26.
27.
Q,
16)+ (a?-3)*=ia
+
16.
^^'
=
9;B=flr'+ 36.
10.
14.
x-5^x-lA
l)+""=23-f2;r.
+
+
2.
0.
=
a^+3aa^
+
=
a~
^,
x-J{x^-l)_
.
b"x
b"x
b
28.
+
x'
6;B2(a-ar)(a"-^(aJ+3a"
=
176
PROBLEMS.
lead to Quadratic
which
ProUems
XXVIII.
Equations.
that their
such
numbers
two
and their productis 54.
Find
258.
is 15,
sum
of the numbers, then 16"
denote
one
Let X
denote the other number; and by supposition
a?
(15" a;)
=
-
=
54 ;
-
/15\^
=
-54+
^
"2
15
therefore
a;
=
x=^
take
we
a?
=9
we
"
2
-
:
*
3
^-
have
=
15
dorQ.
=6, and
-a?
if
we
9.
althoughthe quadraticequation gives two values
yet there is reallyonlyone solution of the problem.
of
have
we
15
"
.r
=
9,
the two numbers
are
Here
X,
take
6 and
6
Thus
'^^
~
""
9
-7-=
3
15
r
therefore
^u
If
225
(f)=
a^-16x-^[" )
therefore
will
54.
a;^ 1 5;r
By transposition,
x
of money
A person laid out a certain sum
goods,which he sold again for ^24, and lost as much
he laid out.
cent, as he laid out : find how much
259.
Let
X
denote the number
the
24 will denote
then a?"
lost. Now
by suppositionhe
per
of pounds which he laid out ,
of pounds which he
number
lost at the rate of
that is the loss was
the fraction
therefore
a?2_ioOa?= -2400.
From
in
"
"
x
per
cent.,
of the cost ; therefore
this quadraticequation we
shall obtain a; =40
Thus all we can infer is that the sum
60.
of money
laid
or
either ;"40 or MO;
out was
for each of these numbers
eatistiesall the conditions of the problem.
178
PROBLEMS.
solvingproblems it is often found,as
In
262.
that results
which
obtained
are
actuallyproposed.
in Art. 260,
do not apply
to the problem
to be, that
the
appears
The
reason
of expressionis more
mode
nary
general than ordialgebraical
language, and thus the equation which is a proper
of the problem will also
of the conditions
representation
Experience v.illconvince the
apply to other conditions.
that he will always be able to select the result
student
it will be
which belongsto the problem he is solving.And
in
the
enunciation of the
often possible,
by suitable changes
form
to
a new
problem correspondingto
problem,
original
to the original
which
was
inapplicable
problem ;
any result
and
will
illustrated
in
Article
this is
we
ther
now
261,
give anoexample.
Find the price of eggs per score, when ten more
crown's worth lowei-s the pricethreepence per
263.
in half
a
score.
Let
denote
X
the number
of pence
of eggs, then each eg^ costs
score
of eggs
the number
be
can
pence
"
-^ ot;
t^^t
^
"
"
bought for
If ^^^"
price
x
less,each
score
per
number
would
of eggs
be
;
price of
a
and therefore
half
a
crown
600
X
is 30
which
in the
"
-
.
e^g
which
would
could
be
cost
"
were
threepence
Z
"
"
bought
pence,
for half
and
a
the
crown
Therefore,by supposition,
."W
600
a;--3
X
therefore
60^=60(a;-3)
therefore
a?-^x=\SO.
'
+
a?(^-3);
shall obtain x=\5
this quadraticequation we
It
the pricerequired is 15"^. per score.
Hence
"12.
or
will be found that 12c;?.is the result of the followingproblem;
ten fctcer
when
find th9 price of eggs per score
in half a crown's
worth raises the price threepence per
From
Boorc.
Examples.
1.
Divide
2.
The
the number
their productmay be 864.
sum
their squares
3.
of two
is 1872:
179
XXVIII.
EXAMPLES.
XXVIII.
60
into two
numbers
is 60, and
find the numbers.
The difference of two numbers
is 720 : find the numbers.
that
parts such
the
of
sum
6, and their product
is
Find three numbers
such that the second shall be
two-thirds of the first,
and the third half of the first; and
that the sum
of the squares of the numbers
shall be 549.
4.
The difference of two numbers is 2, and
their squares is 244 : find the numbers.
5.
the
of
sum
'
the number
their productadded to the
6.
Divide
10
sum
into two
parts such that
of their squares may make
76.
7. Find the number
will make 210.
which
added
to ita square
number is 16 times another;and
of the numbers is 144: find the nimibers.
8.
One
the
root
product
di-saded
One hundred and ten bushels of coals were
of poor persons ; if each person
a certain number
among
he would
have received as
bushel more
had received one
find the number
bushels as there were
persons:
many
9.
of persons.
A
diningtogetherat an inn find their
company
not allowed to
billamounts to "8. 155. ; two of them were
to 10
pay, and the rest found that their shares amounted
all
than
if
had paid: find the number
a man
more
shillings
of men
in the company.
10.
cistern can
by one of them
pir.es;
than by the other,
and
the time in which each
11.
A
be suppliedwith water
it would be filled 6 hours
by
by both togetherin 4 hours
pipe alone would fillit.
12"2
two
sooner
:
find
180
XXVIII.
EXAMPLES.
of pieces of
person bought a certain number
which
he
sold
cloth for "33. 155.,
again at "2. 8*. per piece,
whole
much
in
the
he
and
as a single
gained as
piece cost :
of piecesof cloth.
find the number
A
12.
and B together can
14| days; and A alone can
than B alone : find the time
A
13.
perform a piece of work
perform it in 12 days
in which
alone
A
can
in
less
form
per-
it
14.
18
per
how
A
bought
man
a
certain
quantity of meat for
rise in price one
penny
to
were
shillings.If meat
lb.,he would get 3 lbs. less for the
much meat he bought.
same
Find
sum.
kind of sugar per stone of 14 lbs.
one
than
that
of another kind; and 8 lbs. less of
is \s. 9d. more
be got for "1 tlian of the second: find
the first kind can
the priceof each kind per stone.
15.
priceof
The
of money
spent a certain sum
which he sold againfor X24, and gained as much
16.
in
person
goods cost
the
as
A
him
side of
The
18.
Find
find what
the
goods
cent.
cost.
square is 110 inches long: find the
of a rectanglewhich
shall have its
length and breadth
4 inches longer than that of the square, and its
pei-imeter
4 square inches less than that of the square.
area
a
17.
:
per
goods,
the
worth
shilling's
Two
19.
former
an'ived
a* what
eggs per dozen,when two less in
raises the priceone penny per dozen.
priceof
A
messengers
and
B
were
despatched at
to
a
rate per hour
each travelled.
of
person rents a certain number
land for ."70 ; he keeps 8 acres
in his own
20.
the
place at the distance of 90 miles; the
than the latter
by riding one mile per hour more
hour before him: find
at the end of his journeyone
time
same
a
A
and sublets the remainder
at 5 shillings
per
he gave, and
thus he covers
his rent and
find the number
of
acres.
acres
of pasture
possession,
acre
more
has "2
than
over:
EXAMPLES.
From
21.
A
persons
travelled
A
they
B
day
they met
in a day.
A
in
order
far
how
each
meet
the
half
to
miles,
320
B-, and
equal
Find
of
to
than
more
w^as
two
other.
number
the
each
ct
number
travelled
drew
of pure
of pure
certain
A
solid
square
companies
being four
a
;
of
before
the
water,
number
prcat
as
in like
treated
in the
water
at the
end
It,and
adds
iiis capital at
end
"o382:
the
manner.
vessel
of
the
the
men
the
into
such
equal
talion
bat-
and
C
the
third
the
and
obtained
(7;
and
B,
B
be
put
by dividing
of
is
water
and
C
proportion
of
A
men
formed
square
in the company.
A^
of
of
formed
had
nine
been
brandy
to
C.
remainder
same
another
determine
solid
quantity
contents
Fhid
time.
seven
brandy,
fraction
gallons
each
the
square,
formed
by the
vessels
contents
lends
^"5000
person
he receives
of a year
A
25.
if the
as
of
second
If the
and
water.
brandy
that
the
together, it is found
the
quantity of brandy by
times
square
the
as
equal
the
of
hollow
a
large
as
three
are
contains
into
hollow
Find
There
24.
first
times
company.
consisting
be
he
as
64
drew
can
with
much
that
he
soldiers
of
formed
deep.
much
vessel
as
found
was
how
Find
The
it
full vessel
a
the
up
mixture
the
; and
company
battaUon
be
can
is sixteen
one
wine
from
filled
then
from
remained.
wine
23.
drew
then
of wine
quantity
a
gallons, and
81
He
before
drew
person
held
water.
"he
distauce
a
met.
which
by
a
at
out
set
miles
8
went
22.
a
and
which
days in
miles B
places
two
181
XXVIIL
the
rate
year
rate
at
to
a
certain
rate
of interest
his interest,
spends
his
he
capital;
of
interest
finds
that
of interest.
as
he
then
"25
;
of
lends
at
before, and
has
altogether
182
SIMULTANEOUS
-^XIX.
EQUATIONS
Simultaneous
We
shall
Equations involving Qiindratict,
solve
cixamplesof
ous
simultaneThere
two
are
cases
equationsinvolvingquadratics.
wliich
rules
for
be
of frequent occurrence
can
given ; in
both these cases
there are two unknown
quantitiesand two
will always be denoted
quantities
equations. The unknown
264.
by
the letters
x
now
some
and y.
First Case. Suppose that one
of the equations
is of the first-degree,
and the other of the second degree.
265.
From
Rule.
value
the
equatiofiof the first degreefind the
either of the unknown
quantitiesin terrns
this value in the equation
other, and suhstitute
of
the
the second
of
of
degree.
Example.
Solve Sx
From
first
+
4y--18,
hx^
-
^xy
=
2.
""
the
equation y=
-^
;
substitute
this
4
value in the second
equation;
therefore
therefore
20a^-54x
+
therefore
From
9x^=8;
29;i;2 54a?
-
this
=
8.
find ;r=:r2
quadraticequationwe
or
x^;
"
267
then
in
by substituting
266.
Solve
the value of y
3afl+ 5x-8y
=
36,
wo
find 2/
2x^-3x-4y
^
3
=
or
-"
:
3.
Here althoughneither of the given equations is of the
firstdegree,yet we can immediately deduce from them an
equation of
the first degree.
INVOLVING
For
by
multiplythe
first equationby 2, and
the
second
3 ; thus
1 0.^-15^
therefore,
by subtraction,
that is,
90? +
+
12^^ 72 "9
=
;
19;c-4y=63.
From
this
obtain
equationwe
this value in the first of the
3^
+
5a;
therefore
y
"
substitute
;
"
-
"
givenequations;thus
(19a; 63)
2
-
-
3a;2- 33a; +90
^-2-11^;
therefore
or
133
QUADRATICS.
+ 30
=
36 ;
=
0;
=
0.
shall find that a?=f
From
this quadraticequationwe
in the value of y we find
6; and then by substituting
that 2/
8
=
or
12f.
Second
267.
When
Case.
quantitiesh\ each
which
Art.
is
the terms
equation
an
second
the
of
hoiuog"3Meousand
the
involving
constitute
known
un-
expression
degree;
see
23.
substitute in both
and
Assume
vx,
y
be found.
then by division the value o/y can
Rule.
=
Example.
Assume
Solve
a;^+ a^
2y^ 4:4:,2x'^"a:y+y^=lQ,
+
=
and substitute for 2^; thus
y=vXj
a;2(!+"?+
a;^(2
44,
2??'^)
=
1? +
-
1?^) IX
=
Therefore,by division,
l4-p+2g^
2-T
therefor"
th erefore
t!ieraforo
4(1 4-??
3""
"^
+
-
y
_
_
""
~
"j2
+
4
IQ
.
'
2z"^;
-11(2 -"?
1 5"
tions;
equa-
+
1S
=
"^-6tf+G^0.
0 ;
+
??2);
184
this
From
In
and
=
since ?/
equation put
same
vx,
have
we
y
=
+
r:r,
=
=
i?
the second
By
subtraction
the
2:^:^+4^2=88;
+
2x^
equationis
multiplythe first of
:
we
2^2
=
^2.
3
"
v
=
we
for
3
2otZ.
thus
2"2) 44 put 2 for v;
" 4.
have y
Again, in the
and
since
; thus x="^'2]
might proceed thus
given equationsby 2 ; thus
Or
shall obtain
quadraticequation we
equation ^-2(1+"
the
x='^2;
y
EQUATIONS
SIMULTANEOUS
y^
xy +
"
Sxt/+ Zy^
16.
=
72, therefore 2/^
=
=
Again, multiplythe second
equationby
2
24
"
xy.
and subtract
the first equation; thus
therefore
Sx'^-3xy=-l2;
9^=siyy^4u
Hence, by multiplication
a;-V (24-:py)(^y-4),
=
2a;V
or
28a;y
=
-
96.
-
By solvingthis quadratic we obtain xy
the former in the given equations
;
a;2+
22/2 36, 2a;2+ 2/2
=
=
%
=
we
268.
Solve
Assume
2a;2+ 3^y
y=iDX,
+
may
=
thus
+ 3z? + 2"2) 70, a;2(6
+ tJ-"?2)60.
a;2(2
=
=
Therefore
by
division
2 + 3g +
1?2
6 + r-Zj2
therefore
therefore
therefore
5(2+
3" +
yo
7
~60~6'
"72)7(6 + "-""2);
=
122?'+ 8")-32":0j
3"2
+
take the
2/^ 70, ^x^^xy-^j^-^Q.
and substitute for y\
2"-8
=
0.
stitute
Sub-
24.
find x^ and
can
6.
thus
\j\ Similarly
we
other value of xy, and then find ^ and y^.
Hence
or
186
SIMULTANEOUS
We
have
to find
now
+ y
x
these lead to
;c2+
and y from
x
6y
=
;i?=l
Solve
271.
EQUATIONS
x^
+
a^
+
a; +
And
a;=i5
Solve
x"y=^l.
shall obtain
?/=i4
i4,
^
x^-i-xy+ y^
xy
-
to solve the
now
d?==t3
273.
Solve
is,
that
is,
x^
x*
+
+
a?"-
therefore
6.
=
shall fmJ
we
or
2^^
=
242.
;
+
xy^ + y*= 121,
y* + xy{x^+ y-)+ xY=V2l
a^
"
x'^+
-
"3.
'
2
a?
succer.jively
242
-"
a^y-^ x^
Now
square
xy
2/= "2
x-y
that
obtain
we
"2,
a^-y^
^-=
_.-...
'
=
;c-y=2,
By^ division,
'
US.
;
"
in Art. 271,
or
=
7.
=
y^=-lZ,
proceedingas
y*
x'^"xy-"ry^ 'J.
By addition and subtraction
Then
+
equations
x^-^anj-"y'^=\9,
x'^+
"
"5.
xY'
^
19
y'
+
or
x*
=
"
=
x^
have
roots,
=
"
8ly
=
4l-40=l;
x'^+xy^y^- l9,
tliat is,
We
U+40
=
square
or
By^ division,
'
20.
by the second cast-; or
just exemplified.For
manner
=
1.
or
=
solved
?/="9,
thus finally
we
272.
xy
y^+2xy
y^-2xy
by extractingthe
then
=
=
be
they may be solved in the
deduce from them
we
can
4
j/
2/^ 41,
equationscan
These
simpleequations
y-"3;
a?"
4,
or
the
y
Ixy
=
4-
(i).
2 ;
?/2
y^^2xy
=
4 ;
+ 4
(2)
EXAMPLES.
therefore
X^
Substitnte from
2xhf^+ lQxy-^lQ
=
(3).
(2)and (3)in (1); thus
2x*y"+ 16ary+
that
y*
+
187
XXIX.
16
+
5;cV
is,
therefore
+ 4)+ a?2y2
(2;2?y
xy
=
20^
+
1 05 ;
=
+ 4^2/
ix^y^
121 ;
21,
=
this quadraticequation we
shall obtain xy
Z
with x
7. Take :n/^S, and from this combined
or
y=2,
If
shall obtain a:
3 or
"3.
take
or
we
we
-1, y=l
possible
7, we shall find that the values of x and y are imary
; see Art. 236.
From
=
"
"
=
=
"
Examples.
XXIX.
x^-xy-i-y*=2l.
1.
x-y=l,
2.
2x-5y
3.
x
4.
5{x^-y^) A{x^ + y^,
o.
x-y
6.
4;r-52^
7.
4a?
8.
(x-6y
9.
4a^
+
y
x^+xy
3,
=
x^+y''
7(x-y),
=
20.
=
=
66.
3y^+ 3x-4y=47.
+
{i/-5y+2xy-=60,
+
2xy
+
5y-4x
^+-"{4x y)
+
15
5xyy
15x^4y
=
3
4l,
^
4xy.
11.
Zx
12.
xy+2
13.
8(a:y-"-l) 3.3y, 4(a:y+ l) 33a?.
14.
xy
+
2y
=
=
9y,
xy
x
+
yy
=
+ 2=x.
=
=
=
=
l.
5
^'
10
8,
=
=
ifi
12
y^
y
2x'-i-xy 6y\
9y=12,
+
+
2x^-xy
l,
=
+
x'
3f
+
x
=
=
100:
=
ax=by.
4.c-y=4.
188
15.
18.
XXIX,
EXAMPLES.
1 ^=2,xy
+
^
+
=
ah.
^=1, ^!+ ^^i
19.
^2^.^,y^28, xy-y'^
20.
:c2+
21.
22.
2:cy-y^ 48.
2x2-icy;f:56,
a;2-2.ry=15,xy-ly'^^'J
23.
a;'+
24.
^2+^y_g^^2^21, a^-2y2
25.
a;'^
+
26.
"^
=
2.
4. ^^^3g^
2/2
^y^4g^
=
3^y
28,
=
3xy
54,
=
-'^
+
:^:y+
xy
+
^
+
2'
y
8.
=
4y^
=
4.
n5.
=
a:2.",2_Q/)
^"*
2^
=
,
.c-v
42/"
29.
+ y(:zr-y) 158,
^(:"+ 2/)
1ai{x+ y)=^1iy{x-y),
30.
:c2y(^^y^^g()^^V(2^-3y)
31.
2x'^-xy+ y'^ 2y,
=
=
+
a^.
x
+
+
4xy^by.
X^4-lfi
-0 h^ .
x
+y
"
=
x-y
2x^
,
'
a
y
33.
x'^-\-xya{a + h), a^+y^=a^
34.
;r'+
=
+
2.ry-g/2^^2+2a-l,
{a-\)x{x + y"i a{a+\)y{x-y\
=
35.
80.
=
"-y
=
2.
;i:3_2^_j52.
j)\
EXAMPLES.
36.
x
37.
a;2+
38.
x-y
39.
^
40.
x^
41.
+
y
ip3+ 2/3 i89.
9,
=
=
2/2 20,
asy-x-y=^2.
=
+ y
=
\y
x^-y^
=
ns\.
=
3,
ic"+
=
33.
y'
=
-^
^^=1,
;c"
a; +
y
a?*+ ^V
y'^ yjy
xy +
+
2 +
43.
ic2+ 2/2_i
44.
4.r2+
45.
a;2+ ;Fy
=
8a?4-3,
y' +
46.
lar^"a:y
=
2:c +
xy
47.
2x + y +
48.
18 +
49.
a^'-a;y=a(a;+l)+ 6+l,
50.
5,
"J(2x+
=
12
6^
+
=
x^z=a,
+
18,
+
6.
y'^^^y+
2.
By
=
4x^-ex=y^
4) 23,
=
+
3y.
{x-yf.
=
xy-y'^=ay
+
h.
=1.
"
y'^
by,
=
xy-z
=
=
ay + bx.
xyz^=c.
b,
+ ;2r)(y
+ ;??)h^,{z+ :cX^+ 2/)
a^ (?/
=
=
2y^
2^a;-4a:i/=16,
47/2"
56.
+
3.
=
xy
+ 2/)(a?
+ ^)
(a;
+
y
Q.
=
+ l)
4;i?y(a;y
;P2^
"
20.
2(j7+ 2/)2, Q-{x-y)
9(:c+ y)
63.
3y^
xy{xy+l)
2ary,
=
x^=ax
55.
=
y2+ 2(2;r+ y) 6,
62.
54.
;c2_y"4.^/(^_y2)
34^
=
^
aH
3;z?.
=
2/
a;2+ y2
e
y*=481.
+
3a?y
42.
=
189
XXIX.
;2rx
"
32ra;+ ;C2/=
-
3.C2/=15.
Q{a^+y^+z'^ \3{x+y-{-z)
=
5,
=
'^,
xy=z\
c\
19U
PROBLEMS.
with
added
number.
than
more
lead to Quadratic
unknown
one
Equations
quantity.
of two digits;
is a certain number
the sum
creased
inof
is
the
number
the digits equal to
squares
if thirty-six
bo
; and
by the product of its digits
the digitsare
reversed: find the
to the number
There
274.
of the
which
Problems
XXX
denote the digitin the tens'
digitin the units' place. Then the number
if the digitsbe reversed we obtain lOy +
Let
a;
place,and
is lOx
x.
+
the
y
and
y;
by
Therefore,
have
we
supposition,
a!^+
y^ xy+lOx
I0x+y
d6
+
=
10y +
obtain 9y=9x
(2)we
From
+
=
Substitute in
a?2+
y
(1).
a;
(2).
+
Z6; therefore y=a;
+
4.
(1),thus
+ 4)+
+ 4)2=i"
(a;
(a;
ic=
therefore
-
7a;+ 1 2
=
1 0.'c+ a; + 4 ;
0.
obtain a;=Z or 4;
this quadraticequationwe
From
the required number
and therefore y
7 or 8. Hence
satisfies
must be either 37 or 48 ; each of these numbers
the
all the conditions of
problem.
=
A man
to
starts from the foot of a mountain
275.
His
of
rate
walk
to its summit.
walking during the
is
half
the
half
mile
second
of
distance
a
per hour less than
his rate during the first half,and he reaches the summit
in
form
5^ hour^. He descends in 3| hours by walking at a unithan liis rate
rate,which is one mile per hour more
Find the distance to
during the first half of the ascent.
the summit, and his rates of
Let
suppose
2a! denote
that
walking.
of miles to the summit, and
the first half ol the ascent the maxi
the niunbcr
during
191
PROBLEMS.
walked
y
miles
per
Then
hour.
took
he
-
hours
for the
y
for the second.
hours
ascent, and
firsl half of the
Tlierefore,by substitution,
thercfore
15
(y +
therefore
1 X4y
28y-
-
-
1)
=
44y (2y
89?/ + 1 5
-
1 );
0.
^
5
quadratic equation
this
J%)m
we
5
The
value
"
is
because
inapplicable,
obtain
y
Z
=
or
"
.
by suppositiony
is
28
15
1
^ea.ier
than
-.
Therefore
y=3;
and
then
^
*lie whole
"
.
i
i
that
=
distance
to the
summit
is 15 milea.
so
192
EXAMPLES.
XXX,
Examples.
XXX.
is 170, and
The sum
of the squares of two numbers
the difference of their squares is 72 : find the numbers.
1.
product of
The
2.
is twice their difference
The
3.
:
is 640
:
theur
\k 192, and the
two numbers
find the numbers.
product of
their squares
4.
numbers
is 1 08, and
find the numbers.
two
sum
sum
of
is 128, and the differThe product of two numbers
ence
of their squares is 192 : find the numbers.
and the
6.
product of
The
5.
two
of their squares
sum
numbers
is 325
:
is 6 times their sum,
find the numbers.
ence,
The product of two numbei*s is 60 times their differof their squares is 244 : find the numbers.
and the sum
The
numbers
is 6 times their difference,
their product exceeds their sum
ber*.
by 23 : find the num-
7.
and
of two
sum
Find
two numbers
such that twice the first with
three times the second may make 60, and twice the square
of the first mth
three times the square of the second may
8.
make
840.
9.
Find two numbers
such that their difference multiplied
into the difference of their squares shall make
32,
and their
shall make
multipliedinto the
sum
sum
of their squares
272.
10.
Find
two
11.
Find
two
such that their difference added
numbers
to the difference of their squares
make
14, and their
may
added to the sum
of their squares may make
sum
26.
numbers
to their sum,
and their
equal to 12.
squares
such that their productis equal
added
of their
to the sum
sum
194
EXAMPLES.
whole
days,
time,
have
received
those
proceeds
When
they
town.
miles
108
at the
run
and
towns
4
the
same
uniform
a
of
rate
of the
two
In
A,
hour,
river.
a
the
and
Q,
design
the
nm
between
miles
18
from
man
the
walks
stream
in
than
more
the
apart
^
-5
to
and
the
on
in
walking
first
half
at
being with him, ho
going, and accomplishes
.3^hours.
in
at
the
Find
his
mile
two
^-inning post
his
2
of
rates
in the
and
Q,
as
2
before
miles
per
arrives
then
A
Find
wluit
at
first heat.
-B, set
time;
same
In
course.
minutes
speed
his as much
; and
before B.
minutes
two
direction
a
increases
A
heat
through
round
the
travellers,A
to pass
same
situated
A
race
reaches
ran
man
Two
distance
returning he
a
run
winning post
24.
other
has
train
one
towns,
the
goes
first half of the distance
diminishes
B
each
rate
B
second
the
and
two
trains.
before, but
B
first heat
the
from
days^
rowmg.
and
A
23.
of
towards
that
the
towns
hour
per
whole
distance
walking and
daVvS they would
number
time
rate
the
other, and that if they continue
they will finish the journey in 9 and
miles
1^
worked
the
In
as
six
it is found
meet
are
half.
the
in
at
hours, by rowing the
second
P
B
bank
the
at
same
had
B
the
at the
rates
and
same
rows
Find
start
rate
same
the
A
22.
the
alike.
respectively. Find
hours
16
absent
day.
than
more
If
14",
per
each
and
"2.
been
had
A
trains
Two
21.
to
and
only
exactly
each
was
paid
what
and
receive(i
XXX.
and
A.
A
B
out
starts
starts
When
A
from
two
from
Q and
overtook
places,
with
P
from
B
the
travels
it
was
together travelled thirty miles, that
had passed through Q four hours
A
before, and that B, :it
hours' journey distant from
nine
his rate of travelling,was
P and
between
Find the distance
P.
Q.
found
that
they
had
195
INVOLUTION.
XXXI.
Involution,
We. have alreadydefined a power to be the product
of two or more
equalfactors,and we have explained
the notation for denotingpowers; see Arts. 15, 16, 17. The
of obtaining
powers is called Involution; so that
?rocess
nvolution is only a particular
of Multiplication,
but
case
it is a particular
which
often that it is
occurs
so
case
convenient to devote a Chapterto it. The student will find
that he is alreadyfamiliar with some
of the results which
shall have to notice,
and that the whole of the present
we
Chapter follows immediatelyfrom the elementarylaws of
Algebra.
276.
277. Any even power
of a negativequantityis
is negative.
and any odd power
tive,
posi-
This is a
of the Rule ofSigns.Thus,
"a^a^x
"a=
"a^;
"ax
and
In
tho
so
on.
"ax"ax"ax"a="a^x"a=a^;
when we use the words give the pro;"er
following
Articles,
that the sign is to be determined by tho
sign,we mean
rule of the presentArticle. (See Art 38.)
for
simpleconsequence
example,"ax
-a=a\ "ax
278.
of a power.
a power
obtaining
denotingthe powers for the new
proper sign to the result.
Rule for
the numbers
and give the
Thus, for example,{a^f a'; {-a^f=-a^;
of
This is a simpleconsequence
{"a*f= "a^\
=
powers
which is demonstrated
=a^xa^xa^
(^2)3
For
in Art. 59.
=
02+2+2 =a^
The Rule of the present Article leads
that which we shall now
give.
=
Multiply
exponent,
{a^f^a^"";
the law
of
example,
a\
immediatelyto
Rule for
obtaining
any power of a simpleintegral
pression
expression.Multiply the index of every factor in the exby the n umber denoting the power, and give the
ipToper sign to the result.
279.
13-2
IDG
INVOLUTION,
Thus, for example,
{-a'W(ff= -d''"'c''; (2a62c3)6=2"a"2.iV8=64a"6i"ci
Rule
280.
for
obtainingany
and
power
denominator
Art.
145.
of
fraction. Raise
to that power
and
j
both the numerator
give the proper sign to the result.
This follows from
f^^- ^
V"V "'
( ^-
example,
/2a^*_
V 36 ;
_^'
"
~
281.
For
'
^'
^V
\
a
'
2V
16a8
_
"
examples of Involution in the case
already been given.
expressions have
Some
binomial
Arts. 82 and 8S.
"
816*
3^"*
of
See
Thus
b\
{a+ bY
=
a:'+ 2ab +
(a+ bf
=
a^ + 'Sa'^b
+ 3ab^
S^.
+
for exercise obtain the fourth,
fifth
The student
may
It will be found that
and sixth powers of a+b.
(a + by -=a*
{a+ bf
(a 4- 6/
=
=
4a^b
+
6a:^b^+ 4ab^
+
6S
+
+
6ab^
a^ + 6a5" + 15a^6" + 2QaW
+
15a="* + Qah^
In like manner
+
results may
following
the
{a-bf=a^-2db
=
a'
3"26
+
Sa^J
(a by
=
a*- 4a^b
+
Ga'^b'^
-4aP
{a of
=
a^-
+
lOa^ft^ IQa-b^ + 5ah*
+
Ua^b^-20aW
-
-
{a-by
=
-
5a^b
a^-ea^b
+
6*.
be obtained
:
W,
+
(a bf
-
6*.
a^ + 5a'^b4- 1 Oa^b^-+ 1 Oa^^
-
b^,
+
b\
-
+
-
b^.
15a-b*-6ab^-i-lA
Thus
where
of a"
in the results obtained for the powers
"
of b occurs, the negativesign is prefixed;
any odd power
of a
be inimediatelj
and thus any power
b can
of a + b, by changing tlie
from
the same
deduced
power
Bigns of the terms which involve the odd powers of.b.
"
197
INVOLUTION.
The student will see hereafter that,by the aid
282.
called the Binomial
of a theorem
Theorem, any power
of a binomial
expressioncan be obtained without the
laboui' of actual multiplication.
The
283.
in tlie way
we
formulae given in Article 281 maybe
have alreadyexplainedin Art
84.
of 1x
pose, for example,we requirethe fourth power
In the formula for {a"h)* put 2x for a, and 3y for b;
that we can obtain required
It will b" easily
seen
results in Involution by different processes.
Suppose, for
sixth
of
a-^b.
We
example, that we require the
power
284.
this by repeated multiplication
by a + b. Or
we
may firstfind the cube of a + b,and then the square of
this result ; since the square of (a+ bf is {a + b)^. Or we
may first find the square of a + b,and then the cube of this
obtain
It is found by observation that the square of any
multinomial expression may bo obtained by either of two
rules.
Take, for example, (a 4- 6 + c + df. It will be found
that this
286.
=
a^ + l^ + c^ + d^
+
2ab + 2ac-h2ad
+
2hc
+
2bd+2cd;
followingrule ; the square
i"f
expression co7i,nsts of the square
of any multinomial
each term, togetherwith twice the product of every pair
of terms.
be obtained
and this may
Again, we
(a +
6+
may
by
the
put the result
in this form
c+cO''
=""+
2"i ("+
c +
+
"/)
62 + 2" (c+
+
"/)
c2 + 2"'cf+ fi?^
rule ; the squnre
be obtained by the following
qf
of any 7nultinom.ial expression consists of the square
each teriUy togetherwith twice the product of each term
and
by
this may
t/ie sum
cfall tlie terms
whi^Jt follow it.
Examples.
XXXI.
Find
1.
{2a^2^)\
2.
{-ZaWf.
4.
{-2xhfz^\
/9;".2\2
3.
(^3^).
EXAMPLES.
{^
XX
by.
199
XL
b.
{a-b)\
ia + bf{a-b)\
10.
{l-x)\
(2 + :c)".
12.
(3-2^)".
{l+x)\
14.
(;c-2)*.
(2^
+
3)\
16.
{ax
{ax
+
+ {ax- by)*.
52/)*
18.
(1 + x)'
(1 + iB)4(l ;i;)*.
20.
(1 +
{l-x-hx^f.
22.
(l+.r-ar2)2.
(1 +3^
24.
(1
+
-
(2 +
+
3a? +
2a?2)2,
4^2)2+ {2-3x
+
h'i/f-\-{ax-tyyf.
-
X
-"-
3x
(l-^
{l-i-x-a^^K
29.
(H-3a;
3^3.
(2 +
3a? +
4a?2)3 (2
(l-a?
b
+
{a +
b
+
-
c
+
c-hdf
+
{a-b
+
c-d)^.
+
c-d)\
(l+3a?+3a?2+a;")^
(1 +
4a? +
ea;''+ 4ar" +
il-xy{l+x
+
(l +
33.
(rf-{a-b
3;e2)i.
;c2)s
+
+
2a^.
4a?*)'.
3a? +
a?24-a:')2.
+
+
{a
-
a;)".
4x^f.
+
27.
3a? +
-
a?2)2.
+
{l+x+x-'f.
(1-
(1
-
37.
2a? +
3"2+4a^\
(l-6a?+12a?=-8"")".
x*)K
ay')\
40.
il-x+X'f{l-^-x-i-jrf.
200
EVOLUTION.
XXXII.
Evolution.
387. Evolution is the inverse of Involution; so that
Evolution is the method
of findingany proposed root of
or
a given number
expression. It is usual to employ the
with the
word
and its derivatives in connexion
extract
"word roo#; thus, for example, to extract
the square
root
the same
means
thing as to find the square root.
In the present
shall
beginby statingthree
of the Ruh
shall then
we
of Sigiis,
simple consequences
consider in succession the extraction of the roots of simple
the extraction of the square root of compound
expressions,
expressionsand numbers, and the extraction of the cube
and numbers.
root of compound expressions
Any
288.
either
even
positiveor
Chapter we
of
negative.
root
Thus,for example,a
the square
+ a
or
a
=
a^,and
root of a^ is either
a
or
fore
there-
"a=(f\
"ax
"a,
that
Any odd
the quantity.
is,eithe?
of
root
a
quantity has the
sign
same
Thus, for example,the cube root of a' is a, and the
root of -a'
290.
h"
"a.
289.
as
x
positivequ/intitymay
a
is
There
cube
-a.
can
be
no
even
root
of a negativequantity.
Thus,for example, there can be no square root of -a^;
for if any quantitybe multipliedby itself the result is
a positivequantity.
fact that there can be no even
root of a
such
quantityis sometimes
expressedby calling
The
itnpossible
quantity or
an
negative
a
root
an
imaginary quantity.
Rule
for obtainingany root of a simple integral
the index
expression. Divide
of every factor in the
expression hy the number
denoting the root^ and givt
291.
the proper
sign
to the result.
202
EVOLUTION.
If there
as
has
go
t"M-ni in the
fcae
The
divide
sliould
of
."uni
Then
root.
2{a
b)
+
must
process
the
should
we
its square,
from
the
a;
subtracted
been
aheady
we
terms
more
formerly with
did
we
were
remainder
for
and
the
a
new
new
continued
be
Examples.
4^
12x1/ + 9y^
4^
4j^
-
bx)
-
2Qa^
+
37^" -ZOx
+
9
12^2
+
9
+
9
-20^+25^2
Aai^-lOx-^Z)
_
30^
12a;2-30A-
-4a?3?/2+ 4.rV
that is,a^
wnth
+
a+"
^ab-k-h'^
proposed expression,
by 2(rt+ d)for a now
subtrahend
we
midtiply
term.
term, by the new
until the required root
is found.
295.
proceed
203
EVOLUTION.
of"
2:c'+
2a^
+
2x^)40^
4x^-2j;
J
-10a;3
-4:0:^-100:^
-40?*-
2;p3+
+4x+l
8a;3^.4;p2
2x^-4x^
4a;"-44?-lJ
-
+
4x+l
2a^-Ax^-"Ax
-
already observed
It has been
296.
+4;c+l
+
\
that all
even
roots
sign; see Art. 288. Thus the square
"a"b.
In fact,in
root of d' + 2ah-^b'' is eittier
a + 6 or
root of a^ + 2ab + Ij^,
the process of extractingthe square
we
begin by extractingthe square root of a-; and this
admit
of
a
double
take the latter,and continue
the operationas before,we
shall arrive at the result
a"b.
A
other
holds in every
similar remark
case.
for
last
in
the
of
worked
those
out
Art.
295.
Take,
example,
be either
may
a
or
If
"a.
we
"
Here
may
we
begin by extractingthe
square root of x^: this
take the latter,
and continue
shall airive at the result
be either a^ or "a^.
If we
the operation
as
before,we
-x^-2a^-i-2x
+
l.
fourth root of an expressionmay be foand
by extractingthe square root of the squnre root ; similarly
be found, by extractingthe square
the eighth root may
297.
The
root of the fourth root ; and
298.
Bquare
cannot
cannot
In Arithmetic
of every
find the square
find the square
root
so
on.
that we
cannot find the
know
number
exactly; for example, we
we
root
root
exactly. In Algebra we
every proposedexpression
of 2
of
204
EVOLUTION.
sometimes
exactly. We
find sucli an example as the following
proposed; find four terms of the square root of 1 2;r.
-
a}
5.r*
Thus
have
we
findingfour
terms
/
know
that
remamder
"
"
~
"o
a?
""
T
of the square rc"ot of I -2;2;; and
of
^x*
x^
x*- a^^
"^
=1-2^-^
"^
^)
(^1-^-2
"
'
'*'^*'"''
so
we
^
T
2
*
4
of the square root of
precedinginvestigation
Algebraicalexpression will enable us to demonstrate
rule which is given in Arithmetic
for the extraction of
299.
an
a
the
the square
The
root of
a
number.
square root of 100 is 10, the square root of 10000
is 100,the square root of 1000000
is 1000, and so on ; hence
less than lOt)
it follows that,the square
of a number
root
consist of only one
must
figure,the square root of a
The
205
EVOLUTION.
between
of two placesof figures,
100 and 10000
of
between
and 1000000
of three placesof
10000
a number
and so on.
If then a point be placed over
figures,
every
second figurein any number, beginningwith the figurein
of pointswill shew the number
the units' place,
the number
of figuresin the square
root.
Thus, for example, the
and the square
square root of 4356 consists of two figures,
root of 6il^2i consists of three figures.
number
300.
Suppose the
square root of 3249
required.
Point tne number
accordingto the
rule ; thus it appears
that the root
must
consist of two placesof figures.
Let a + b denote the root,where a is 100
the value of the figurein the tens'
place,and h of that in the units' place.
Then
must
a
be the
of
+ 7
324"(^50
2500
+
7J749
749
"
greatest midtiple
ten,which has its square
to be 60.
Subtract a\ that
less than 3200 ; this is found
is,the square of 50, from the
remainder
Divide
is 749.
this remainder
given number, and the
by 2a, that is,by 100, and the quotient
which is the value of b. Then
that is,107
(2a4-6)",
749, is the number
no
remainder,we
square
is 7,
x
7 or
is now
subtracted ; and as there
conclude that 50 + 7 or 57 is the required
to be
root.
It is stated above that a is the greatestmultiple
of ten
which has its square less than 3200.
For a evidently
canbe a greater multipleof ten.
ii-;"t
If possible,
it
suppose
less than this,say x; then since
is in the
b in the units' place,a? + 6 is less
X
than a ; therefore the square of x + b is less than a?,and
consequentlyx + b\s, less than the true square root.
to be
some
multipleof ten
tens' place,and
let a reIf the root consist of three placesof figures,
present
the hundreds, and b the tens; then hfving obtained
and
and
b as
tens
a
before, let the hundreds
together be considered as a new value of a, and find a new
value of b for the units.
206
301.
EVOLUTION.
bo omitted
for the sake of
Tlie cyphers may
tlie followiDg
rule may be obtained from the
and
brevity,
process.
Point every second figure,beginning
with that in the unit^ place,and thus
into periods.
divide the whole number
324"
(^57
25
the greatestnumber
Find
whose
107 ) 749
square
is contained in the firstperiod; this
*j^q
is the first
figurein the root; subtract its
square from the firstperiod,and to the
the next period. Divide this
remainder
bring down
omittingthe lastfigure,by twice the part of the
quantitij,
the result to the root and
annex
root alreadyfound, and
also to the divisor; then multiply the divisor as it note
stands by the part ofthe root last obtained for the subtrahend.
there
be
more
If
periodsto be broughtdown, the
operationmust
302.
be
repeated.
Examples.
Extract the square root of 132496,and of 5322249.
In the firstexample,after the firstfigureof the root is
found and we have brought down
the remainder,we ha\e
to the rule we
divide 42 by 6 to give the
424 ; according
in the root : thus apparently7 is the next
next figure
obtain the product
67 by 7 we
figure.But on multiplying
tliat 7 is too
is
This
shews
which
than
424.
469,
greater
of the root,and we accordingly
largefor the second figure
in this
try 6, which succeeds. We are liable occasionally
the
at
to try too largea figure,
especially
early
manner,
of
of
the
root.
extraction
a
stag^
square
SO"?
EVOLUTION.
Tn the
example, the
cypher in the
second
of the
occurrence
student
should notice tb"
root.
The rule for extractingthe square
root of a
must
serve,
obdecimal
follows from the preceding rule. We
however, that if any decimal be squared there will
of decimal placesin the result,and
number
be an even
be an
therefore
there cannot
exact
root of any
square
has
odd
of
decimal which in its simpleststate
number
an
decimal places.
303.
of the square
of 32'49 is one-tenth
also
the square
So
root of 100 X 32-49 ; that is of 3249.
of the square root of
of '003249, is one-thousandth
root
Thus
is
of
deduce
3249.
lOUOOOOx
we
-003249, that
may
this rule for extracting
the square root of a decimal.
Put
a point over
everi/ second figure^heginning with that in
The
square
root
the uniis^place and
the left of it; then
continuing both to the right and to
proceed as in the extraction of the
root' of integers,and mark
decimal
off as many
square
qf periods in the deri^
placesin the result as the number
mal
part of the proposed number.
In
this rule the student
should pay particular
attention to the words
with that in the units' place.
heginning
In the extraction of the square root of an integer,
if there is stilla remainder
have arrived at the
after we
figurein the units' place of the root, it indicates that tlie
We may
proposed number has not an exact square root.
if we pleaseproceed with the approximationto any desired
extent, by supposing a decimal point at the end of the
304.
of cyproposed number, and annexing any even number
pherSjand continuing the operation. We thus obtain a
decimal part to be added
to the integral part already
foimd.
if
Similarly,
root,we
may
annex
a
decimal
number
cyphers,and
mati:"n to any desired
extent
has
no
exact
proceedwith the
square
approxi-
208
EVOLUTION.
305.
of '4 to
Tlic
seven
is the extraction
following
decimal places:
of the square
root
0-4606...
(^-6324555
36
123^ 400
3G9
1262
J
3100
2524
12644; 57600
50576
126485; 702400
632425
1264905; G997500
6324525
12649105; G7297500
63245525
4051975
We
306.
cube root of
proceed to the method
compomid expression.
now
a
of
the
extracting
The cube root of a^ + 3a-'j+ 3a"2 + 53 jg a+h\
and we
shall be led to a generalrule for the extraction of the cube
root of any compound expression
by observingthe manner
in which a + h may be derived from a^ + M'^b + ^db^ + h^.
Arrange the
terms
acdimensions
cording to the
the
of one
letter "; then
first term is a',and its cube
root is a, which
is the first
terra of the required root.
its cube, that is
Subtract
a',from
the whole
a^
+
M^h
+
Zdb'^+ l"^{a
+
b
^s
Sa^JZaP'b+
3ab^+l^
^^^j^^ ^j^2
expression,and bring
do\vn
^
yt
the
re-
210
EVOLUTION,
In continuingthe operationwe must add such a
308.
the
firstcolumn, as to obtain there three times the
tenu to
This is conveniently
part of the root already/found.
effected
column
have
thus; we
already in the first
d and
3a + 5; place 2" below
add;
3a
obtain 3a + 35, which
is three times
a + ", that
is,three times the part of the root
alreadyfound. Moreover,we must add such a
to the second
term
column, as to obtain there
thus
+
we
times tJie square
of the part of the
found. This is convenientlyeflFectedthus ; we
three
"
1
25)
3a
root
have
+
35
already
already
5)5,
and below
in the second colunm (3a+
that 3a2 + 3a5 + 52; place " below, and
add the expressionsin the three lines;
obtain 3a^ + 6a5 + 352,which
is
tlius we
(3a+ 5)5
3^2
+
3^
+
5^3
times (a + 5)^ tliat is three times
the square of the part of the root already
found.
three
309.
Example.
E3^ract
8a;" 36;"* + 102;c*
-
-
3^3
1 Tla:^+ 204^
+
144;" + 64.
-
5i;z;=-36aj+ 16
8.r"-36;2:"+102;i;*-l7l^+ 204:c2_i44^
+
+ "
(j4(^2.r2-3J;
8.f"
36^
+
-36a?"+
102;c*
-
1 71^
+
204a?"
-
144:b + 64
54.^'*- 21 a^
48^
-
g^j ^ 351
the cube root of
12;r*-36ar"
-
+
144.^3+ 204:^2- 1 44^;
4"i?*- 144a:*+ 2()4;j;2 144^
_
+
64
+
g4
211
EVOLUTION.
The cube root of 8.r*is 2a^,which will be the firstterm
of the requiredroot ; put 8a^ under the given expression
in the third column and subtract it. Put three times 2i(^
in
in the first column, and three times the square of 2^
the second column; that is,put Qa^ in the first column,
Divide "ZQx^
aiid 12:c* in the second column.
by 12.r^,
and thus obtain the quotient Zx, which will be the second
term of the root; placethis term in the firstcolumn, and
in the first column, that is
multiplythe expressionnow
"
Gx^
Zx, by "Zx; placethe productunder the expression
in the second column, and add it to that expression
; thus
this by
obtain 12a:* ISx^ + ^x^ ; multiply
we
3:r,and place
and subtract. Thus
the productin the third column
we
have a remainder in the third column, and the part of
"We must
the root alreadyfound is 2x^"2x.
now
adjust
the first and second columns in the manner
explainedin
Art. 308. We put twice
3^',that is Qx, in the firstcolunm,
and add the two lines;thus we
obtain Qa^
dx, which is
three times the part of the root alreadyfound.
We put
tlie square of
in the second column, and
3a',that is dx"^,
add the last three lines in this column ; thus we
obtain
\2a^"'ZQa? + '2.*!x^,
which is three tunes the square of the
part of the root alreadyfound.
"
"
"
-
"
"
"
Now
divide the remainder
in the third column by the
expressionjust obtained,and wo arrive at 4 for the last
term
of the root, and witii this we
proceed as before.
this
Place
in the first column, and multiplythe
term
in
the first column, that is Qx-"dx+A,
expressionnow
by 4 ; place the product under the expression in the
second column, and add it to that expression;thus we
obtain l^x^-ZQofi-hblx'^-ZGx+lQ', multiply this by 4
and placethe product in the third column
and subtract
As there is now
conclude that 2j^"Zx+
4
no remainder
we
is the requiredcube root.
The preceding investigation
of the cube root of
an
Algebraical
expressionwill suggest a method for the
extraction of the cube root of any number.
310.
The cube root of 1000 is 10, the cubo root of 1000000 is
the cube root of
100, and so on; hence it follows thatf,14"2
212
a
EVOLUTION.
number
less than
1000
the cube
two
consist of only one figure,
between
and 1000000
1000
of
If then a pomt be placed
on.
must
root of a number
and so
placesof figures,
every third figurein any number, beginningwitli the
of pointswill shew
figure in the units' place,the number
the number
of figuresin the cube root. Thus, for example,
over
of 405224
consists of two figures,
the cube root^
and
cube root of 1281^904 consists of three figures.
Suppose the
cube root of 274G25
180 + 5
10800
the
required.
274625 C60 + 5
925
216000
11725
68626
58G25'
Point the number
accordingto the rule ; thus it appears
that the root must consist of two places of figures.Let
the root, where a is the value of the figurein
a + h denote
the tens' place,and h of that in the units' place. Then
a
be the greatest multipleof ten which has its cube
must
Place the cube
less than 274000 ; this is found to be 60.
of 60, that is 216000, in the third column
under
the given
Place three times GO, that is 180,
number
and subtract.
in the first column, and three times the square
of 60, that
is 10800, in the second column.
Divide the remainder
in
in the second
the third column
column,
by the number
that is,divide 58625
by 10800; we thus obtain 5, which
is the value of b.
Add
5 to the first column, and multiply
thus formed by 5, that is,multiply185 by 5; we
the sum
and
thus obtain 925, which we place in the second column
obtain
add to the number
11725;
alreadythere. Thus we
the
third
column,
multiplythis by 5, place the product in
therefore
is
remainder
65 is
The
and subtract.
zero, and
the requiredcube root.
The
cyphers may
and the probe ^emitted for brevity,
cess
will stand thus;
185
108
925
11725
274625 (65
216
68G25
68625
213
EVOLUTION.
Exampic.
311.
1271
14
Extract the cube root of 109215352.
10921535^(^478
48
f
889 1
1418
64
6689 "
45215
49
39823
6627
5392352
5392352
11344
674044
of the root,namely
obtainingthe first two figures
47, we adjustthe first and second columns in the manner
explainedin Art. 308. We place twice 7 under the first
column, and add the two lines,
giving141 ; and we place
tlie square of 7 under the second column, and add the last
three lines,
giving6627. Then the operationis continued
After
as
before.
The
In the
course
imagined that
cube root is 478.
working this example we might have
of the root would be 8 or
the second figure
of
trial it will be found that these numbers
of the square root, we are
too large. As in the case
are
liable occasionally
at the
to try too largea figure,
especially
even
9 ; but
on
earlystagesof
1
10)
6153
operation.
Extract the cribe root of 8653002877.
Example.
312.
605
the
"653002877 (,2053
1200
3025)
123025
(
25^
126075
18459
8
653002
615125
37877877
37877877
12625959
In this example the student
of the cypher in the root.
fence
should
notice the
occuiv
214
EVOLUTION.
If the root have
of decim:;! places,
any number
the cube will have thrice as rufiny ; and therefore tiic number
of decimal places in a decimal
number, which is a
and
in its simpleststate,will necessarily
be a
perfectcube,
and
of
the
decimal
number
of
placesin the
multiple three,
313.
cube root will necessarily
be a third of that number.
Hence
be a decimal,we placea point
if the given cube number
the figure in the units' place,and over
over
every third
figureto the rightand to the left of it,and proceed as in
the extraction of the cube root of an integer; then the
of
number
pointsin the decimal
will indicate the number
number
cube root.
part of the proposed
of decimal
Example. Extract the cube
314.
256
1
1456
y
16
7236
root of 14102'327296.
l4lO^-327296(,2416
12
I
placesin the
8
J
278327
1728
173521
7211
173521
r
104806296
104806296
174243
43416
17467716
315.
cube
If any number,
root,
we
may
annex
integralor decimal, has no exact
cyphers,and proceed with thr
approximationto the cube
root
to any desired extent.
The following
is the extraction of the cube root of '4 to
four decimal places
:
EXAMPLES.
Examples.
Find
XXXII.
the value of
I.
^/(9a*6^).
2.
4/(8a""').
4.
4/(16a^6"0-
5.
4/(-a"6V).
Find
the square
roots
16""
13.
36^-"+12:c34.i.
+
25a"4-20a^
+
4y
25a"
+
4c"'
+
of the
40"zd + 25fe2.
II.
*"''"
215
XXXIL
20flc
3.
A"'(-64a"5").
followingexpressions:
12.
49a*-84a26
14.
64a"
+
+
366".
48a5c+96*c*
9^7*-240^^+16
4;B2-12a?+9
'
216
EXAMPLES.
XXXII.
17.
x^ + ^x^ + ^x'^+ ^x+l.
19.
a?*+ 6a^ + 25a;'+48.r
21.
l-Ax-^lOa^-lla^
22.
4i^r' 4a;^-7^4-4;"' + 4.
23.
a?* 2a;r3+ ba^a^
24.
^
25.
a:^
26.
a^^- 4ax^
27.
l-2.^
18.
20.
64.
+
+
l-2a;
6;"2_4^^4^
+
a?*-4a?3 + 8;r+4.
^ai^,
"
-
-
-
2"a?3+ {a"+
1 2ar"+
+
4a^x + 4a\
"
2W')a^ ^aVx
60a;^-1 60a?'+ 240.?:=- 192:1?+ 64.
4a*;i?
+ a*.
1 0"';i;"
+
-
h\
+
-
3^-4^
5:c*-4^
+
+
3a?^-2:"'+ ;i?".
4x*
X
16^
gy'
exy
16a?*
9y*
;^
15yz
16z*
5z^
Ihs?'
^*
:
following
expressions
Find
the fourth roots of the
29.
l+4a;
30.
+ 216a?V-216a?y3+ 81y4^
16a?*-96a?3y
31.
l-4a?+10a72-16a?3+19a?^-16a?^+10a?"-4a?^
32.
+
6a?'+ 4a?'+ a;*.
+ (a2+
{a?* 2(a+ 6)a;3
Find
-
the
eighthroots
4a6
of the
+
+
a?".
+ a^ftsjs.
V^o^ 2ah{a+ ")a?
-
following
:
expressions
33.
x^ + 8a^ + 28afi+ 56a^ + lOx* + 56a;3+ 28a?a+ So? + 1.
34.
{x*-2a^y+ 3a^^-2xy^
Find
the square
35.
1156.
36.
2025.
39.
7569.
40.
9801.
43.
165649.
46.
-835396.
roots of the
44.
47.
+
y*}\
numbers
following
37.
41.
3080-25.
1522766.
3721.
15129.
45.
48.
:
38.
42.
5184.
103041.
41-2164.
29376400.
218
INDICES.
Indices.
XXXIII.
have defined an index or exponent in Art. 16,
index has hitherto
an
atid,accordingto that definition,
been
number.
We
w
hole
about
a positive
are
now
always
the definition of an
to extend
the
index,by explaining
of
and
of
fractional
indices
negativeindices.
meaning
We
316.
and
a"*"*"".
If
317.
a"*
X
a"
=
are
n
m
any
positivewhole
nwnbers
truth of this statement
has alreadybeen shewn
in Art. 59, but it is convenient to repeat the demonstration
here.
The
arz=a-Kaxax
a" =a
X
X
a
a X
to
m
to
w
factors,
by
factors,
by
Art. 16,
Art. 16 ;
therefore
a"*x""=axaxax
""+",by
=
a'^xa'^x
so
...
m
+ n
factore
Art. 16.
whole number,
positive
aP=o'""^" x aP
a"*+*+';
ifjo is also
In like manner,
and
to
...xaxaxax
a
=
on.
318.
If
greaterthan
and
m
n,
we
n
whole numbers, and
positive
by Art. 317
are
have
"""""
X
a*
=
"""""+"
=
m
a"";
"=""-".
therefore
a
This also has been
As
not yet been
319.
tioiii
we
alreadyshewn
; see
Art. 72.
negativeindices have
libertyto givewhat defiiii-
fractional indices and
we
defined,
please to them
are
;
at
and
it is found
convenient
U)
219
INDICES.
givesuch definitions
relation a"
he.
may
x
a"
to them
the
trite,whatever
"""*""always
=
will make
as
important
and
m
n
example; requiredthe meaning of a*.
a^
Thus a 5
to have a^xa^
a.
we
are
By supposition
be such a number
that if it be multipliedby itself
must
the result is a; and the square
root of a is by definition
For
=
such
number; therefore a^
a
square
root of o, that
=
of
Again ; requiredthe meaning
Hence,
of a, that is
root
=a
before, a^
as
i+i+i
i
y.a
"Ka
a^.
to have
are
i
i
a
equivalentto the
s/a.
is,a'^
we
By supposition
be
must
=
,
=a
be
must
=a.
equivalent to the cube
a^= ^a.
Again ; requiredthe meaning
I
a
By supposition,
xcr
xa
I
therefore
2
I
3
a
=
of a*.
=a';
xa
"i/a\
These examples would enable the student to understand
what is meant
by any fractional exponent ; but we
will give the definition in generalsymbols in the next two
Articles.
1
320.
Required the meaning
of
a"
tchere
n
is any
whole number.
positive
By supposition,
111
1
h
1
a*xa*xa*x
"
...to
n
factors
=
a"
"
"
_^i_^j
"
1
therefore a* must
that
is,
be
equivalentto the n"*
a*=
;ya.
root
of a,
220
INDICES.
Required the meaning qf
positivewhole numbers.
321.
any
where
a"
o,nd
ra
art
n
By supposition,
xa"
a"
a"x
m
m
m
i"
X
...
therefore "" must
that
to
be
is,
factors
?i
a''
=
m
"
"
="'";
the n^^ root of t"",
equivalentto
a"
m
V^"*-
=
m
the n*^ root of the
that is,in a fractional index the numerator
and the denominator
a root.
Hence
a"
have thus
We
322.
index, whether whole or
to
meaning
For
fractional
;
a?
x
a~^
a^-^
=
a~^
therefore
givethe
now
=
"
=
-s
.
generalsymbols.
the meaning of a""/
whole or fractional.
a"*
X
m,
"-"
m
may
suppose
and' then, by what has gone
X
a"
=
a*" x
therefore
"
a-
a~''=
be, we
we
before,
I
n
therefore
where
are
n
\"f
any
to have
"""-".
=
and
""" J
may
positiveand
we
Therefore
to
or
definition in
whatever
By supposition,
a*"""
positive
assigna
",
=
Required
positivenumber
Now
power
of ar\
a^
=
a^
323.
a
to any
it remains
example,requiredthe meaning
will
denotes
assigned a meaning
of a\
power
negativeindices.
By supposition,
We
m**
means
"
.
greater than
have
a"*"*
=
-^
"
w,
221
INDICES.
In order
this
to express
reciprocal. One
word
of another
will define tho
we
quantity is said to be the
the product of the two is
when
unity; thus,for example,x
Hence
words
in
is the reciprocalof
reciprocal
equal to
-
.
is the reciprocal
of a" ; or we
may
result symbolically
in any of the following
ways,
a~"
a-"=\,
a"=^,
a"
a"
a^xa"
It will follow from
given to
than n,
m
negativeindex
a
as
well
as
n
we
less than
;
when
meaning
that "'"
"
a"
has been
a"""" when
~
is gi'eaterthan
m
which
n,
is less
m
For
suppose
have
a"* -^ a"
=
=
"-
-"
=
-
then
"-("-"")
a"*~*.
=
""
a"
a"
Suppose
the
this
1.
=
'
324.
put
is
obviously
and
"""-"="'*.
The last symbol has not hitherto received a
meaning, so that we arc at libertyto give it the meaning
which naturallypresents itself;hence we
may
say that
m
w
=
;
a"
-^
a"
=
1 ;
In order to form a complete theory of Indices it
325.
would be necessary to give demoiistrations of several propositions
which will be found in tlie luri^er
Algebra. But
these propositionsfollow so naturallyfrom the definitions
and the propertiesof fractions,
that the student wOl not
find any difficulty
in tho simple cases
which will come
fore
behim.
We
shall therefore
refer for the completetheory
only give here some
examples as
to the
largerAlgebra,and
specimens.
If
326.
that
hold
m
and
n
(""*)"a"""; see
=
when
m
and
n
are
whole
positive
Art. 279.
are
not
Now
numbers
this result
positivewhole
we
know
will also
numbers.
For
example,
{a^ a^.
=
Por
let
fourth power
(a^)*x
^
we
;
have
by raisingboth sides to tDG
a^ .tr*j
then by raisingboth sides
than
=
222
INDICES,
to the tliird power
to be shewn.
was
is a positivewhole number
(a6)". This result will also hold
If
327.
a"x6*
a
have a=;c'';therefore aj=aT7, which
we
=
n
know
when
n
if
each
that
is not
For
number.
example, a^ x 6* {cib)^.
raise each side to the third power, we
obtain in
db ; so that each side is the cube root of ab.
positivewhole
"oY
we
case
In like
=
manner
have
we
1
ill
a"
a,
we
X
"" X c"
X
(a"c...)".
=
...
that there are
Suppose now
6, c,...,and that all the rest
of
m
these
equal to
are
quantities
a;
thus
we
obtain
1
1^
(a")"*
[arY; that is,(IJaT J^JdT.
=
=
the m* power of the w*^ root
n^ root of the wi"" power of a.
Thus
of
a
is
equal to
the
fraction may take different forms without
expect to be able to give
may
any change in its value,we
out
different forms to a quantity with a fractional index,withalteringthe value of the quantity. Thus, for example,
Since
328.
2
since
k
4
=
2
we
^
expect that
may
a^
4
=
a^ ; and
this is the
o
o
For
case
a
obtain o^
;
if we
raise each side to the sixth power,
that is,each side is the sixth root of a\
fraction can
be transformed
into
w
ith
the
surd
equivalent
expression
part integral.
A
335.
y?
y?^
'^S^'^axQ
=
yi" -^
^^27" 3
=
an
_
*
Surds which have not the same
index can
be
transfoimed into equivalent
surds which have ; see Art 327.
336.
For
4/11: ^5
and
example,take J6
=
5^*4/11 (11)*
;
=
6^ 5^= 4/53=^125, (1I)*11^= 4/(11)'
=4/121.
=
=
of the preceding
notice an application
Suppose we wish to knew wliich is the greater,
have
reduced them
to the same
we
,^/5or 4/11- When
ir.dex we see that the former is the greater,because 125 is
greaterthan 121.
337.
Article.
We
may
Surds are said to be similar when theyhave, or
iiTatioual factors.
be reduced to have,the same
338.
can
and bjl
4/v/7
Thus
are
also similar surds,for
To add
surds; 5,^2 and 44/16
similar
are
4^^16 84/2.
=
subtract similar surds,add or subtract
their coefficients,
and affix to the result the common
irrational factor.
339.
For
or
example, ^/12+ ^75
=
2
a
73
^2
=
(2 + 5-4)^3
7256
1
^/48
-
=
2
^^3+
6
J3
3^/3.
"/12 1 '/64xl2
^~27"
8 -^-i
2
"'4-^"9=3 ^
^2^12
3
2
1
-
44/12^24/12
"^4
'
3
3
4
J3
227
SURDS.
multiplysimple surds which
Index, multiplyseparatelythe rational
To
340.
have the same
factors aud tho
irrational factors.
For
example,3 ^/2x ^3
2^4x3 4^2
6
=
3
=
"y6
; 4
4/8
^5
7 ^6
x
=
28
^/30;
6x2=12.
=
To multiply
simplesurds which have not the
surds which have the
index,reduce them to equivalent
index,and then proceedas before.
341.
For
By
example,multiply4 ^/S by
Art. 336
Hence
342.
the
^5
=
same
same
4/11.
2
^\25, 4/11=4/121.
that is,84^15125.
x 121),
productis 8 4/(12o
The multiplication
of compound surds is performed
like the multiplication
of compound algebraical
expressions.
For
example,(6^3 5 ^y2)x (2^3
36 + 18 ;v^6-10^6 -30
-
=
=
+ 3
6 +
v/2)
8^/6.
Division
may
by a simple surd is performedby
that for multiplication
by a simple surd; the
be simplified
by Art. 335.
For
example,3v2^4V3
343.
like
4
=
nile
result
a
^ ?y^ ^y^^f;
=
=
yi2^2iill)
/5^2Vll-^^-^^^-2 -2V y^^-2
121-2V
4^5.2^11-2"-
^^^^
121
x(ll)*
_2 4/lS30125
11
The student will observe that by the aid of Art. 3:15 the
merical
convenient for nuresults are put in forms which are more
the
have to find
mate
approxiapplication
; thus, if we
metliod
easiest
is
numerical value of 3^^/2^4^3. the
to extract the square root of 6, and dividethe result by 4.
1.-)";.
228
1
SURDS,
of division by a compound surd
The only case
is of any importanceis that in which the divisor is
surds
or difference of two quadratic surds,that is,
344.
which
the
sum
effected
involving
square roots. The division is practically
by an importantprocess which is called rationalisingthe
denominator
of a fraction. For example,take the fraction
4
5
/04.0
/Q
"
^^
^"
multiplyboth
nator
denomi-
and
numerator
of this fraction by 5 J'2-2sj3,the value of the fraction
is not altered,
while itsdenominator is made rational;
4
J3)
4(5^2-2
thus
6j2-t-2V3 (5^/2+ 2^3)(5^/2-2V3)
4(5v/2-2v/3) 10s/2-4j3
==
^
60-12
19
s/3+ V2
Similarly,
"^'2 3-^2
(,/3W2)(2^3-f^2)_
(2^/3-^2) (2^3 +^2)
^8+3^/6^8+ 3^6
*
"10
12-2
shew how to find the square root of
We shall now
binomial expression,
of whose terms
is a quadratic
one
345.
a
surd.
root
requirethe
Suppose,for example,that we
of 7+4^3.
Since ija;+Jyy
square
aj + y-h2 J{xy), it is
and y from x + y
7,
root of 7 + 4^/3 will be
=
obvious
find values of x
we
and 2 J{xy) 4jS, then the square
to/x+ ^y. We may arraage the whole process thus
that if
=
slO + 4.JZ)= pjx+ Jy
Suppose
7+
square,
Assume
4^^3
=
(x +
square, and subtract,
that
^
;
Ji^xy).
2
+
:
JS;
thcn2j(xy)=4:
y=T,
x +
+
^
=
y)'^4xy
is,(a!"yf ljtherefore
=
Sincea;+y=7 and
x-y
=
x"y
49
=
"
^(7 +
4
v/3) ^4
Similarly,
n/C7
4
^/3) 2-^3.
-
=
+
=
1,
have x=4j y
therefore
=
48
l.
=
l, we
"
^3
=
2 +
V3.
=
3;
229
XXXIV,
EXAMPLES.
Examples.
XXXIV.
Simplify
1.
3^/2 + 4^8-^/32.
3.
2V3
+
2^4
2.
3^/(H)-V(5i).
+
54^32-^/108.
Tj^-^Q-
4.
Multiply
5.
^/3.
N/5+V(li)-;j5by
^/6- J2.
7.
l+^/3-^2by
8.
-L.
^/3+^/2by-^
+
Rationalise
Q
of the
the denominators
v/3+ n/2
x/3-^2-
3+V2
^*
2-^2*
2V5+
+
13.
14
16.
4-^/15.
+
'
2;^3*
the square
Extract
6^/5.
3^2
3^3-2;V5'
2^/3
V3
n
3"/5
followingfractions:
root
of
14.
16-6J7.
+
15.
Bimplify
17.
^
"..^
.,.,.
18.
8
+
4^1^3.
230
\
TIO,
RA
XXXV.
Ratio.
Ratio is the relation which one
quantitybears
to another with respect to nvi^itudc, tlie comparison
34(".
what
made
beiii"?
by considering
is of the second.
the \\\-"i
multiple,
part,or parts,
Thus, for example,in comparing 6 with 3, we observe
that 6 hus a certain magnitude with respect to 3, which
it contains twice ; again,in comparing G with 2, we see that
different relalice- magnitude,for it contains
has now
a
2 throe times ; or 6 is greater when compared with 2 than
it is when compared with 3.
6
The
347.
by two
expressed
usually
them, thus,a : 6 ; and the former is
of the ratio,
and the latter the consequent
ratio of
pointsplacedbetween
called the antecedent
of the ratio.
to " is
a
A ratio is measured by the fraction which has for
the antecedont
of the ratio,and for its
its numerator
dwiominator the consequent of the ratio. Thus the ratio
348.
of
a
to
" is measured
say that the ratio of
Hence
349.
to the ratio of
If
350.
hy the
351.
c
;
then for shortness
to " is equalto
a
may
the terms
t
or
is r
say that the ratio of
d^ when
to
,
of a
r
we
may
.
a
to 6 is equal
=
t
.
ratio he
or
multiplied
divided
quantitythe ratio is not altered.
same
xn
we
by
ma
a
We
,.
compare
the fractionswhich
Gtruuxiiiaatur.
.
,"""
two
measure
Thiia,
suppose
or
more
these
cue
ratios
ratios
by reducing
to
a
common
ratio to be that of
a
to
d,
231
RATIO.
and {inot!^crratio to be that of
f
J,,and
"
the second ratio
bil
b
Hence
than
the
c
=
^
a
d; then the firstratio
to
r^*
ha
the first ratio is gi-eater
than,equal to,or less
second ratio,accordingas "w? is gi'eater than,
to,or
etjual
be.
less than
of
ratio is called a ratio of greater inequality,
less inequality^
or of equality^
according as the antecedent
A
352.
to the consequent.
than,less than, or eqwd
is greater
ratio of greater inequalityis diminished^
A
ratio qf less inequalityis increased^by adding
a
number
to both terms
of the ratio.
353.
and
any
the ratio be
Let
adding
to both
x
^,
and
terms
let a
of the
new
ratio be formed
originalratio;then
by
,^
o-\-x
is
greateror less than ^, accordingas 6(" + a;)is greateror
that is,
a(ft+a?);
accordingas bx is greateror less
than "M?, that is,accordingas " is greater or less than a.
less than
354.
ratio
ratio of greater
inequalityis increased^and
inequality is diminishedyby takingfrom
A
of less
both tenns
of the ratio
a
each
of those
any
number
is less than
which
terms.
Let the ratio be
takingx from both
v
and
let
,
a
of the
terms
new
ratio be formed
original
ratio;then
^
0
is greater
or
or
less than
less than
,
than
"
"
"
X
x) is firreater
is,accordingas "a; is less or
that is,according
6 is less or greater
as
a{b"x)\
gi'caterthan ax,
accordingas b(a
by
that
a.
If the antecedents of any ratios be multiplied
together,and also the consequents, a new ratio is obtained
which is said to be compounded of the for'ner ratiua. Thud
355.
232
EATTO.
the ratfo
a : h and
5c?is said to be
d.
"c
c
:
of the two ratiw
compounded
:
the ratio a : 6 is compounded with itself tlie
resultingratio is a^ \W-\ this ratio is sometimes called the
duplicateratio of a : d. And the ratio a? \W\" sometimes
ratio of a : 6.
called the triplicate
When
The
356.
follomng is
a
importanttheorem
very
cob-
ccrningequalratios^
Supjyosethat ^=
tlicn each
=
of thew
ratios
fpa'^+ qc'^+ r^^X
_
wherejt?,
g,
r,
rorletA;
n
any mmibcrs
are
whatever.
^ ^=-^;then
=
=
oaf
kh
therefore
p
=
kd
a,
c,
=
+
+ q {kd)"
{kb)"
r
/cf=e
;
{kff =pa*
+
go"4- 2r"*;
therefore
U%"3pl^^'.
\pb* qd'* rJ^J
therefore
+
+
of demonstration
mode
there
similar result obtained when
The
a
ratios
same
^
po
p
=
a^c
b
+
a
examx"Iewo
particular
that if
^
"
+
qd
+
q^r,eo
+
are
applied,and
than
more
threa
givenequal
As
see
be
may
e
d+f'
r
=
-,
0
a
";
and
=
",
each
may
suppose
of these
1, then
?*=
ratios is
equal
we
to
J
then
as
a
rf*
that each of the
specialcase
we
may
givenequalratios
is
suppose
equal to
234
PROPORTION.
XXXVL
Proportion,
said to be proportional
Four numbers
when
are
the firstis the same
multiple,
part,or parts of the second
357.
y"
ft
as
the third is of tlie fourth ; that is when
i
0
numbers
ft,
c, d
",
the
terms
a
and d
is to 6
a
a
is
c
a
called the extremes, and
are
usually
is to d; and
:b^c : d.
as
d^ ov thus
:
the four
-,
called,
proportionals.This
are
expressedby saying that
thus a : b :: c
represented
The
=
it is
h and
c
ineang.
when
two ratios are equal,
the four riunibers
which fonn the ratios ai-e called proportionals
; and the present
devoted
the
to
is
of
two
ratios.
subject
equal
Chapter
358.
Thus
359.
When
of the
Let a,
four
nuinhers
equcd to
is
extremes
proportionalsthe product
the prodtictqfthe means.
are
ft,
;
c, d be proportionals
a
c
,.
then
1=3;
o
a
multiplyby hd; thus ad=bc.
three terms in a proportion
are
given,the fourth
from the relation ad=be.
be determined
If
may
any
If ft
c
=
second
as
is
extremes
When
tinued
between
have ad=b^; that is,if the first be to the
tfte second is to the third, the product of the
we
a
to
eqital
:b ::b
:
the sqvAxre
d then
proportion; and
a
and
a,
of the
d
ft,
msan.
said to be in
are
ftis called the
co7i-
2?roportiondl
m,ean
d.
be equal to the
If the product qf two numbers
the four are
proportionals, the
product of two otJiers,
mid
terms
of either product being taJcen for the m^an^
the terms of the other product for the extremes.
360.
Fur let xy=ab\
divide
by
ay, thus
=
-
a
or
X
: a
lib :p
-
;
y
(Art.357).
235
PROPORTION.
If
361.
For
a
6
:
and
=
Y
:: c
-,
J,
=
b
For
: a
thus
he proportionals,
they are
b
each
unity by
divide
d
::
: a
taken
'.:h:d.
: c
For^|;multiplyby
=
or
a
T=
'.
a
add
i;
,
1
=
:
c?,
these
of
equals;
=
:'.b : d,
c
proportionals,the first
are
1, that
Also
the second
as
unity
second
the
:
as
the
fourth;
third
IS
^
the excess
the excess
+
=
"
._
; OTa
thus
equals;
d
"^
is
that
d.
these
to
bc
+
..a
.,
,
+
^
365.
pro-
:: c
c
c
+
"
thus^^;
togetherwith the second is to the
the fourth is to
together with
i( a :b ::c : d, then a + b:b :: c + d
r
:
: c.
;
-
numbers
If four
364.
a
a
they are proproportionals,
that is,if a : 6 :: c : rf,
alternately;
when
poi-tionals
For
is,if
be
If four numbers
363.
a
;
c
a
then
t;
\f.
v.e
:/.
:: e
inversely; that
taken
or
-:
=
-
=
r
\h
a
: c.
;i ;
=
V
d
::
:h
a
numbers
If four
poi'Honalswhen
thou
therefore
;
^
or
362.
then
and c\d\\e:f,
rf,
:
b:0::c
+
,,
+
d:d.
of the firstabove the second is
of
the third
the fourth
above
to
is to
the fourth.
For
"
=
"
J
r
b
1
,
=
-"
d
subtract
unity from
^~^
c
,
"
;
"
J.I-
J.
that
1,
'
c
d
"
"
-^"
b
"J-
d
or
a"
b
,
-
,
"
IS
:
thus
equals;
these
b
v. c-a
,
:
a,
536
PROPORTION.
366.
second
Also the firstis to the excess of th^ firstabove ths
the third is to the excess
of the third above thsi
as
fourth.
the last Article
By
a
-
..
therefore
b
"
b
x
-
,"
"
r-
b
d
c"d
=
-
=
",
"
a
b
"
: a
:: c
"
d
also
^
b
d
x
"
=-,:
d*
a"b
c"d
or
-
,
bade
or
;
'
=
.
a
:
therefore
c;
: a
a
b
"
'
c
:: c
: c
d.
"
When
are
four numbers
proportionalsy tliesum
t}ie
and
second
is
to
their
of
first
differenceas the sum
the
third
and
is
to
their
of
difference; that is,if
fourth
b ;: c + d : c"d.
a : b :: c : dj then a + b : a
367.
"
By
Arts. 364
and 365
=
"
i"
"
j-
0
,.
a
-
therefore
"
+
j"
a"b
b
+
"
b
j"
c +
=
:,"
b
or
a+b
d
"
d
:
a"b
and
,
c
+
a
^,
--3"
d
::
"
d
0
d
"
=
-5"
a
c+d
,
'
that
.
^
is
+
a-b
:
b
"
^
j-
;
c+d
=
:,"
c-d%
c-d.
It is obvious from the precedingArticles that
four numbers
derive from then^
are
we
can
proportionals
also Art. 356.
other proportions;
see
many
368.
Proportionit is supposed that
determine what multipleor what part one quantity
we
can
kind.
is of another
But wo
cannot
quantityof the same
if
the side of a
always do this exactly. For example,
the
noted
inch long the length of
diagonalis desquare is one
by J 2 inches ; but J 2 cannot be exactlyfound,so
that the ratio of the length of the diagonalof a square
to the length of a side cannot be exactlyexpressed by
numbers.
Two
quantitiesare called incommensurable
when
the ratio of one
to the other cannot be exactly
pressed
exby numbers.
will sugThe
student's acquaintancewith Arithmetic
gest
to him
that if two quantities
are
reallyincommen369.
In the definition of
sm'able stillwe may be able to express the ratio of one to
the other by numbers
as
nearlyas we please.For example,
find two mixed numbers, one less than J2, and the
we
can
other greater than ^2, and one diflfering
from the other by
as small a fraction as we
please.
237
PROPORTION.
with respectto the
"We will give one
proposition
quantities.
comparisonof two incommensurable
370.
Let
known
another
:
then
x
integerq
that both
such
integer p
and
-
quantities;and
and y denote two
that however
great an
X
and y
are
it
be we can
find
y lie between
may
and
x
suppose
equal.
q
q
the difference between
For
as
;
-
and
x
and
by taking q largeenough
y cannot
can
-
be
so
be made
great
less
But if x and y were
unequal their difference could not be made less than any
assignedquantitywliatever. Therefore x and y must be
than any
assignedquantitywhatever.
equal.
371.
It will be useful to compare
the definition of portion
proin this Chapter with that
of Euclid.
Euclid's definition
used
has been
which
which is given in the fifth book
are
proportionals
may be stated thus : four quantities
be taken of the first and the
wlien if any equimultiples
of the second
and the
third,and also any equimultiples
fourth,the multipleof the third is gi-eaterthan, equalto,
less than, the multipleof the fourth,according as the
or
multipleof the first is greater than,equal to,or less than
the
multipleof
372.
We
the second.
will first shew
the
definition
algebraical
Euclid's.
satisfy
For
suppose
that
a
:
b
that if four
of
::
c
satisfy
quantities
will
also
proportion,they
:
d; then
"
=
0
j.=
j^
whatever
nmnbers
p and q may
-j :
therefore
a
be.
Hence
ji7Cis
is
greater than,equal to, or less than qcl,according
as pa
greater than, equal to, or less than qb. That is,the four
Euclid's
quantities
a, ",c, d satisfy
definition of
proportion.
"We shall next shew that if four quantities
satisfy
Euclid's definition of proportionthey will also satisfy
the
373.
definition.
algebraical
For suppose
whatever
that a, h, r, d are
cumbers p and q may
four
quantitiessuch
be, pc
is greater
that
than,
238
PROPORTION.
equal to,or less than qd,accordingas pa is greater thail
equalto,or less than qh.
First suppose that c and d
p and q such that pc=qd; then
1
=
--
^;
=
therefore
Next
Then we
1=3.
b
qd
qb
are
commensurable
by hjTpothesis
pa
Therefore
a
:b
;
=
takcSI
qb: thus
:: c
d
:
d.
j
m
that c and c? are
incommensurable.
cannot find whole numbers
p and g', such that
pc=qd. But we may take any multiplewhatever of d, as
of c,
qd, and this will lie between two consecutive multiples
suppose
say between
pc and
^
and
is greaterthan
,
is less than
"^^
Thus
-
-,
"
.
must
unity,and
^
^^
is greaterthan
"
,
both greaterthan
^
And
since this is true howeyer
are
and
,
be
Tliat
equal:
see
^
Art.
is less than
unity,
unity. Hence, by hypothesis,
and
infer that
be, we
{p + l)c. Thus
cannot
370.
be
-,
unity.
and both less than
great p s6idq
may
unequal;that is,they
Therefore
a
:
b
:: c
:
d.
the algeis,the four quantities
braical
a, b,c, d satisfy
definition of
proportion.
stated that the Algebraical
It is usually
definition
be used in Geometry because tliere is
of proportion
cannot
374.
of representing
geometricallythe result of the
operationof division. Straightlines can be represented
no
method
number which expresses
how often one
straightline is contained in another. But it
should be observed that Euclid's definition is rigorous
and
to incommensurable
to commensuras well as
able
applicable
definition is,strictly
quantities
; while the Algebraical
Hence
this consideration
confined to the latter.
speaking,
alone would
for the definition
furnish a sufficient reason
adopted by Euclid.
but
geometrically,
not the abstract
EXAMPLES.
Examples.
Find
ca'
the value
I.
4
:
3.
6
: 0?
5.
d? +
4
: a;
6.
a; +
4:
:
7.
3a? + 2
8.
a;2+ ic+l
9.
aa?+"
7
10.
If j9g
II.
If
:
W
12.
+ 2
7
+
::
::
5a?+a
:: mor
:: c
If
a
:
"
:
6
::
:
+
n
J, and
:
c"^ : d(^ and
::
:
7
:; a?
4.
x
;
9
::
+
a"'
:
c, then
42.
:
16
:
a?.
5.
3.V + 2.
5a? + 8.
+ m.
: war
a'
"'"
::
63(.^"-l).
l :
+
qt=su, then p
r*, and
6
3
:
9iP-2
followingproportions.
2.
: w
2a;-l
of the
G2{x + l) ::x^-,x
;
:
+ S
x
::
2a? + 8
=
in each
a;
45.
:
: a?
XXXVI.
: a?.
:: ^
:
a
8
::
of
239
XXXVI.
^
::
cd'
:
:
::t
: r
c'
:
:u.
then
c?',
"/";?.
{" + c") (ab+ lc)\
(a^+ 6-^)
=
in contimied proportion;
three numbers
the middle number
of the others is 120 :
is 60, and the sum
tiiidthe numbers.
13.
There
are
Find three numbers
in continued
tliat theu' sum
be 19, aud the sum
may
14.
such
proportion,
of their squares
133.
If
a
:
6
:: c
:
relations
rf,shew that the following
are
true.
15.
17.
+ h).
a{c-k-d)=c{a
16.
aj{(^+ d^=c^{a^
(a+ c)(o"+ "^) {b+ d){"-^d^)
^
{a-c){a:^-c?)~
{h-d){W-d^y
18.
pa* + gab +
pc^ + qcd +
rd^
_
Id^ + mab
-k-rd/" Ic^+ mcd
19.
"^
a
20.
rV^
a
'.
~
2ft
3c
b
iimaf
::
-i-nd^
4"/
+
a"i ( 4
3
'
2"*"J
ncf) : ^ {mhP -^ nd^).
+
lr^
240
VARIATION,
XXXVII.
Variation.
The present Chapter consists of a series of
connected with the definitions of ratio and propositions
portion
stated in a new
w
hich
is
convenient
phraseology
for some
pui-poses.
375.
One quantityis said to tary directlyas another
376.
when the two quantities
depend on each other,and in such
that if one
be changed the other is changed in
a manner
the
same
proportion.
Sometimes
and say
for shortness
simplythat
the
omit
we
quantityvaries
one
as
word
directly
another.
Thus,for example,if the altitude of a trianglebe
varies as the base ; for if the base be
the area
invariable,
from Euclid that the
kicreased or diminished,we
know
is increased or diminished
in the same
area
proportion.
We may express this result with Algebraicalsymbols thus ;
which
let A and a be numbers
representthe areas of two
triangleshaving a common
altitude,and let J3 and b bo
which represent the bases of these triangles
numbers
re377.
B
A
then
spectively;
A
^
=
"
v-.
from
And
this
we
deduce
a
=
r
,
by
If there be
Art. 363.
a
third
triangle
havingthe
altitude as the two alreadyconsidered,then the ratio
whidi
of the number
which representsits area to the number
same
representsits
then-^=m,
base will also be
and
A
=
mB.
Here
equal to
A
may
Put
t.
i
=
*^"
represent the
B
area
as
which have
series of triangles
altitude,and B the correspondingbase, and
that the area
Hence
the statement
constant.
oi any
one
the base may
of
a
also be
expressedthus, the
area
a
mon
comm
mains
re-
varies
has
a
242
VARIATION.
quantityis said to vary directlyas a second
it varies jointlyas the
and inversely as a third, when
of the third.
second and the reciprocal
One
382.
Or if .4
=
-77-
and
as B
directly
IfK
383.
Qc
where
,
is constant,A
m
is said to vary
as C.
inversely
B, and
B
C, then
oc
A
C.
oc
For let A
nC^ where m and n
mB, and B
then A
m7iC; and, as mnis
constant,A
are
=
=
=
If
384.
V(AB)
For
Also
a
A
x
G, and
J(J.B)
=
and
B=nC, where m and
(m " n)C ; therefore A^B
=
A
jC/"
a
BC, "A"w
(7 x
Similarly,
A
J[;r
let
For
therefore AC
oc
A
=
cc
-CrA
X
For let A=mB,
then
-^
and
n
are
stants
con-
C
x
B
-B
^,"wc?
oc
=
d-
C
oc
4.
therefore 5
^
oc
.
.
B, and
mB,
C
and
D, "A^
x
AC
C=nD;
x
then
BD.
AC=mnBD;
BD.
Similarly,if A
and
ACEccBDF;
387.
C,
oc
^J{mnG^)=C^{mn);iheYefore ^{AB)a:C.
For let A =mBC,
386.
A=tB
then
"C,
a:
C.
let A =mC,
; then ^ " i5
385.
B
stants;
con-
C.
cc
so
B, and
0"xDy
and
EccFt
then
or.
B, then
A'
x
B".
then A''=m'B':
therefore A*
x
jB".
243
VARIATION.
//" A
388.
B,
X
quantity variable
For letA
B
AP^
B "rAew
then A
invariable,
7/"A
389.
is
cc
BP, where
P
is any
invariable.
or
mB, then
=
AP
then
X
mBP
C
is
x
BG
; therefore
AP
invariable, and
when
both
A
B
^P.
x
C wjA^w
x
C
and
are
The
variation of A depends on the variations of the
B and C ; let the variations of the latter
two quantities
B is changed to b
take placeseparately.When
quantities
A
let A
be
changed
let C be
Now
changed
to
A
B
-7
a'
X
=
"
a'
^
"
a'
;
then, by supposition,
-/
to c, and
in consequence
then, by supposition,
-
G
0
a
changed
a ;
X
to
that^. IS,
c
^,
=
"
a
-r
.
let a' be
Therefore
-
.
BC
A
.,
;
=
B
=
"1"
be
;
.
therefore ^
.
x
"^
BC.
this proposition
is furnished in
Geometry. It can be shewn that the area of a triangle
the height is invariable,
and that
varies as the base when
the base is invariable.
the area varies as the heightwhen
both the base and the heig:ht
when
Hence
vary, the area
which representthe
varies iis the productof the numbers
A
very
base and
good example of
the
height.
this proposition
are
suppliedby the
in Arithmetic
under the head of the
which occur
questions
"For
instance suppose
that the
Double Rule of Three.
which
be
work
varies
can
as
accomplished
quantity of a
of workmen
when the time is given,and varies
the number
is given ; then
of workmen
the number
as the time when
the quantityof the work ^^^llvary as the product of the
and the time when both vary.
number of workmen
Other
390.
examples of
In the
same
of
manner, if there be any number
varies as another
each of which
B, C, D,
quantities
A
quantity when the rest are constant,when they all vary
A varies as their product.
...
16"2
244
EXAMPLES.
XXX
Examples.
varies
A
1.
value of A
If A^
2.
varies
as
B
when
^
"
yl varies
and C= 1
:
varies
A
2, and C- 2
+
SB,
8.
B^b,
9.
when
C^b.
+
B
when
:
B
A -4
when
jB
C=S:
find
1, and
=
n.
A
l
=
2 and
=
B
and
C jointly;and A
12
find the value of A
when
(7=4
: B
and
find the
=
when
C= 2.
and
directly
C=c:
=
as
C
are
."112 and
inversely
; and
find the value of A
Charitable
partly vary as the
expenses
of
when
C jointly
A
when
a
; and
value of A when
B=h^
and
^
constant, and
the inmates
When
'2;
=
as
as
=
=
A
The
A
5
and
C jointly
^
8 when
; and
find the value of BC when ^
10.
i?
varies
B=h, and
A^5
B
and
(7 jointly
; and
find the value of A when
/?
as
10.
the
^
varies
A
and
(7=c:
"
and
as
A varies
2, and (7=3:
=
find
that
as
7.
S
jS=1;
B", shew
-
=
6.
=
A"
as
5A
as
=
^=1,
when
2
A varies as nB + C; and
and
^
7 when -S 2, and
;
5.
^
varies
varies
: B.
ZA + 5B
find the ratio A
C=2
=
-B.
3.
4.
A
2.
=
B^
+
XXXYII.
B, and
as
VIL
a
A=a
^=c
and
when
Institution
number
are
of
partly
inmates.
and 3000 the expenses
are
spectively
reFind
the expenses
for 1000
"180.
960
inmates.
11.
find how
The
many
wages
men
of 5
can
for 7 weeks
being ."17. 10*.
be hired to work 4 weeks for "30.
men
If the cost of making an embankment
vary as the
lengthif the area of the transverse section and heightbe
constant,as the height if the area of the transverse section
and lengthbe constant, and as the area
of the transverse
section if the length and height be constant, and an embankment
cost
1 mile long,10 feet high, and 12 feet broad
12.
find the cost of an
"9600
embankment
16 feet high,and 15 feet broad.
half
a
mile
long,
ARITHMETICAL
24^
PROGRESSION,
Arithmetical
XXXYIII.
Progression,
Quantities are said to be in Arithmetical
when they increase or decrease by a common
by subtractingany
follows
it. In the first
term from that which
immediately
difference is 3 ; in the second series it is
series the common
2 ; in the third series it is h.
The
common
"
denote the
Progression," the common
Let
392.
terra is
a
+
a
6, the
third
term
Thus
of
first term
Arithmetical
an
diflference;then the second
is a-}-2",the fourth term
is
the n^
36,and
an
To find the sum
of a given number of terms nf
Arithmetical
Progression, the firsttei^m, and the convdifferencebeing supposed known.
so
on.
is
1)6.
a+
term
("
+
a
"
393.
mon
denote the first term, " the common
of terms, / the last term, s
the number
Then
terms.
Let
n
difference,
a
"
=
a +
(a+ 6)+ (a+2") +
And, by writingthe series in the
the
+
sum
1
order,we have
reverse
idso
*
=
/+
(Z-6)+ (/-2")+
+a.
by addition,
Tlierefore,
1s
=
{l+ a) + {l+d)+
=
thetefore
s
=
to
n
terms
n{l+a)\
-^{l-^(i)
of the
(1).
246
PROGRESSION,
ARITHMETICAL
Also
thus
/-a+(n-l)6
(2),
(n-l)"}
"=|{2a
(3).
+
equation(3)gives the value of s in terms of the
which were
Equation (1) also
quantities
supposed known.
gives a convenient expression for s, and furnishes the
in
number
of term^
of any
followingrule: the sum
Arithmetical
Progression is equal to the product of the
number
into half the sum
of the first and
of the term^
The
Now suppose that r is less than unity; then the larger
is,the smaller will r" be, and by taking n largeenough
If we neglectV*
be made
can
we
as small as
please.
n
r"
obtain
we
a
and
enunciate the result thus.
In a Geometrical
ratio is numerically
Progression in which the common
than
less
number
of terms
unity,by taking a sufficient
the sum
be
to
made
little
can
as
we
differas
please
we
may
from
.
1"r
example,take
For
409.
Here
a
=
therefore
l, r=-;
sufficientnumber
a
little as
as
the
terms
we
sum
of terms
pleasefrom
is 2
"
-
"
,
if we
if
-
8
2
the series 1,
,
take six terms
-r-"
,
2.
In fact if
the
sum
can
we
five terms
is 2
...
by taking
be made
sum
take
,
,
Thus
=2.
the
we
-
-
-
"
"
,
to differ
take four
the
and
so
sum
is
on.
result is sometimes expressedthus for shortness,
the sum
of terms of this series it
of an infinitenumber
is 2.
to infiniiy
2; or thus,the sum
The
252
EXAMPLES.
XXXIX.
Recurring decimals
410.
of wliat
Progression.Thus for examp^
called infmite Georaetricul
24
3
"3242424... denotes
examples
are
24
24
-+_+--
+
+
_
...
3
Here
the
after
terms
form
"
a
Geometrical
Progw
24
sion,of which
is
Hence
7-5.
is
the firstterm
we
may
and
--3
that the
say
the
,
of
sum
rati
common
an
iufinit
10
number
of this series is
of terms
TTp
^^^^^
(1 "77:2)
"^
"
24
Therefore
-"
the
value
the
of
.
recurring decimal
2"
1
\
10"'"
990*
value of the
thus:
tically
The
recurringdecimal
Let
-32424...;
5=
then
10
and
1000
be found pmo
may
3-2424...,
5=
5=324-2424...
(1000 10)5
Hence,by subtraction,
-
=
324
-
3
=
323 ;
321
therefore
And
*
any
other
=
"
.
example may
be
manner.
XXXIX.
Examples.
Sum
the
series
following
:
1.
1, 4, 16,
to
6 terms.
2.
9, 3, 1,
to
5 terms,
3.
25,10,4,
to
4 tei-ms.
4.
1, ^2, 2, 2^2,
...
to 12 terms.
treated
in
a
simil jr-
t
EXAMPLES.
3
1
1
g,
gj
g,
X
r
5.
2
^*
terms.
6
to
3
6.
3"
7.
253
XXXIX.
to7terms.
-1"2'
1,
-3,
to
infinity.
infinity.
g,
8.
1,5, Y^,
to
9.
If
to
infinity.
to
infinity.
-X,
-7,
2
10.
6, -2,
-,
o
Find
the value of the
followingrecurringdecimals
;
11.
-ISlSlo...
12.
"123123123...
13.
-4282828...
14.
-28131313...
15.
Insert 3 Geometrical
means
between
1 and
16.
Insert 4 Geometrical
means
between
5^
and
40|.
17.
Insert 4 Geometrical
means
3
and
-729.
between
25 G.
of three terms in Geometrical
The sum
Progression
thii'd
is 45:
difference
of
the
fii-stand
terms
the
and
Is 63,
find the terms.
18.
of a Geometrical
of the first four terms
of the first eight terms is
Progression is 40, and the sum
19.
3280
:
20.
is
terms.
The
sum
find the
The
21, and
Progression.
of three
sum
the
sum
terms
in Geometrical
of their squares
is 189
sion
Progres:
find the
254.
Harmonical
XL.
Progression.
C
A, B,
quantities
when
A : G
Progression
Three
411.
Any number
Harmonical
A
::
"
monical
said to be in HarB : B
G.
"
said to be in Harmonical
three consecutive quantities
in
are
every
Progression,
Hairmonical
in
reciprocalsof qiiantities
The
412.
are
are
quantities
of
Progressionwhen
in Arithmetical
Progressionare
A
PROGRESSION,
HARMONTOAL
Let A, B,
v.A-B
: G
C
be
:
B-G.
Therefore
A{B-G)
Diyide
ABG
by
Harmonical
in
Progression; then
G{A-B).
=
; thus
=
Ti~'n
the
This demonstrates
Progression.
~l
j"~
proposition.
The property established in the precedingArtich
Lrticleal
will enable us to solve some
monical
questions relatingto HarFor
insert
five
Harmonical
Progression.
example,
413.
2
between
means
8
and
-
Here
"
.
we
3
Arithmetical
(2)of
have
to insert five
15
6
between
means
15
-^
and
-
Hence, by
.
Art. 393,
3
therefore
6?"
=
1
therefore
-
h
=
"
.
,
8
16
3
Hence
TF
lo
"
t;;
16
,
J
tion
equa-
the
r^
lo
,2
gressionis-,
"
Arithmetical
^
"
Progressionis
^""^^ therefore
the
16
-,
-,
16
16
16
-^,
-,
-,
,
r^
Harmonical
o
16
25
8
^5'
26
-
,
:j-^,
Pro-
EXAMPLES.
$55
XL.
Let a and c be any two
quantities;let A
their Arithmetical mean,
G their Geometrical
mean,
their Harmonical
Then
mean.
414.
A-a
c"A
=
a
:
G
a
:
c
G
:.
:
Continue
the
^{ac).
^"
therefore H=
H-c,
:
.
Examples.
1.
-{a + c).
=
c; therefore G-=
a-U
::
therefore A
;
be
H
XL.
Harmonical
Progression6, 3,
for
2
three terms.
2.
Continue
the Harmonical
Progression8, 2, \\
for
three terms.
3.
Insert 2 Harmonical
4.
Insert 3 Harmonical
The Arithmetical
the Harmonica!
is 8
mean
5.
6.
The
Geometrical
the Harmonical
7.
Find
two
8.
Find
two
4
between
means
numbers
of two
find the numbers,
:
46^
:
such that the
Arithmetical
and Harmonical
of the
Arithmetical
mean
and
"
.
is 9, and
is 48, and
of two numbers
find the imiiibers.
of their Arithmetical,
is 9f,and the
numbers
such that the sum
Geometrical,and Harmonical means
is 27.
product of these means
numbers
and 2.
-
mean
mean
is
mean
between
means
product of
is 27, and
means
above
the
the
Harmonical
their
excess
mean
isli
9.
If a, fc,
c
a
10.
and
+ c
If three
"
26
:
a"c
numbers
are
is increased
each of them
thHt the results are
shew
Progression,
in Harmonical
are
:: a
"
c
:
a-vc.
in Geometrical
by the middle
in Harmonical
that
Progression,
number, shew
Progression.
256
PERMUTATIONS
XLI.
Thus
at
a
called their
are
and
Combinaiiona.
a
permutationsof the three
time,are o",ba, ac, ca, be,cb.
combinations
of
a
letters a, h,c, taken
of
formed
a
ab,
time, are
permutations^form
ca, and
and
417.
T at
a
of
out
things
are
them,
placed.
of the three letters a, b,c, taken
ac^ be ; ab and ba, though different
the combinations
Thus
at
the
things are
set
be
different collections which
can
the order in whiijh the
without i-egarding
two
thingscan
set of
permutations.
the
The
416.
COMBINATIONS,
different orders in which
be aiTanged
two
Permutations
The
415.
AND
same
also do
eombination,so
a"c
be and cb.
The number
time
the
is
n(n
"
of permutations of n things taken
(n" r + 1).
l)(n" 2)
Lot there be n letters a, b,c, d,
shall first find
; we
of permutationsof them
taken two at a time.
the number
thus obtain
Put a before each of the other letters; we
stands first. Put b bcforo
1 permutationsin which
""
a
thus obtain n"
other letters;we
1 permutations
in which
b stands first. Similarlythere are
n"l
Thus,
permutationsin which e stands first. And so on.
there
of
letters
the
are
n
on
n{n-l) permutations
whole,
of
We
taken two at a time.
shall next find the number
of n letters taken thj^^e at a time.
It has
})ermutations
just been shewn that out of n letters we can form n (n l)
permutations,each of two letters;hence out of the n"l
form {n-l) {n 2) permutations,
letters b,c,d,
can
we
letters: put a before each of those, and
each
of two
each of three letters,
have (n l)(n 2) pennutations,
we
each of the
"
"
"
"
stands first. Similarly
there are
{n
\){n 2)
in which b stands first.
permutations,each of three letters,
first. And
there are
in which
c stands
as many
Similarly
in which
80
on.
a
"
Thus, on
of
n
the
whole, there
n
are
letters taken three at
a
{n
"
time.
l)(w" 2)
"
mutations
per-
258
COMBINATIONS.
AND
PERMUTATIONS
For the number of permutation* of n thingstaken r M
a time is n{n
l)(n 2)...{7ir+l)hyAn.411; and each
combination produces [r permutationsby Art 420; hence
/
"
"
"
of combinations
the number
be
must
n{n-l){n-2)...{n r-^\)
"
If
multiplyboth
we
expressionby
this
*'
^
"aloe of
In
"
r
and
numerator
it takes the form
"
denominator
,"
.^"^=
"
-
[r [n~^
,
of
tlie
beingimchanged.
course
To find the mimler
422.
which are
taken all togetlier
of permutations of
not
all
n
things
different.
letters ; and suppose p of them to be a,
q of them to be S,r of them to be ";,and the rest of them
then the
to be the letters d, ",-.., each occurringsingly:
of permutationsof them taken all togetherwill be
number
Let there be
n
[p\q\r'
For
iV to represent the
suppose
If
penuutations.
'm
any
of the
one
required number
of
permutations the
p
and diflei-cnt letters,
letters a were
changed into p new
tlien,without changlHg the situation of any of the other
could from the singlepermutation produce J7?
letters,
we
if the p letters a were
changed into p new and different letters the whole number
if the q letters
would be iV x |^).Similarly
of permutations
also
b were
whole number
bo iVx 1^ X
into
r
thus
penuutations:and
different
new
^
.
And
if the
r
and
different letters the
could now
obtain would
also changed
letters c were
changed into q new
of permutations we
of perdifferent letters the whole number
mutations
this
number
be iV X [^ X [^ X [r. But
woidd
and
of
of permutations
letters taken all together,
that is to [".
nuist be
equalto
the number
n
different
\n
Thus
And
iVx["x '^x [r tn; therefore
=
similaily
any other
case
may
^^^J^^-
be treated.
259
XLI.
EXAMPLES.
method of
The student should notice the peculiar
demonstration which is employedin Art. 417. This is (railed
mathematical
induction^and may be thus described: Wo
is true in one case, whatever that
shew that if a theorem
case
may be,it is also true in another case so related to the
former that it may be called the next case ; we also shew
is true in a certain case ;
in some
that the theorem
manner
hence it is true in the next case, and hence in the next to
the theorem
be true
must
that,and so on; thus finally
in ever}' case after that with which we began.
423.
induction is
of mathematical
The method
used in the higherparts of mathematics.
Examples.
Find how many
1.
from
a
company
of
of
parties
24*men.
frequentiy
XLI.
each
6 men
can
be formed
be formed
can
many permutations
taken all together.
letters in the word company,
of the
be formed
of the
2.
Find how
3.
Find
how
many
lettersin the word
combinations
can
taken four
longitude,
at
a
time.
Find how many permutationscan be formed of tba
lettersin the word consonant, taken all together.
4.
of a set of things
of the combinations
number
taken four at a time is twice as great as the number
taken
three at a time : find how many thingsthere are in the set.
The
5.
6.
and
word.
how
words each containingtwo consonants
many
vowel can
be formed
from 20 consonants
and one
5 vowels, the vowel being the middle letter of the
Find
Five persons
7.
find in how
often
are
to be chosen
by lot out
be done.
many ways this can
assigned*
person would be chosen.
an
Find
of
twenty:
also how
"
boat's crew
and a
consistingof eight rowers
is to be formed out of twelve persons, nine of
whom
but cannot steer,while the other three can
can
row
steer but cannot
find in how many
row:
ways the crew
be formed.
Find also in how many
cjin
ways the crew
could be formed if one of tiiethree were
able both to row
and to steer.
S.
The number
of terms on the right-handside is one
plied
than the number
of binomial factors which are multi-
together.
II. The exponent of x in the firsttdrm is the same
as
a
nd
the
number
binomial
in
other
of
terms
the
factors,
is
less
than
term
that of the preceding
each exponent
by
unity.
III. The coeflScient of the first term
is unity; the
coefficient of the secdnd tei-m is the sum
of the second
letters of the binomial factors ; the coefficient of the third
is the sum
of the products of the second letters of
tcmn
the binomial factors taken two at a time; the coefficientoi
the fom-th term is the sum
of the products of the second
letters of the binomial factors taken three at a time ; and
is the productof all the second letters
so
on ; the last term
of the binomial factors.
shall show that these laws always hold,whatever
We
factors.
be the number
of binomial
Suppose the laws
that is,
to hold when n-i
factors are multiplied
togctlier;
BINOMIAL
there
guppose
and
are
261
THEOREM.
factors ic+a,
n-\
J, a!+c,...a?+Ai^
^7+
that
where p
the
sum
the
sum
two
at
the
sum
=
g
=
r
=
of the letters o,
of the
a
three at
.
.
A?,
.
products of these letters taken
time,
of the
products
of these letters taken
time,
a
productof
M=the
6,c,
all these letters.
sides of this identityby another factor
arrange the product on the righthand according
of x ; thus
Multiplyboth
x
+
l,and
to powers
+ a)(;?+6)(a?
+ c)
(dj
.
.
.
+ 0
+ *) (ar
(a;
{q+pl)^-'
-r
Now
p + l=a+b
=
the
+ c +
sum
...+k
+
=
"" +
+
Cp + /)a?"""
+...
(r+ ql)x''-^
l
"+...
-^-k)
of the products taken
the sum
time of all the letters ot, 6, ";,...
A, / ;
=
g; r+/(a6+ac
=
td.
of all the lettersa, b,Cy..k,
I;
+ h +
q+pl =q-\-l{a
r +
+
+
two
at
a
6c"4-...)
of the products taken three at
=the sum
of all the letters a, b,\...ky
I;
a
time
tt^=the product of all the letters.
factors are multiplied
when
n-1
together,they hold when n factors are multiplied
together;but they have been shewn to hold when /our
fectors are multiplied
therefore they hold when
together,
and so on: thus tiiey
""5 factors are multipliedtogether,
hold universally.
Hence,
if the laws hold
262
BIl^OMIAL
THEOREM.
shall write the result for the
factors thus for abbreviation :
We
of
multiplication
-\-Jix''-^
+
Now
n
in
P
is the
Q is the
number;
letters two
of the letters
sum
and
sum
of
are
products of these
the
"^
^
are
of these
q"" products;and
"
"
so
See Art. 42t.
on.
I each equal to
Suppose b,Cf..,kf
n(n-l)
,
^
na,
Q becomes
and
so
/
V.
...+
k, I,which
a,b,c^...
that tliere
two, so
products; R is the
of
sum
n
V
Thus
on.
Nn
on.
or, R
n{n
becomes
P
,
^
n(7i-\)
7i{n-l){n-2)
"_,
,
1
1,2
^
becomes
l)(n-2)
^a^;
^\
"
finally
n-,
-
^
Then
a.
.
J
.
.
,
o
^
''^
,
^'*-
1.2.3.4
justobtained is called the Binomial
Theorem; the series on the right-handside is called the
and when
we
+ a)",
expansion of (a?
put this series instead
said to expand {x + ay.
of {x + aY we
The theorem
are
discovered by Newton.
was
The
426.
formula
the theorem
that we have demonstrated
is
in which the exponent "
a positiveinteger;
have used fn this demonsti-ation the method
It will be
seen
in the case
and that we
of mathematical
427.
Take
induction.
for
example {x4- af.
) 6^
n(n-l)_6.5
"1.2"^^'
_
1.2
_
Here
n
=
6.5
n{n-l){n-2)
?i(?^-l)(yi-2)_
6^
6,
.4
~1.2.3"^""
~1.
_
1.2.3
w(w-l)(n-2)(n-3)
6.5.4.3
1.2.3.4
1.2.3.4
15,
n(n-l)(n--2)(n--3)(n-4) 6.5.4.3.2
"1.2.3.4.6' 6;
1.2.3.4.5
_
^03^
{x-h"if^3^-\-
1 5aV
+
26S
THEOREM.
BINOMIAL
20aV
+
1 5a*:c*+ 6"*" + efi,
+
requirethe expansion of {l^-\-cy)*:
Agrain,
suppose we
have onlyto put "* for x and cy for a in the preceding
we
identity;thus
+ M'
4- eieyfb^
15("^)*(52)2
+
Again, suppose
put
must
";
"
+ 15";V=**
Gq/ft^o
+ (fl/^.
+ 6c'y'b^
+ I5c*y^*
200^1/^1/
+
.
^'* +
=
for
requirethe expansionof (x"e)'; we
we
in thd result of Art. 425
a
thus
;
n{n-l){n-i)
*"
1.2.3
Again,in the expansion^f (x + a)" put
/.
\-
n{n-l) _'
\"
n(w-l)
'^2
{l+xy=l-hnx+
1
l)(n
"
2
1
and ais thiols true for allvalues of
/,
n(n
\ ^^
,
{l+aT=l+na+
a
we
1 for ar; thua
may
"
2),
^"^""'
3
put a; for
w(/i l)(n 2)
"
,
"i ;
thua
"
"^
,
+
"^
I
2
"
3
to expand
apply the Binomial Theorem
For example,
than two terms.
expressionscontainingmore
requiredto expand {i+ 2x"a:^)*. Put y for 2x"x^; then
have (lf2x-^-)* {l^yy=l + 4y + 6y^+ 4y^-k-y*
we
We
428.
may
=
=
4- 6 {2x
4.(2^ a;2)
1+
-
Also (2x
{2x
-
x^f
=
=
-
-
+ 4 (2a; a^' + (2"
a?^*.
aj2)2
-
-
+ (a^')2
=4a^^4x^-\-x*,
.7r*)*(2jr)2 2 (2;c)a;*
=
-
(a;2)2{a^^
{2xf Z[2xfa^+ 3(2a;)
8a;3-12.r*4-6ar"-ic",
-
-
4 (2a;)(a:2)'
4- (aj""
+ 6 (24;)2(a^)2
(2"-"**)* (2a;" 4 (2a;)3a;"
=
=
-
i"ir*
-
-
32j:" + 24j;"
-
"c7
+
a?8.
264
BINOMIAL
THEOREM,
tlie temia, we
Hence, coTlecthig
429.
terms
are
the
{\+
2x-
a^*
the
of
expansion of (I+x)" the coefficients
and
the end
equally distant from the beginning
In
"
M^
same.
coeflRcient of the r^ term
The
'
^^
-^^
obtain
the
from
beginning is "'
both
'; by multiplying
numerator
^~
_
In
and denouirnator by
fn
"^
"
1
r+
this becomes
-"
yj'-i]ji-r^i
from the end is tlie
the beginning,and its coefficientis
The
r"" term
\n-r
by multiplyingboth
+
.
,
L
y
*
\
nnmerator
(a
-
r 4-
term
2)**
from
that is
[n
dentmiinatMr
and
by [y" 1
in
tkis also becomes
V"l
\n
"
r-i-l'
Hitherto in speakingof the expansion of (x + a)*
430.
that n deiwtes sr)me
have assumed
we
positiveinteger.
is
also
Theorem
J3ut the Binomial
applied to expand
when
is
or
a
w
positivefraction, a negative quan{x + "V"
tity
whole or fractional. For a discussion of the Binomial
with any exponent the student is referred to the
be a useful exercise to
lai-gerAlgebra; it will however
from
the general formula.
obtain various particular
cases
Theorem
Thus the student will assume
bo the values of x, a, and n,
for the presentthat whatever
n{n-l)(n-2){n-S)
"^"^
^
^
1.2.3.4
If n i" not
a.
positiveintegerthe aeries
never
ends*
26(J
",
Thus
EXAMPLES.
.
,
.
(1 + y)"
=
,
I
XLIT.
m{m^-\)
m(.m4- l)(m +
y^
"jTy^
^ 3
,
-
my
+
m(m4-l)(m
+
2)(m4-3)
^
^
1.2.3.4
As
a
case
particular
suppose
Again,expand (1+2^"^^)^in
a"-4?^; thus
have (l+
powers
2^-";2)"
=
1 + ?a
of
ar.
Put y
(l+y)^
(2;?;-;c2j3^j^^i collect
expand (2jp-a^)2,
Now
:
thus
we
shall obtain
3
\s
Examples.
1.
"
1,1,5^
,1
terms
we
._
1 ; thus
m=
be verified by dividing1 by
This may
2^
-
Write
XLIL
down
the first three and the last three termg
down
the
QiicL-x)^,
of
2.
Write
3.
Expand (l-2y)^
4.
Writ"
down
expansionof (3 2a^)'.
the first four
ix-+2yY.
6,
Expand {l-^x-x^\
e.
Expand U+.f;-*-*'^/.
"
terras
in the expansios
267
XLII.
EXAMPLES.
^)*
7.
Expand
8.
Find
the
coefficient of
ar' in
the
expansion
9.
Find
the
coefficient of
3?
the
expansion of
(1-2^
+
in
of
(l-2a; + 3a,'2)5.
10.
If the second
term
X, y, and
sixth,seventh,and
eighth
in the
terms
{x + 7jY be respectively112, 7, and
of
and
lOSO, find
term
n.
If the
11.
the fourth
720, and
240, the third term
+ y)*be
expansion of {cr.
in the
pansion
ex-
find x, y,
-,
n.
12.
Write
down
the
first five tenns
of the
expansion
of(a-2x)K
13.
Expand
14.
Expand
15.
Write
(1
to four terms
~
^
^)
"
(l-2^)-\
down
coefficient of a?' in the
the
expansion
rf(l-*:)-2.
16.
the
doAvn
Write
sixth term
in the
expansion
of
Sx-y)-l
17.
z
-
1 and
Expand
6
=
-
five terms
to
the fourth
(a-36)""V:shew
is
term
greater than
that
it
either tho
o
third
or
18.
the fifth.
Write
down
the coefficient of x*" in the
expansicm
of(l-;c)-^
11^ Expand (l+x
+
x^)-
Expand {l-x
+
x^)~^
to four
20.
to
four terms
terms
in powers
of -u.
in powers
of
".
268
OF
SCALES
XLIII.
NOTATION
Scales of Notation.
have learned from
The student will of course
of expressiiijj
method
the
in
that
Arithmetic
ordinary
the number
whole numbers
representedby each
by figures,
figureis alwayssome multipleof s'tme power often, Thiw
in 523 the 5 represents 5 hundreds, that is 5 times 10';
the 2 represents 2 tens, that is 2 times 10^; and the 3,
which represents 3 units,may be said to represent 3 times
432.
10'^;see
Art. 324.
common
or
radix
is called the
numbers
ten is said to be the base
(rf expressingwhole
This mode
and
scale of notation,
scale.
of the common
shew
We
shall now
that any positiveinteger
of 10 for the radix ;
be
than
used
instead
unity may
greater
be
and
then explain how
a
given whole number
may
433.
proposed scale.
The figuresby means
of which a number
When
called
we
are
digits.
speak in future
that this radix is
shall always mean
we
integergreater than unity.
expressed in
434.
any
To shew
in terms
that any
whole
number
is expressed
of any
radix
positive
some
he expressed
may
radix.
of any
Let N denote the whole number, r the radix. Suppose
that r" is the highest power of r which is not greater than
mainder
N; divide iV by r"; let the quotientbe a, and the reP : thus
N=ar*+P.
is
a is less than
Here, by supposition,
r, and P
than *-*. Divide P by r*-^-let the quotientbe b,and
remainder
Q
:
less
tl\e
thus
in this way until the remainder
Proceed
thus we
find N expressed in the manner
is less than
shewn
+
"r*"^
i
by the I
"
identity,
following
iV="zr"
r :
+
cr"~2+
+hr
+
k.
Each
one
any
h, k is less than r; and
digitsa^b,Cj
dfter
the first may
of them
happen to be
of the
or
260
NOTATION.
OF
SCALES
more
zero.
To
scale.
435.
given
a
express
whole
number
in any
posed
pro-
number
we
mean
a whole
By a given whole number
expressed in words, or else expressedby digitsin some
scale
assignedscale. If no scale is mentioned the common
is to be understood.
Let N be the given whole number, r the radix of the
scale in which it is to be expressed. Suppose k, h,..c, b,a
the requireddigits,
with that
n + 1 in number, beginning
the righthand : then
on
N=ar^+br^~^
+
cr''~^+...+hr
+
k.
be the quotient;then it is
Divide iVby r, and let M
obvious
that M=af'''^
+ br*~^-h
that the
+ /*,and
is k.
Hence
the first digitis found
remainder
by this
rule: divide the given number
by tlis proposed radix,
is the firstof the required digits.
and the remainder
J/ by r ; then it is obvious that the
is /t; and thus the second
remainder
of the required
digitsis found.
Again,divide
By proceedingin this
all the requireddigits.
436.
We
Transform
shall
now
32884
way
solve
we
some
shall find in succession
examples.
into the scale of which
the radix
is
seven.
7 132884
7 46!)7...5
7|67J_...0
7[Q-^...6
7J23...4
1 ...6
Thus 32884
1 .7"+ 6.7* + 4.7' + 6.7' + 0. 7^ + 5,
80 that the number
expressed in the scale of which
radix is seven
is 164605.
=
the
270
SCALES
Transform
twelve.
74194
NOTATION.
OF
into tiie scale of which
12
tlie radix w
I74194
{6 182... 10
12|_515...2
12
12 j42. ..11
3...G
Thus
74194=3.
12^
+
12^+11.122+2.12
6.
+ 10.
In order to express the nuiuber in the scale of which
che radix is twelve in the usual manner,
we
requiretwo
new
symbols,one for terijand tlieother for eleven: we will
and e for the latter. Thus tlio number
t for the former,
use
in
the
scale of which the radix is twelve is
expressed
36^2^.
is expressed in the scale of
the radix is nine,into tliescale of wluch the radix is
Transform
which
645032, which
eight.
8
[645032
72782...
The
division
4.
by eightis performed thus
:
First
eight is
have to find how
often eight is
not contained in 6, so we
contained iu 64; here 6 stands for six times nine, that is
often is eight conthat the question is how
so
tained
fifty-four,
is seven
and the answer
times with
in fifty-eight,
Next
two over.
have to find how often eight is conwe
tained
in 25, that is how often eightis contained in twentythe answer
is twice with seven
Next
over.
we
have to find how often eightis contained in 70, that is how
often eight is contained in sixty-three,
and tlio answer
is
Next
have to find how
times with seven
we
seven
over.
often eightis contained in 73, that is how often eight is
is eighttimes with
and the answer
contained in sixty-six,
two over.
Next we
have to find how often eight is contained
in 22, that is how often eightis contained in twenty,
and the answer
is twice with four over.
Thus 4 is tlie first
of the required digits.
three,and
of the process ; the
"We will indicate the remainder
student should carefully
work it for himself,
and then com-
verifying
examples. Thus, take two numbers, expressed in
the common
stale,and obtain their sum, their difference,
these into any proposed
and their product,and transform
the
into the proposed
numbers
transform
scale; next
and their proscale,and obtain their sum, their difference,
duct in this scale ; the results shouUl of
with those already obtained.
spectively
Examples.
1.
2.
3.
4.
5.
radix
agree
twelve; transform
XLIII.
them
to
the
scale aud
common
together.
becomes
6.
Find
in what
scale the iinmber
4161
7.
Find
in what
scale the ninnbcr
.^261 becomes
Express 17161
divide it by te in
S.
9.
ro-
Express 34042 in the scale whose radix is five.
Express 45792 in tlie scale whose radix is twelve.
Express 1866 in the scale whose radix is two.
Express 2745 in the scale whose radix is eleven.
Multiply eAt by te-,these being in the scale with
multiplythem
and
course
Find
in the scale whose
that scale.
the radix of the scale in which
radix
is
lOIOl.
4020.5.
twelve,
13, 22, 33
are
in geometricalprogression.
10.
whose
Extract
the square
radix is twelve.
root
of eetOOl, in the scalo
272
INTEREST.
XL
IV.
Inter esU
The
subjectof Interest is discussed in treatises
on
Arithmetic; but by the aid of Algebraicalnotjitiou
the rules can
be presentedin a form easy to understand
438.
and to remember.
Interest is money
paid for the use of money.
The money
lent is called the Principal
The Ainount
at
end
of
the
of the Principal
a given time is the sum
and the
Interest at the end of that time.
439.
Interest is of two kinds,simple and compound.
When
interest is charged on the Principal
alone it is called
simple interest ; but if the interest as soon as it becomes
due is added to the principal,
and interest charged on the
whole,it is called compound interest.
440.
The rate of interest is the money
paid for the use
for a certain time.
In practice the sum
of a certain sum
is
and
the
time
is
and
when
one
we
usually^100,
year;
say
that the rate is "4. 5s. per cent, we
that "4, 5*.,that
mean
is "\\, is paid for the use of "100 for one
year. In thcor-ij
shall see, to use a symbol to denote
it is convenient,
as we
the interest of one pound for one year.
441.
442.
time at
To find the amount
of a given
in any
sum
given
simple interest.
P be the number
of pounds in the principal,
n the
of years, r the interest of one pound for one
year,
of
expressed as a fraction of a pound, M the number
of
in
the
the
amount.
Since
is
interest
one
r
pound
pounds
for one year, Pr is the interest of P pounds for one year,
and wPr is the interest of P pounds for " years; therefore
Let
number
M==
443.
the four
found:
P
+
Pnr
=
P{l
+
7ir).
nr\ if any three of
equationM=P{1+
M, P, n, r are given,the fourth can be
quantities
From
the
thus
r"
^*lT^'
M-P
_M-P
^
^~
Pr
'
^"
Pn
'
\
274
At
EXAMPLES,
XLIV.
ample interest
M^P{\
^
At
In
interest of a
the discount
448.
stead of
mediate
442;
D=M^P=^.
interest
compound
therefore
Art
""nr\ by
Jf=Pi2-,
byArt444;
P^g;
D^M-P=^^^^^.
practiceit
is
sum
as
allow the
of money
paid before it is due instead of
here defined.
Thus
at simple interest in-
the
^
payer
to
common
very
would
be
allowed
Mnr
for im-
payment
Examples.
At what
interest in one
rate
1.
yeur
XLIV.
will "a produce the same
per cent
"h produces when
the rate is "c
as
per cent 9
2.
Shew
that
a
of
at
money
given rate per cent
sum
becomes
greater at a
than
it does at twice
of years
of years.
that number
3.
Find
itself at
4.
interest
years.
a
that
interest
for
number
rate
a
given
per
of money
many
years a sum
rate of simple interest.
in how
given
compound
cent
for half
will double
nomial
Shew, by taking the first three terms of the Bithat at five per cent, compound
series for (1 4-r)'*,
a sum
of money
will be
more
than doubled
in tiiteen
Examples.
Miscellaneous
1.
Find the values when
a^ + Za^h + Mh'^
+
275
EXAMPLES.
MISCELLANEOUS
6 and 6
a=
h^,of a2+10a"
+
=
4 of
9"2,of(a-6)",
and of {a+ 96)(a" 6).
2.
5:c
Simplify
3.
Square Z
-
3
+ 9y
[2;p
-
2
{3a; 4 (y a;)}].
-
-
5x + 1x\
"
Divide 1 by l~x
+ x^ to four terms
to four terms.
1" arby l-j^
4.
^
^-
Q.
6.
Find
,.^^^P^y
:
also divide
4;c3-17:"+12
6:^-m^l2-
the
42?^- 9,
of
l-C.M.
6jr--5.r-6, and
6a;24.5;|._6.
a
X
7.
Simplify
"
a
1
solre
8.
^
a
+
-
-
+2
^
+
^-2
"
X
-2
-+-
a? +
+
"
/
7a;-6
5
=
-5-.
The first edition of a book had 600 pages and was
divided into two parts. In the second edition one quarter
of the second part was
omitted,and 30 pages were added
the two parts of the
to the first part; this change made
same
length. Find the number of pages in each part in
the firstedition.
9.
exceeded
the
other by one third of the less,
the changeout of a "5 noto
difference
bills
of eacli
half the
of the
was
: find the amount
bill
10.
In
11. Add
paying two
one
bills,
together2/+-2r--aj,
and from the result subtract
of which
z +
^s-^y, x+^y-^z;
-x"y^-.z.
18"2
27()
MJSCELLANEOUIS
12.
If
a
=
1, "
EXAMPLES.
3, and
=
c
5, find the value of
=
2a"4-fe34.c3+a2(6--c)
+ 62(2a^c)-K;2(2a.^")
2a3-63+c8+a2(ft-c)-62(2a-c)4-c2(2a+")*
13.
14.
SimpUfy(a+")"--(a+""X"-*)-{a(26~2)-.(6"-2a
Divide
by a^"ie^-"2t^.
2ar*-"V-4^pV"''5^2/'-42/'
a?*-2a;3+^_l
15.
Reduce
16.
Find
19.
Solve
to its lowest terms
it^-k-a^+l
the
of
L.C.M.
i(a:-3)
?(a:-l)-|(a?+2)
4.
+
^
20.
=
4
o
persons A
Two
a;"-9;"--10,a^-7a:-30,
and
jB
own
together175 shares in
divide,and A takes 85
They agree to
railwaycompany.
shares,while B takes 90 shares and pays ;"100
a
the value of
21.
Add
22.
Find
a
to A.
Find
share.
together a+2;p-j^+246, 3a-4j"-2^-8l6,
2/-2a4-55";
and subtract the result from 3a+"+3a?+2y.
^
+
when
.d?h
the
value
"=3, 6=2^, and
of
-(2a-36)",
y^+^7a6(2c2-a5)
c=2.
23.
Simplify{x{x-^a)-a{x-a)}{x{a}-a)-a{a-'x)}.
24.
Divide
X^
"X^
1
X
""-
"
-
gg"
"
result by multiplication.
g
26.
Findthea.c.M.
4
^7
1
X
*"^
-
3
2
'
^^^^^ ***"
ofa?4+3;c2_10andar*-34:24.2.
MISCELLANEOUS
26.
SimpUfy
27.
Find
28.
Solve
^^^2
-
the
+
t:Tir
of
l.o.m.
_--
277
EXAMPLES.
+
^
tt^
"
a^-4:, 4a:'-7a?-2, and
S__=4.
A
29.
man
bought a suit of clothes for M. Is. Gd.
The trowsers cost half as much again as the waistcoat,
and
the coat half as much
again as the trowsers and waistcoat
together. Find the priceof each garment.
sells
number
of bushels of
and
bushels
of barley at
200
wheat at 75. 6d. per bushel,
and receives altogether
4s. 6d, per bushel,
as much
as if he
wheat
and
the
at
rate
both
had sold
of 5*. 6d. per
barley
much wheat did he sell1
bushel.
How
farmer
piece of work in one hour,B and G
hours: how long would
each in two
A, B, and C take,
working together?
39.
A
do
can
a
having three times as much money
pounds to B, and then he had t^vice as
40.
two
had.
A
How
Add
42.
Find
together
2x
+
3y
+
4Zf x-2y
44.
Simplify
2a~3("~c)
Find
as
B
5Zf and
product of
-
the
{a~2(6-c)}-2{a-3("~c)}.
+
of
g.c.m.
a;^+ 67a?2+ 66
45.
+
the sum, the difference,
and the
3a?2 Axy + 4y^ and 4;i;2
+ 2xy
'Sy\
-
43.
gave
much
had each at first?
much
41.
B
as
and
x*-^-2x^ + 2a^ + 2a}+\.
^^
a^^-i
x
l.o.m.
of ;c2 4^ ^^2 5^ ^ g^ ^^^ x^-9.
^'^^
Simplify
x* + 2j^'+2x^+2x
+
1'
46.
Fiod the
47.
Reduce
48.
Solve
3(aj^l)-4(a?-2) 2(3-a:).
49.
Solve
J{9
_
_
""
to its lowest terms
~'^~
=
+
4x)
5-'2^x.
=
How
much
be mixed
tea at 3^. 9c?. per lb. must
with 45 lbs. at 3*. 4c?. per lb. that the mixture
be
may
worth 35. ed. per lb.?
50.
Multiply3(x2+
producthy a + b.
51.
the
Find the
2ar'-5x^'-6x+15.
62.
63l
6'
a"-
g.o.m.
of
hy a?-2ab-2b\
l+x
divide
+ 3(a?-6|)+ 15 and
2a;(;r-3)
Simplify
i__L_
*
and
i__L
1-
"4.
Simplify "^"-l-^-^-^.
56.
Solve
-+-
66.
Solve
ar +
=
68.
Solve
3^--
=
-.
2,
2(;c-3)-
3,
^(y-3)
3(y-5) +
|^"-2)
=
10.
=
Solve
7yz=lO{y
r"
59.
-26
"
=
y
67.
2
7
1^(1+2^
=
--
-^,
3
279
EXAMPLES.
MISCELLANEOUS
.
Solve
+
z), Zzx=4{z
-
+
x\
9xy
=
20{x+y%
ha
b
a
+
=w,
-
=n.
of a certain fraction exceeds the
denominator
be increased by 5 the
numerator
by 2 ; if the numerator
fraction is increased by unity: find the fraction.
60.
The
61.
Divide
^--5
by
a?
X
33a:"-49a;-10
to its lowest
62.
Reduce
63.
Simplify
Solve
^.
a
65.
1
Solve
2-^^_-^^_^_-^,
Ja-^w(f.l-l).
^
64.
terms
x'
3(a;-l)+ 2(a?-2)=;r-3.
"-l
^-
y+l
66.
Solve 5:r + 2
67.
Solve
-^
L,
=
2;c-3
5
=
3y,
13-2y
=-^.
6xy-l0x'^^^^
S.
3- -3,
=
-4"+
g
=2*
280
MISCELLANEOUS
68.
Solve
^
Solve
"
69.
x-i-A.
sJ{x'^-^AO)
=
x^ + ^x
,
EXAMPLES
+ 2
5x
X'-x-6
-^^-j
=
J72-
Y-
A father's age is double that of his son ; 10 years
three times that of his son : find
ago the father's age was
the presentage of each.
70.
Find the value whence
71.
=4
of
^(2x.l)-(,.-l)-(3-;^
72.
^"^ '*^ ^''''^^
Reduce
2a^-Ux^+na!~6
and find its value when
73.
Resolve into
and x'"6x
a;
=
""''
3.
simplefactors x^"
3x
+
2,a^
"
*Jx+ lOf
+ 5.
^_3^^^^4.^_/^^^^-^_
7^-
Simplify
75.
Solve
76.
Solve 9^' -63;c
|(5;r-l).
j^(3;r+^^)-^(4;c-2|
=
+ 68
0.
=
and a boy beingpaidfor certain days'
work,
77. A man
been
had
who
the man
and
received 27 shillings the boy
absent 3 daysout of the time received 12 shillings:
had the
instead of the boy been absent those 3 daystheywould
man
both have claimed an equalsum.
Find the wages of each
per
day.
Extract the square root of 9^-*-6a^ + Tx^-2x+l;
and shew that the result is true when ;c
10.
78.
=
79.
If
a
:
b
:: c
a^c + ac"
80.
a*-"d}
81.
by 8.
:
:
d, shew
Vd+hd^
::
that
{a-"cf: {h+ df.
If a, b,c, d be in geometrical
shew that
progression,
\9"
greaterthan b^ + c'.
If
n
is a whole
nunSber 7'""*''
+ 1 is divisible
positive
282
MISCELLANEOUS
94.
Solve
95.
Solve
"y +
EXAMPLES.
+
^^
3
=
-y-
",
20(a?-2/) 0, y5r4-30(t^-2r)
0,
=
96.
=
Solve 2a^-1x+
V (3.^2 4:r
_
3dy-2xf=0.
6)
1 8 + 2a:.
=
-
He leaves
at the rate of 8J miles an hour.
time that B leaves Ely. A spends
Cambridge at the same
in Ely and is back in Cambridge 2 hours and
12 minutes
at the rate of 7^
after B gets there. B rows
20 minutes
Find the distance
miles an hour; and there is uo stream.
97.
from
sold
rows
Cambridge
Ely.
to
apple woman
findingthat apples have this
become
so much
cheaper that she could sell 60 more
lowered her priceand
she used to do for five shillings,
them one penny per dozen cheaper. Find the price
98.
year
than
A
An
per dozen.
99.
Sum
and to
to 8 terms
12
infinity
+ 4 +
1J +
...
Find
three numbers
in geometricalprogression
if
such that
1, 3, and 9 be subtracted from them in order
is 15.
whose sum
they will form an arithmetical progression
100.
101.
'iUvMiyl^x^-a^-^-x^-x^
x^-^-i;
+ x^-x+x^-lh^
and divide \
102.
Find
a^
103.
x*
"
"
by
the
1
+
.
of a?
L.C.M.
ax'"a^x
Simplify
ar
-
-a*,
a^
+
a^,x*
+
o^,B.ndai^ + aa^-a'x"d^,
^52"
"
"
a +
6
,
1+
~^,
a
104.
x+5
105.
"
b
Solve
l/x
Solve
2\
-1x~l
+
4;"-14
2/^
-^
x~5
=;
a?+10
,
"
-^l -i^,
5
+
x
+
x
+
aW
+ a
106.
283
EXAMPLES.
MISCELLANEOUS
aj2+y3+ 5?2=50,
Solve
xy^yz"zx^^l.
and
B
travel
miles
rail. B
back again takes a return
ticket iot
intending to come
much
A
and
which he pays half as
again as
they find that
;
for
than
^d.
A by 4s.
B travels cheaper
every 100 miles.
the
of
A'e,
ticket.
Find
price
107.
A
108.
Find
between
3
third
a
the
the harmonic
mean
between
geometric mean
_,
and
togetherby
to
proportional
and
and
120
2
18-
109.
yj
y\
110.
If
111.
Dividea;a
112.
Reduce
a
6
:
"
::
:
x\
3^
^
X""
"
.^'
fl:"-
,a;"
.
20
to
"
Xr
r^
its lowest
terms, and
1.2
~~
^=2,
a?+2
13
x-S
Solve
113.
4
Find
114.
+
by
-! ^.-7
n
n
"
3^7'
+
j
yj
that h^=
8n
-a?"
"
find its value when
X
c, shew
8n
m^x^
root of
Exta*act the square
3(6-a:)'
3
for which the equation
of m
will
have its roots equal to one
0
the values
(m' + 7n)ax-i-a^
=
another.
115.
116.
Solve 3^:2/
+ :f*= 10,
Solve
-
X
+
-
y
=
5,
'
-
y
5xy -2x1^=2.
+
^
=
X
.
21.
""
Find the fraction such that if you quadruple the
117.
and add 3 to the denominator
the fraction ia
numerator
and quadruple
doubled ; but if you add 2 to the numerator
the denominator
the fraction is halved.
284
EXAMPLES.
MISCELLANEOUS
118.
{-(-a;)-'}*.
Simplify{-(x')M"*x
The third term of an arithmetical
18 ; and the seventh term is 30: find the sum
119.
120.
shew
If
"
6,
X-,
that a, ",c
of 17 terms.
be in harmonica!
"^
in
are
progressionis
progressioi
geometricalprogression.
1
121.
Simplifya"
+
a-h
122.
the square root of
+ 9:c* 20a;y"
^Ixhp- 30a:"y
+ Ay^,
Extract
-
-
123.
Resolve
,".
1
a
Solve
124.
S^r*- I4a^-24ca! into its simple factors.
^
3(5a:+l)
-^
1
+ 5
j;
2a;-l
5ii?+ 4
125.
Solve
^+J5=y.
126.
Solve
a;^-y^ 9,
127.
Solve
y
128.
If a,
=
+
^(^^
6,c, d
a
are
:
4
=
r
6
+
a; +
2a;-l
4
=
-24.
*
3(i/-l).
i) 2, J{x + 1)
=
_
-
^(a? 1) s/V-
-
in Geometrical
fl?:: "?
:
Progression,
c^d+(P.
difference in an arithmetical proThe common
gression
is equal to 2, and the number
of terms is equsd to
the second term : find what the firstterm must be that the
sum
may be 35.
129.
130.
Sum
to
n
terms
Find the o.c.m.
132.
24a:* + 14a^-48a!"-32"r.
the series whose
wi*^ term
of 30a?*+ 16a?3-50j;'-24:c
is
MISCELLANEOUS
133.
Solve
a^-x-l1
134.
Form
a
3 and
-
shall be
roois
2.
^+^ ^^^"
136.
Solve
137.
Having given ^3
"
Q.
quadraticequationwhose
Bolve
-
1^o
=
135.
-7T
28A
EXAMPLES.
"x=l4-
o
.,
=
1-73205, find the
value
of
to five placesof decimals.
1
138.
Extract
139.
Fiud
root of 61
the square
-
28
JZ.
x
the
-\-v
between
proportional
mean
aud
-
_
If a, 6,c be the first,
second and last terms
arithmetical progression,
find the number
of terms.
of the terms.
find the 8um
140.
A herd cost "180, but on 2 oxen
the
147.
beingstolen,
first
than at
rest average "1 a head more
: find the number
of oxen.
Find
148.
numbers
two
their
when
is 40, and the
sum
5
sum
of their
Find
149.
third
ia
reciprocals
a
--
.
to 2^ and 6f ;
proportional
mean
100. and
to
proportional
and
130.
If 8
gold coins and 9 silver coins are worth
much as 6 gold coins and 19 silver ones, find the ratio
the value ol a gold coin to that of a silver coin.
150.
Remove
151.
a
as
of
the brackets from
{x-a){x-b){x-c)-\pc{x-a)--{{a-\'h-"c)X''a{h-^c)]x
and
^{a^h)+
162.
Multiplya
+ 2
153.
Find
g.c.m.
4d:"
-
the
48.cS+ 1^^
_
234.
2
^/"by a
-
2
*J{a%)+
of a^-16aj'4-93a?'-234.r-f
2
^b.
216
MISCELLANEOUS
164.
Solve the
^^
3oJ +
Solve the
3o?+l
28-50?
_
9~3or-l"3*
3C4)=n(i-|).
(3, .-.=3,
155.
287
equations:
following
130?- 1
",
EXAMPLES
equations:
following
+ ^/(2o:)=7.
(1) ^/(o;+l)
(2) 7o:-20.v/o?3.
=
(3) 7o?"/-5of2=36,4ory- 3^2=105.
in oranges ; if he had
money
for
his
they would have averaged
bought 5 more
money
more
: find how
an
half-pennyless,if 3 fewer an half-penny
156.
much
he
157.
boy spends his
A
spent
Potatoes
6 lbs. for bd.
:
are
find the
sold
so
gain per
as
to
cent,
gain 25 per cent,
when
they are sold
at
at
5 lbs.for e^;.
A horse is sold for "24, and the number
the profitper cent, expresses also the cost
of the horse : find the cost
158.
159.
pressing
ex-
price
Sunplify
^{Aa^+ ^{IQa^x^4- Saaj*+ or*)}.
160.
If the
161.
the following
:
Simplify
expressions
of two fractions is unity,shew that the
firsttogetherwith the square of the second is equalto the
second togetherwith the square of the first.
sum
a~\h--{a+ (Jb-a)]l
26a-
19"-[36-{4a-(56-6(j)}]-8a,
[{(a-)-}-^]-[{(""-r"nn
288
MISCELLANEOUS
EXAMPLES.
162.
Find
the
g.O.m.
of 18"z'-18a2;c+
163.
Find
the
L.C.M.
of
24(^
+
and
6^^2-6:^3,
\S{a^-y^l I2{x-y)\ and
2/3).
164.
Solve the
equations:
following
(1)
(2)
(3)
(4) 2(a?-y) 3(a;-4y),
+ y) ll(a?4-8)
14(a?
=
165.
Solve the
(1)
=
following
equations:
32;c-5a?2=l2.
(2) ^{2x + ^)J{x-2)=\5.
(3)
a;2+
y2=290,
;ry
=
143.
(4) 3;B2-4y^=8, 5a^5-6;C2/32.
=
-4 and B togethercomplete a
would
have occupied A alone 4
it employ B alone ?
166.
which
would
work
in 3
days: how
days
long
2
167.
Find
two
of their squares,
96
times
the
numbers
whose
productis
-
of the
sum
and
is
the difference of their squares
quotientof the less number divided by the
greater.
168.
Find
a
fraction which
becomes
-
on
its
increasing
o
numerator
nator.
by 1, and
-
on
its denomiincreasing
similarly
290
MISCELLANEOUS
176. "When are
after 12 o'clock ?
EXAMPLES,
clock-hands at rightanglesfirst
the
divided
gireaas quotient2, and the
27 : find the number.
177.
number
A
by the product of its digits
digitsare inverted by adding
A bill of "26. 15". was
and
paid with half-guineas
exceed^ the numand the number
of half-guineas
ber
crowns,
of crowns
by 17 : find how many there were of each.
178.
to six terms
179.
Sum
180.
Extract
181.
If
182.
Reduce
183.
If two
and to
12
infinity
the square root of 55
"=^|^,and y=4lT-i
^^^
"
3^2
same
digitsin
the numbers
184.
to its lowest terms
a
Solve the
5^ +
..
..
_
t^e yalue of
\Qx"
12
a^-^a^-Ux+l'k'i'
equations:
following
(1)
21-4a?
3^-4
3a;-3
9
3
4
2a; +
185
+
be expressedby the
of two digits
numbers
reversed order,shew that the difference of
be divided by 9.
can
(2)
8
/v/24.
7
-
+
3y
"
g"
ly-^^r.
x_
+
3-8,
2
2/-11.
Solve the following
equations:
24.
(1) ^(a;-|.3)xV(3a?-3)
=
(2) ^(a?+ 2)+ ^(3;c+ 4)
=
(3) a^'-x^{2x-2)
=
8.
2x-\-^
of 9 to 7
Find two numbers
in the proportion
shall be equal to the
such that the square of their sum
cube of their difference.
186.
291
EXAMPLES
MISCELLANEOUS
traveller sets out from A for B, going3^ miles
Fortyminutes afterwards another sets out from
A
187.
hour.
B for Ay going 4^ miles an hour,and he goes half a mile
beyond the middle pointbetween A and B before he meets
the first traveller;find the distance between A and B.
an
188.
Two
189.
li
A bets B
persons A and B play at bowls.
four shillings
to three on every game, and after playinga
certain number
of games A is the winner of eightshillings.
The next day A bets two to one, and wins one
more
game
and
that
he
out of the same
finds
has
to
receive
number,
three shillings.
of games.
Find the number
shew that
m
mn
Sum
190.
=
+
x"
x-'^ and
y~^,
n=y"
J{{m^+ 4)(n'+ 4)}
2
=
9
to nineteen terms
Multiply--3
4
4
o
Reduce
.
t
2
+
....
4
3-2.
+
58^-39
-
Find
193.
^
4
a;* 9ar*+ 29a;2- 39;c
the
Solve the
L. o. M.
*
+
18
of a^-k-2^V+
^'^+ 8t/^and
:
following
equations
(1)
l(a;
+ e)-^{16-Sa:)
4l
(2)
'^^Z^
^?t^
(3)
j
to itslowest terms
4^ -27.^2
194.
+
-
+
"
^-4i!'+^*"-V"'-t*+27
by ^-*+a
DiTide
192.
jby-
+
+
3
3
:, +
4
191.
/^
=
=
s.-20,
4), l{a^-y)^l{a:l{x+y)=l{2x
+
19"2
292
EXAMPLES,
MISCELLANEOUS
195.
equations:
following
Solve the
(1)
\{x-^-^)=\{m-Z),
(2) J(a;+ 3)+ V(3iC-3)
(3)
a? +
=
10.
+ 2^)
(a;2+2,2)(^
y=6,
=
i44o.
and CamLondon
bridge,
The express train between
which travels at the rate of 32 miles an hour, performs
the
in
less
hours
than
the journey
parliamentary
2$
train which travels at the rate of 14 miles an hour: find
the distance.
196.
of two digits,
which
Find the number, consisting
and is
is equalto three times the product of those digits,
of the digits
the
also such that if it be divided by the smn
quotientis 4.
197.
of a certain
The number
of resident members
in
ber
college the Michaelmas Term 1864, exceeded the numin 1863 by 9.
If there had been accommodation
in
in
number
1864 for 13 more
students in college
the
rooms,
in lodgings,
collegewould have been 18 times the number
and the number
in lodgings would have been less by 27
Find
the
than the total number
of residents in 1863.
of residents in 1864.
number
198.
199.
the square
Extract
a"
and of
-
la^h
+
root of
ZaW
-
2at^
+
h\
(a+ 6)^-2(a2+ 62)(^+ 5)2+2(^4+ 2,4).
Find a geometrical progressionof four terms
Buch that the third term is greater by 2 than tlie sum
of
the first and second, and the fourth term
is greater by 4
of the second and third.
than the sum
200.
201.
Multiply
8-3x+'^-"^-'8
T-2x
7^' -55
by
^-2x
+
30a;
+
6-Zx
202.
Find the
g. c. m.
of x*
+
4:X^+ 16 and x*-x^
+
8x-S.
I
MISCELLANEOUS
203.
Add
together^^,
Take
\"x
\+x+x'^
204.
^-5^..,^^^^.
from
r
,
293
EXAMPLES
Solve the
+
x^
following
equations:
(2) {a+ h){a-x)^a(p-xl
^^
205.
Solve the
(1)
^
'
*'
12
16
a
"
following
equation*:
6x+
=
44.
X
+ Sx)-2^{x^
(2) 4:{x^
(3)
;r2+ :cy
=
+
3x)
15, y^+ xy
=
=
l%
lQ.
A person walked out from Cambridge to a viTIago
at the rate of 4 miles an hour, and on reachingthe railway
for the train which was
station had to wait ten minutes
which were
then 4^ miles off. On arrivingat his rooms
the Cambridge station he found that he had
a mile from
Find the distance of the village.
been out 3j hours.
206.
of a number
is less by 2 than the
The tens digit
207.
and if the digits
units digit,
number
ai*e inverted the new
is to the former as 7 is to 4 : find the number.
of money
consists of shillings
and crowns,
A sum
208.
of crowns
and is such, that the square of the number
Is
also the sum
is
equal to twice the number of shillings;
of
florins
there
worth as many
as
are
: find
pieces
money
the sum.
209.
Extract the square root of
4x*
210.
firstterm
+
8ax^
+
4a2:"2+ isb^x^ + 16ai^x
+
16ft".
Find the arithmetical
progression ol wmcn
is 7, and the sum
of twelve terms is 348.
ine
294
MISCELLANEOUS
Divide 6a?"
211.
-
by 2;"2-7^
+
25^y
4- 47^V
-
49.iV + 62:ry* 45y"
-
92/2.
Multiply
212.
12
"
EXAMPLES.
+
^
2 + ^^
213.
41a: + 36;i:?
u
4T7^"
Reduce
26j7-8:p"-14
^y^-^^^
"
o
'
3-4;r
to its lowest terras
4ar3-45.g^-4-162.r-185
ar*-15;c* + 81^''- 185.C+
214.
Solve the follo^Ning
equations:
^^'
11
6
(2) a?+-y=17,
^^^
216.
160'
t/+^^.^=8.
i^y~2' ^"^^"9' y'^^'lS'
Solve the
following
equations:
^^
a;
ar +
3
6
(2) lO;Fy-7;i?27^ 5y2_3ary
=
=
(3)
a?4-2/
=
6, a;*+ 2^
=
20.
272.
Divide "34. 4*. into two parts such that the number
in the one
of crowns
of
may be equal to the number
in the other.
shillings
216.
217.
A
of three digits
whose
number, consisting
sum
is
equalto 4*2 times the sum of the middle and left-hand
of the
digits;also the right-hand digitis twice the sum
9, is
otiier two
:
find the umnber.
of railwayshares when
person bought a number
at a certain price for i,'2U25,
and afterwards
they were
when
the priceof each shar^ was
doubled,sold them all
but five for "4000 : find how many shares he bought
218.
A
in arithmetical
productof the second
numbers
is 50, and the
their sum
156 : find the numbers.
219.
Pour
Extract
221.
Divide ^-
root of 17 + 12
by or*- 1
1
;
and
Find
21;c2_26^
the
Solve the
7:5^-4*2-214?+
12
and
fdlowingequations:
?^-?^=7.
(2) 17^-13^
/3)
^ '
144, 23;p + 19y
1
i
=
y
Solve the
'
=
1-1
X
^
of
l. c. m.
8.
+
(i)
225.
mx-n,
Simplify
223.
224.
and third ia
J%
Tn{qji^-rx)+p{Tnx^-njF)-n{qx"r)hj
222.
progressioa;
are
the square
220.
2%
EXAMPLES.
MISCELLANEOUS
100
%'
X
+
l
i
1"
89a
=
i
A
=
=
^'
z
z
y
72*
following
equations:
l^x
(2) "0075;J"+
4
-75:1?=150-
(3) J{x-^y)-"j{x-y)
6("-a)
+
=
j",
a("-y)=0.
A person walked out a certain distance at the
226.
rate of 3^ miles an hour,and then ran part of the way back
tance
at the rate of 7 miles an
hour, walking the remaining disHe was
in 5 minutes.
out 25 minutes : how far did
he run?
leaves his property amounting to "7500
227. A man
to be divided between
his wife,his two sons, and his three
(Laughtersas follows: a son is to have twice as much as
296
a
MISCELLANEOUS
EXAMPLES.
than all the fivechilthe widow ^500 more
chO-fl
dren
much
each person obtained.
: find how
together
daughter,and
be
filled by two pipesin 1^ hours.
than
The larger pipe by itself will fill the cistern sooner
Find what time each will sepathe smaller by 2 hours.
rately
228.
cistern
A
can
take to fillit
The third term of an arithmetical progression
is
and
first
the
sixtli
term is 17: find
four times the
term;
the series.
229.
230.
Sum
231.
:
Simplifythe followingexpressions
to
3^ + 2^ + If+
terms
n
6
a +
6
2a
a'-ah
+
+
.
..
o2+"2
h
2a(a-")*
b''-
a^-b^
X
a?2+ lia?+30
to its lowest terms
232.
Reduce
233*
Solve the following
equations:
^^
x'^2x
(2)
^ '
^
1 +
Solve the
(1)
^
+
'
(2)
3*
3x
-I=
8.
\-x
a;
2x"y
5y
4a; +
-".
234.
9a;3+ 53a;2-94?-18
1
^
following
equations:
J8_^_16^_
x
'S
+
aa^
+
a?+10
b^+c^=a^
+
2bc
+
2(p-c)x^a,
(3) ^/(^+y)+^/(a;-y)=4, x' + y^=4l.
298
MISCELLANEOUS
EXAMPLES
"
the
bought
at
borrow
355.
they at
244.
number
leaves
a
to
p^ce 350 lbs.,but
complete the payment.
same
I to
fA
obliged
was
How
much
had
first?
two digitsof a number
are
is subtracted
thus formed
from
of the
remainder equal to the sum
The
Find three numbers
the third of which exceeds
the first by 5, such that the product of their sum
plied
multi245.
by the firstis 48, and
by the third is 128.
the
product of
their
sum
tiplied
mul-
lends
A
at a certain rate ot
"1024
person
interest ; at the end of two years he receives back for his
of "1150:
capitaland compound interest on it the sum
find the rate of interest.
246.
of money
I take away
247. From
"50 more
a
swn
than the half,then from the remainder
than the
"30 more
from
than
the
second
remainder
the
then
"20
more
fifth,
find the original
fouith part; at last only "10 remains:
sum.
248.
numerator
Find
such
a
fraction that when
its value becomes
-,
2 is added
and when
1 is
to the
taken fron
o
the denominator
its value becomes
-
.
4
the smaller of two numbers
by the
is
'04162 ; if
greater,the quotientis '21,and the remainder
I divide the greater number
by the smaller the quotientis
4, and the remainder is 742 : find the numbers.
249.
(3) a-x=J{a'^-xJ{Aa^'-*la^).
Divide the number
such that
208 into two parts,
the sum
of one quarter of the greater and one third of the
less when increased by 4, shall equal four times the difference
of the two parts.
255.
Two
A
men
purchase an estate for ^9000.
if
the
whole
B
him
while
B
half his capital,
pay
gave
could pay the whole if A gave him one-third of his capital
:
find how much
each
of
them
had.
money
256.
could
piece of ground whose
breadth by 6 yards,has an area
of
257.
A
length exceeds the
91 square
yards : find
its dimensions.
258.
A man
buys a certain quantityof applesto divide
his children. To the eldest he giveshalf of the whole,
8 apples;to the second he gives half the remainder,
among
all but
all but 8 apples. In the same
also does he treat the
manner
and
third
fourth child. To the fifth he givesthe 20 apples
which remain.
he bought.
Find how many
259.
The
their squares
260.
"\^A,
and
A
of two
horse-dealer
gainsjustas
had cost him.
261.
is 13, the difference of
numbers
is 39 ; find the numbers.
sum
Find
buys a horse,and sellsit againfor
x)oundsper cent, as the horse
many
what
he gave
for the horse.
SimpUfy
("n-6)(a-6)-{a+ "-c-C6-a-c)
+
(t+c-a)}(a-"-c)t
300
MISCELLANEOUS
Multiply a?+af^-\-a:^-"x^+\ by x^-\)
262.
^
X
EXAMPLES
-1
by
+
and
1.
''ax
a
multipliedby
263.
"What
quantity,when
264.
the following
expressions:
Simplify
Za^-lZx'^
r a^h
a?"-,
+ 2Zx-'2.\
2"2
a-h
\a-b
2b
\2{fl-h)
2(o ")'^a2-"2/
'
+
265.
Solve the followingequations
:
"1) ^31^2^31
(2) ^(3 +
(3)
266.
+ 90?=
167.
following
equations:
a^-x-Q
^+1
(9\
^^
.)W-=^.
f +9y=91, ^
Solve the
(1)
6.
=
=
a? +
Q.
2_2^+13
a?-l"'".'c-2~
^p+r*
(3) x^"xy
+
y'^=7,
x
+ j/=
5,
The ratio of the sum
to the difference of two
is that of 7 to 3. Shew that if half the less be
added
to the greater,and half the greater to the less,
the
ratio of the numbers
formed will be that of 4 to 3.
so
267.
numbers
The
priceof barleyper quarter is 15 shillings
less than that of wheat, and the value of 50 quarters of
barleyexceeds that of 30 quarters of wheat by "7. I0s,i
find the priceper quarter of each.
268.
Shew
269.
that
cda
(pcd+
dab
+
dbcf -{a-^l
+
Extract
"
"
the square
root of
5x^
,
.
ar-\-ar
and of
1- -.
33-20^2.
=
272.
Divide
273.
Add
ar*-21^
Za-k-x
974.
MultiplyBx-
,
from
1-^
^r"
-,
-"k
"^i"
^
a?
+
4;g^^_^
+
rf
^^-^"^^
a^
ax
?_0^7.
a;
(2) 5y-3a?=2,
(3) -^
by
following
equations:
'
X
.
1+^-^.
by
(1) l-L2
=.
-^
7a;2
+
Simplify
Solve the
and
,
^^_^^
c
276.
l-3;p+a".
27a*+3"^
.
^r
275.
x-2if, and (;=a?^y-a2r,
+
by
8
+
together
Take
Divide
9'
3
b
271. li a=y-k-z-2x,
z
find the value of b'^+ c^ + 2bc"a\
,
1
X
,
12
_,
dfabcd
+ c-^
{he ad){ca bd){ab cd),
-
"
270.
301
EXAMPLES
MISCELLANEOUS
8y-5a?=l.
2"1,
3+2-4,
a^
.
302.
MISCELLANEOUS
EXAMPLES,
Solve the following
equations:
277.
(1) a?{x-af
+ a)\
J"^{x
=
5x+l
x
,^.
^
(3) ^{nx-\)-sJ{2x-\)
5.
=
at the
person walked to the top of a mountain
the same
rate of 24 miles an hour, and down
at the
way
miles an
and
rate of
out
5
hours: how farfl
was
hour,
did he im^k altogether]
A
278.
3|
^
that the difference between the square of a
number, consistingof two digits,and the square of the
number
formed by changingthe placesof the digits
sible
is diviShew
279.
by
99.
li
280.
'.h :: c
a
:
d, shew that
+ h^): U{(^
:: ll{a?
+ "2): J^c'-^d^)
+ d^).
^/(a2
W'hen
a
the value of
Fmd
281.
3, 6
=
/,
,
+
.
^^-^^
V
.o,
4.
=
Subtract (""a){c" d) from {a b){c-d): what
the value of the result when a= 2b, and d=2c'l
282.
"
Reduce
283.
to their
x^"2ax-24a*
"
-=
7a^"
284.
=
:
x
y
.
1
x-vy
x"y
,
y"x
equations:
_^_i
(1)
'
^'
x-y
and
44a'
Solve the
^
simplestforms
,
--"
=
ar"
-y
Z^-x
=
5
(3) J{2x-\)
A.
Ix
X
2
+
~^'
3^2"^-
s/i^x+lO) J{nx-^%),
=
ia
285.
Solve the
SOS
EXAMPLES.
MISCELLANEOUS
equations:
1-^
9.
(1)
10;" +
"^)
(I-?-0(-M)--.
^
'
=
l-x
(3) iiP'-xy-vy^=*ly5:c-2y=9.
286.
In
a
time
boat is rowed
one
race
the
; another
over
course
at an
moves
yards per second
average pace
of
the
rate
the
at
the
half
of
first
course
over
3^ yards per
half
and
last
at
the
over
second,
\\ yards per second,
seconds
later tluin the firsts
15
the
reaching
winning post
Find the time taken by each.
of 4
rectangularpictureis surrounded
by a narrow
hnear
feet,and costs,
measures
altogetherten
three shillingsa foot,five times as many
as
shillings
A
frame, which
287.
at
there
are
square
length and
288.
feet in the
breadth
li
of the
of the
area
picture. Find
the
picture.
a-.hv.c'.d, shew
that
'b
a-\-l)-\-c-\-d\a^'b
"
c
"
d\\a
'b-\-c-d'.a
"
"
"
c-\-d,
The volume
of a pyramid varies jointlyas the
of its base and its altitude.
A pyramid, the base of
area
is 10
which
is 9 feet square, and the height of which
feet is found to contain 10 cubic yards. Find
the height
of a pyramid on a base 3 feet square
that it may contain
2 cubic yards.
289.
290.
Find the
1
gression
291.
6=1,
l+a;'
Find
c=07.
gum
1
\-3^'
of
n
terms
of the arithmetical pro-
1
\-x
the value of a'
-
fr^
^. ^3 +
3^5^^^hen
^
-.
.93^
304
EXAMPLES,
MISCELLANEOUS
{ac-hdf+{ad-^lcyo^iand
Simplify
c' d^
292.
shew
that
+
293.
If
294.
Reduce
"i +
"+
c
=
0, shew
that a^^'b^^"?
=
Zabc,
to its lowest terms
a:^+ 2^
+
6;c-9
+ 2)^/(4:^?-3)
20.
(2) ^/(2a;
=
+ l)-^/(2;c-l)=l.
(3) ^/(3a;
A siphon would
297.
empty a cistern in 48 minutes,
it is empty both
fiU it in 36 minutes ; when
a cock would
begin to act : find how soon the cistern willbellied.
298.
and he
A
time
up and down.
the
same
299.
and
hack in 12 hours,
in
5 miles "vith the stream
row
can
3 againstit. Find the times of rowing
Algebra for Beginners
Description
Isaac Todhunter's Algebra for Beginners: With Numerous Examples is a mathematics textbook intended for the neophyte, an excellent addition to the library of math instructionals for beginners.
To...
Isaac Todhunter's Algebra for Beginners: With Numerous Examples is a mathematics textbook intended for the neophyte, an excellent addition to the library of math instructionals for beginners.
Todhunter's textbook has been divided into 44 chapters. Early chapters highlight the most basic principles of mathematics, including sections on the principal signs, brackets, addition, subtraction, multiplication, division, and other topics that form the foundation of algebra. Simple equations make up the large majority of the material covered in this textbook. Later chapters do introduce quadratics, as well as other more advanced subjects such as arithmetical progression and scales of notation. It is important to note that Todhunter sticks very much to the basics of algebra. The content of this book lives up to its title, as this is very much mathematics for beginners.
The content is provided in an easy to follow manner. This book could thus be used for independent learning as well as by a teacher. A great deal of focus has clearly been given to providing examples. Each concept is accompanied by numerous sample questions, with answers provided in the final chapter of the book. The example questions are every bit as important as the explanations, as one cannot begin to grasp mathematical concepts without having the opportunity to put them into practice.
The basics of algebra are explained in an easy to follow manner, and the examples provided are clear and help to expand the knowledge of the learner. If given a chance, Isaac Todhunter's Algebra for Beginners: With Numerous Examples can be a valuable addition to your library of mathematics textbooks. | 677.169 | 1 |
Mathematical skills and concepts lie at the heart of chemistry, yet they are an aspect of the subject that students fear the most.
Maths
Opening
With its modular structure, the book presents material in short, manageable sections to keep the content as accessible and readily digestible as possible. Maths for Chemistry is the perfect introduction to the essential mathematical concepts which all chemistry students should master.
The new edition of Maths for Chemistry extends, expands and improves on the excellent first edition. Changes include new chapters covering more introductory material as well as more advanced material, making this the most comprehensive mathematics text book specifically written for chemists that I have seen. REviews, The Higher Education Academy UK Physical Sciences Centre
The book is well written throughout and has an admirable step-by-step approach to teaching ( Chemistry World)
Maths for Chemistry is the best mathematical toolkit currently available to chemists. ( Matthew Ryder, student, Heriot-Watt University)
L'autore:
Revd Dr Paul Monk is Team Vicar at Medlock Head Parish in Oldham and was Senior Lecturer in Physical Chemistry in the Department of Chemistry and Materials at Manchester Metropolitan University, Manchester, where he lectured and researched since 1991. Dr Monk gained a BSc (Hons) in Chemistry and a PhD on electrochemistry from the University of Exeter. His research investigates electrochromism and the development of electrochromic materials, a subject area in which he has published over 40 journal articles and several monographs. He is the author of two textbooks, Electroanalytical Chemistry: Principles and Fundamental Applications (Wiley, 2001); and Physical Chemistry - Exploring our Chemical World (Wiley, 2004).
Lindsey J. Munro received both an MA in Natural Sciences and a PhD in Theoretical Chemistry from the University of Cambridge (Downing College). Postdoctoral Research Fellowships at the University of Pittsburgh and in industry in Switzerland followed. After having compared life as a researcher in both industry and universities, she opted for the freedom of academia and returned home to start life as a lecturer at the Manchester Metropolitan University. She currently lectures in Thermodynamics and Quantum Mechanics, whilst pursuing her research into understanding flavour using computational chemistry. AOP9780199541294 Codice libro della libreria AOP9780199541294
Descrizione libro Oxford University Press. Paperback. Condizione libro: new. BRAND NEW, Maths for Chemistry: A Chemist's Toolkit of Calculations (2nd Revised edition), Paul Monk, Lindsey J. Munro, Opening B9780199541294
Descrizione libro Paperback. Condizione libro: New. Not Signed;. book. Codice libro della libreria ria9780199541294_rkm | 677.169 | 1 |
Features a variety of functions, including equation solving.
Can handle calculus, hyperbolic, logarithm, matrix, statistical and trigonometry functions, as well as polynomial roots and regression analysis. | 677.169 | 1 |
Why choose the Clickable Multivariate Calculus Study Guide?
An interactive e-book designed to help you succeed in your Multivariate Calculus course.
You'll never again think you've misunderstood a concept when really you just lost a minus sign. While you are learning, you can leave the mechanics of the computations to Maple so you can focus on understanding the new ideas.
The many plots and animations help you actually see what is going on in your problem and what your solution means.
With over 700 worked problems taken from more than 50 topics from multivariate calculus, you will always find the help you need. This guide completely covers all topics from a standard multivariate calculus course.
You'll learn how to easily visualize and solve your own multivariate calculus problems in Maple, so you can check your homework and get extra practice.
This study guide makes extensive use of Maple's Clickable Math approach. Almost all the examples in this guide are solved using interactive point-and-click techniques – no commands or syntax required. This means you can let Maple take care of the details while you stay focused on the math. You can easily apply these same Clickable Math techniques to solving other math problems, whether they are from calculus or your other technical courses.
And for those who prefer to use commands, command-based solutions are also provided for almost every example in this book. The choice is yours!
Topics
Vectors, Lines and Planes
Cartesian Coordinates and Vectors
Vector Arithmetic
Dot Product
Cross Product
Applications of Vector Products
Lines
Planes
Space Curves
Position-Vector Representation
Arc Length as Parameter
Tangent Vectors
Curvature
Principal Normal
Binormal and Torsion
Frenet-Serret Formalism
Resolution of R" along T and N
Applications to Dynamics
Functions of Several Variables
Functions and Their Graphs
Limits and Continuity
Quadric Surfaces
Partial Differentiation
First-Order Partial Derivatives
Higher-Order Partial Derivatives
Chain Rule
Directional Derivative
Gradient Vector
Surface Normal and Tangent Plane
Approximations
Unconstrained Optimization
Constrained Optimization
Optimization on Closed Domains
Differentiability
Double Integration
The Double Integral
Iterated Double Integrals
Regions with Curved Boundaries
Changing the Order of Iteration
Numeric Evaluation of Iterated Integrals
Changing Variables in a Double Integral
Double Integration in Polar Coordinates
Applications of Double Integration
Area
Volume
Surface Area
Average Value
First Moments
Second Moments
Triple Integration
The Triple Integral
Iterated Triple Integrals
Regions with Curved Boundaries
Integration in Cylindrical Coordinates
Spherical Coordinates
Integration in Spherical Coordinates
Applications of Triple Integration
Volume
Average Value
First Moments
Moments of Inertia (Second Moments)
Changing Variables in a Triple Integral
Vector Calculus New!
Vector Objects
Differential Operators
Differential Identities
Line Integrals
Surface Integrals
Conservative and Solenoidal Fields
Divergence Theorem
Stokes' Theorem
Green's Theorem
Recorded Webinar: Reviewing the Multivariate Calculus Study Guide
This webinar explores the Multivariate Calculus Study Guide, showing how it uses all the best tools Maple has available for mastering the material of the multivariate calculus course | 677.169 | 1 |
Introduction to Ordinary Differential Equations is a 12-chapter text that describes useful elementary methods of finding solutions using ordinary differential equations. This book starts with an introduction to the properties and complex variable of linear differential equations. Considerable chapters covered topics that are of particular interest in applications, including Laplace transforms, eigenvalue problems, special functions, Fourier series, and boundary-value problems of mathematical physics. Other chapters are devoted to some topics that are not directly concerned with finding solutions, and that should be of interest to the mathematics major, such as the theorems about the existence and uniqueness of solutions. The final chapters discuss the stability of critical points of plane autonomous systems and the results about the existence of periodic solutions of nonlinear equations. This book is great use to mathematicians, physicists, and undergraduate students of engineering and the science who are interested in applications of differential equation | 677.169 | 1 |
Introduction to Probability designed for an introductory probability course at the university level for sophomores, juniors, and seniors in mathematics, physical and social sciences, engineering, and computer science. It presents a thorough treatment of ideas andMore...
Book details
This text is designed for an introductory probability course at the university level for sophomores, juniors, and seniors in mathematics, physical and social sciences, engineering, and computer science. It presents a thorough treatment of ideas and techniques necessary for a firm understanding of the subject. The text is also recommended for use in discrete probability courses. The material is organized so that the discrete and continuous probability discussions are presented in a separate, but parallel, manner. This organization does not emphasize an overly rigorous or formal view of probability and therefore offers some strong pedagogical value. Hence, the discrete discussions can sometimes serve to motivate the more abstract continuous probability discussions. Features: Key ideas are developed in a somewhat leisurely style, providing a variety of interesting applications to probability and showing some nonintuitive ideas. Over 600 exercises provide the opportunity for practicing skills and developing a sound understanding of ideas. Numerous historical comments deal with the development of discrete probability. The text includes many computer programs that illustrate the algorithms or the methods of computation for important problems. The book is a beautiful introduction to probability theory at the beginning level. The book contains a lot of examples and an easy development of theory without any sacrifice of rigor, keeping the abstraction to a minimal level. It is indeed a valuable addition to the study of probability theory. --Zentralblatt MATH | 677.169 | 1 |
Production Algebra : A Handbook for Production Assistants
Description
Have you ever wanted to know what it's like to work on a film set? This handbook is packed with information that will help you get into the film and video industry. Mark Adler provides detailed descriptions of the many roles that a Production Assistant plays on the set. Chapters include types of Production Assistants, specific duties, a day in the life on a set, terms and lingo, as well as examples of forms. This handbook is your launching pad into the film and video industry.show more | 677.169 | 1 |
With this seventh volume, as part of the series of yearbooks by the Association of Mathematics Educators in Singapore, we aim to provide a range of learning experiences and teaching strategies that mathematics teachers can judiciously select and adapt in order to deliver effective lessons to their students at the primary to secondary level. Our ultimate... more...
"The US National Science Foundation (NSF) Research Experiences for Undergraduates (REU) program in mathematics is now 25 years old, and it is a good time to think about what it has achieved, how it has changed, and where this idea will go next.". This was the premise of the conference held at Mt. Holyoke College during 21–22 June,... more...
Forget everything you?ve been taught about math. In Burn Math Class , Jason Wilkes takes the traditional approach to mathematics educationwith its unwelcoming textbooks, unexplained rules, and authoritarian assertionsand sets it on fire. Focusing on how mathematics is created rather than on mathematical facts, Wilkes teaches the subject in a way... more...
This book is a rare resource consisting of problems and solutions similar to those seen in mathematics contests from around the world. It is an excellent training resource for high school students who plan to participate in mathematics contests, and a wonderful collection of problems that can be used by teachers who wish to offer their advanced students... more...
The methods for teaching mathematics usually follow the structure of mathematics. The problem with this is that the structure of mathematics took centuries of elaboration to develop and is not the same as how one originally experiences mathematics. Based on research of how mathematics is actually learned, this book presents an innovative approach for... more...
University of Toronto Press, Scholarly Publishing Division1974;Not Available
For over eighty years this delightful classic has provided entertainment through mathematical problems commonly known as recreations. This new edition upholds the original, but the terminology and treatment of problems have been updated and much new material has been added. more...
The Process of Learning Mathematics is a collection of essays from a two-term course of intercollegiate lectures for students of B.Ed. degree. This collection starts with two different views on the nature of mathematics. One essay discusses the role of intuition in understanding mathematics, while another paper expounds on the role of logic. This... more...... more... | 677.169 | 1 |
Check-Up: Relations and Functions
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Students are asked to identify whether a given relation is a function. The relation could be in the form of a mapping, a graph, or a set of ordered pairs. Students must also identify the domain and range from a set of ordered pairs and then match a given graph to a story problem about speed. This is a formative assessment to be given after students understand how to determine if a given relation is a function | 677.169 | 1 |
This manual contains fully worked-out solutions to all of the odd-numbered exercises in the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer.
Biografía del autor:Descripción Cengage Learning, Inc, United States, 2011. Paperback. Estado de conservación: New. 11th. 274 x 216 mm. Language: English . Brand New Book. This manual contains fully worked-out solutions to all of the odd-numbered exercises in the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer. Nº de ref. de la librería BZV9780840053886 | 677.169 | 1 |
Availability 123The fastest, easiest way to brush up on your algebra Quick Algebra Review Need to hone your algebra skills? This bestselling review course in intermediate algebra gives you all the concepts, procedures, and problem-solving methods you need to succeed. * Each chapter begins with an easy-to-use chart that zeroes in on your problem areas. Now you can avoid wasting hours rehashing familiar concepts. * Every key algebraic concept is covered thoroughly--including positive and negative numbers, fractions, rational numbers, factoring, linear equations, quadratic equations, and word problems. * Hundreds of questions, answers, review problems, and quizzes help you to test your progress every step of the way. Now updated and revised to be even more relevant and accessible than ever, Quick Algebra Review is packed with practical examples drawn from real-life situations. Cover Design: Donald Munson Algebra Review: A Self-Teaching Guide (Wiley Self-Teaching Guides) by Peter H. Selby & Steve Slavin today - and if you are for any reason not happy, you have 30 days to return it. Please contact us at 1-877-205-6402 if you have any questions.
More About Peter H. Selby & Steve Slavin
PETER H. SELBY is the author of two other Wiley Self-Teaching Guides: Practical Algebra and Geometry and Trigonometry for Calculus, both revised by STEVE SLAVIN, PhD, who is Associate Professor of Economics at Union County College, Cranford, New Jersey. Dr. Slavin is also the author of All the Math You'll Ever Need: A Self-Teaching Guide and Economics: A Self-Teaching Guide, both available from John Wiley & Sons, Inc.
This book has a disturbing number of errors in its calculations. They may be simple typos (negative signs omitted, etc.) but they can be very confusing for someone trying to validate his/her understanding. The authors do a pretty good job of outlining and presenting the basic concepts of algebra, but skip over some important fundamentals, such as derivation of the quadratic formula.
great review of algebra Jan 5, 2006
I havent had algebra in over 30 years and am getting certified to teach high school math. This book takes you through the review in a very thoughtful sequence. I did all the exercises in this book in just one week, completely refreshing my knowledge. HOWEVER, I was very frustrated by numerous mistakes in the solution sets. I wasted much time trying to figure out how they could have gotten their answer only to discover they were wrong (typically a typo). This book would not be good for people who are learning algebra for the first time, or did not really understand it the first time around. But it is VERY GOOD for refreshing knowledge buried deep inside your brain.
Good refresher, but stay on the look out for errors Nov 24, 2005
Quick and easy way to refresh yourself on algebra, even college algebra. But watch out errors, I have probably found about 10 or more, alot of them in chapter 7. Other than that its good.
The Best Algebra Review Around May 7, 2005
I returned to school after 7 years to attempt a Masters Degree. I needed to take Calculus and I didn't want to start over at Math. I tried a lot of books on Precalculus and Algebra to get up-to-date before moving to Calculus study guides. This one was the best out of all of them. I ended up using it throughout the Calculus class each time I needed to review key algebraic concepts.
Algebra Review May 26, 2001
I had to take the GMAT in order to get into business school. Although I had a good foundation in Algebra, I hadn't used or studied it in over 20 years. I had to re-learn Algebra and practice those horrid word problems. "Quick Algebra Review" was fantastic. I was amazed at how much I had forgotten and at how fast it came back. It wasn't easy though; I put in some long study hours and it paid off. If you have to re-learn algebra, then this is the book | 677.169 | 1 |
Related Subjects
12th Grade Math: Regents
Regents High School examinations, or simply The Regents, are exams given to students seeking high school Regents credit through the New York State Education Department, designed and administered under the authority of the Board of Regents of the University of the State of New York. Regents exams are prepared by a conference of selected New York teachers of each test's specific discipline who assemble a "test map" that highlights the skills and knowledge required from the specific discipline's learning standards.
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12th Grade: Regents Word Problems
Tom drove 290 miles from his college
Tom drove 290 miles from his college to home and used 23.2 gallons of gasoline. His sister, Ann, drove 225 miles from her college to home and used 15 gallons of gasoline. Whose vehicle had better gas ...
Regents
A group of friends decides to buy a vacation home for 120,000 dollars sharing the cost equally. If they can find one more person to join them, each persons contribution will drop by 6,000 ...
regents homework help word problems for 12th grade math students.
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I hope it will teach me the fundamentals for 11th grade and 12th grade Math. In algebra, and for my regents. By helping me prepare for the regents. Regents review - SAT, ACT. With the regents. I need help for regents maths B. I want to passss the regents. Help me prepare my students to pass regents exams using practice. Regents prep | 677.169 | 1 |
Statistical diagrams
About GCSE Maths
GCSE maths has a grading system that uses numbers 1-9 to identify levels of performance with 9 being the top level. A student will either enter for the Higher tier covering grades 4-9 or for the Foundation tier covering grades 1-5.
There will be three papers at each tier, two papers where candidates can use a calculator and a third paper where calculators are not to be used.
The syllabus is divided into six sections that test
1. Number
2. Algebra
3. Ratio, proportion and rates of change
4. Geometry and measure
5. Probability
6. Statistics
The main examination boards are Edexcel, AQA and OCR.
On this site you will find plenty of video tutorials which hopefully will give you the confidence and support you need to tackle your maths GCSE. | 677.169 | 1 |
This project prompts students to build and work with catapults in order to model a real world parabola. This is a great project to facilitate in any Algebra 1 classroom because it will offer a tangible connection to quadratic functions. This product contains information relating to four main components of the project:1. | 677.169 | 1 |
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A must see for any math teacher who cares about reaching ALL types of learners.
This video is an introduction to an alternative method for presenting operations with whole numbers. Prof. Hoff's method for teaching mathematical operations ties whole numbers, fractions, scientific notation, and even polynomial arithmetic into the same simple set of procedures.
Prof. Hoff has had success with adult and young learners alike over 17 years of teaching experience. His method has seen tremendous success with learners who have failed to learn math in a traditional classroom. The method combines showing students how with an explanation of why (Explanations that many college math majors have never seen before.).
This first video shows Prof. Hoff introducing some of the basic concepts of his method. Further videos go into more details on how the method works with larger numbers, variables, polynomials, and any other form of numbers in mathematics | 677.169 | 1 |
Latest
Download Understanding Calculus II: Problems, Solutions, and Tips PDF free. Understanding Calculus II: Problems, Solutions, and Tips
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Department of Mathematics
Five Colgate students took part in the 2015 Putnam competition. They placed 54th out of 447 teams, up five spots from last year. Haochuan Wei '16 also won first prize in RIT's 2016 Probability/Statistics competition.
Academic Program in Mathematics
Our academic offerings will challenge and engage you as you develop your understanding of mathematics. READ MORE
About Mathematics
As a department, we meet the needs of students like you in a number of ways. Each year, we prepare students to go on to graduate school in mathematics, applied mathematics, operations research, or other related fields. We also prepare students for careers in finance, business administration, law, medicine, education and scientifically-oriented industries.
We also have plenty to offer, even if you major in a different area. Non-majors often require mathematical skills to carry on work in other disciplines. Regardless of your interests, you can also use the study of mathematics to assist you in forming habits of precise expression, in developing your ability to reason logically, and to learn how to deal with abstract concepts.
Not to mention that mathematics is an art form in its own right -- to be studied for its own intrinsic beauty! After all, three of the original seven liberal arts are logic, arithmetic and geometry.
Recently Published Research
Our departmental faculty are not only excellent teachers, but are productive researchers as well. Check out a list of our recent publications: READ MORE.
Success after Colgate
Alumni who majored in the mathematics are pursuing careers in a variety of fields. Check out just a few of the possibilities out there for our majors. READ MORE | 677.169 | 1 |
Discrete Mathematics and Its Applications Discrete Mathematics and its Applications is a focused introduction to the primary themes in a discrete mathematics course, as introduced through extensive applications, expansive discussion, and detailed exercise sets. These themes includeMore...
Discrete Mathematics and its Applications is a focused introduction to the primary themes in a discrete mathematics course, as introduced through extensive applications, expansive discussion, and detailed exercise sets. These themes include mathematical reasoning, combinatorial analysis, discrete structures, algorithmic thinking, and enhanced problem-solving skills through modeling. Its intent is to demonstrate the relevance and practicality of discrete mathematics to all students. The Fifth Edition includes a more thorough and linear presentation of logic, proof types and proof writing, and mathematical reasoning. This enhanced coverage will provide students with a solid understanding of the material as it relates to their immediate field of study and other relevant subjects. The inclusion of applications and examples to key topics has been significantly addressed to add clarity to every subject.True to the Fourth Edition, the text-specific web site supplements the subject matter in meaningful ways, offering additional material for students and instructors. Discrete math is an active subject with new discoveries made every year. The continual growth and updates to the web site reflect the active nature of the topics being discussed.The book is appropriate for a one- or two-term introductory discrete mathematics course to be taken by students in a wide variety of majors, including computer science, mathematics, and engineering. College Algebra is the only explicit prerequisite | 677.169 | 1 |
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I designed these lessons to teach my students about quadratic sequences (the 6th in a series of 7). This lesson can be purchased as a complete bundled unit at a discounted price under the listing Quadratic Sequences Complete Bundled Unit.
This lesson focuses on: determining which graphs and which table of values represent a function, interpreting information from 3 table of values, and given a graph--determining the domain and range in interval notation | 677.169 | 1 |
Gsce maths coursework
We will talk about GSCE Maths. GCSE Maths is taken as a GCSE exam. The students' interest to GCSE Maths is determined by the. writing a GCSE Math coursework. Coursework for GSCE (17 Posts) Add message | Report. Swanhildapirouetting Sun 14-Dec-14 19:18:34 - History 20% coursework - Drama - all coursework - Maths . Find great deals on eBay for GCSE Maths Textbook in School Textbooks and Study Guides. Shop with confidence. (As exercise, as it is only a gsce if you use an option), Maths. and do coursework for 1 gsce or. GSCE Options, Have I Made The Right Choice. The Maths GCSE course gently guides the student through basic mathematical skills Coursework. For examinations in 2009 and onwards, no coursework is required. GCSE Tuition at Kip McGrath Bangor Our GCSE service includes: If your child needs help to prepare for GCSE Maths or English coursework or exams, the Kip McGrath.
GCSE Science/Coursework advice; Common coursework experiments. GCSE Science/Osmosis in potato slices coursework; GCSE Science/Temperature. For all other Edexcel GCSE subjects, the maximum mark and grade boundaries for individual units are shown for both the raw and uniform (UMS) mark scales. Get all the latest, breaking GCSEs news on ITV News. Videos, stories and updates. The Maths GCSE course from Oxford Home Schooling progresses the student's advanced mathematical skills over thirteen tutor marked assignments and a written. As Education Secretary Michael Gove sets out his plans to reform GCSEs in England's schools with tougher exams and a move away from coursework, the BBC News. Functional skills - alternative to gcse?. but also to begin a slow approach to coursework for. We have previously used FS English & Maths. GCSE Maths Tutor - a free resource for aspiring mathematicians. Interactive - introducing a new section to the GCSE Maths Tutor site. Below is an excerpt from. STATISTICS 4040. GCSE Statistics Revision notes Collecting data Sample – This is when data is collected from part of the population. There are different methods for. Interactive Worksheets to help your child in English, Maths and Science.
Gsce maths coursework
GSCE statistics coursework So ive dont quite a level english language used by level pe coursework help from the aqa maths statistics coursework. Boys have leapfrogged over girls in maths GCSE results, bagging more of the top grades for the first time since 1997 after the government scrapped coursework last. In addition to the video footage we also prepare a written report which is submitted as part of the GCSE or A Level coursework to your school and then to your. GCSE Maths Revision Papers. Scroll down for the Practice Papers. Teachers – check out our new website and take advantage of the free trial to help your students. International General Certificate of Secondary Education. IGCSE can be taken with or without coursework or IGCSE, in maths, English, the sciences.
Watch video GCSE overhaul in England made final by Ofqual and coursework is being scrapped for most subjects English and maths will be the first subjects. GCSE (in Britain) abbreviation for (Education) General Certificate of Secondary Education: a public examination in specified subjects for 16-year-old schoolchildren. Coursework (GCSE Essentials) Fiona Mapp: GCSE Maths: Maths Dictionary Age 11-14 (Letts Key Stage 3. A GSCE Maths dictionary which gives explanations of key. GCSE English Language. Subject areas: English and Maths; GCSEs Introduction;. GCSE English consists of 60% coursework - written work (40%). GCSE Statistics Revision Exam Questions. Save for later. by zakir. 4.9 52 ratings;. I created this code breaking maths lesson for an Ofsted observation. A GSCE Maths coursework student's guide which aims to help students achieve higher grades by ensuring they understand the key content and skills required. The Pearson Edexcel Level 1/Level 2 GCSE in English is designed for use in schools and colleges. It is part of a suite of GCSE qualifications offered by Pearson.
Statistics coursework help and general. General Certificate of Secondary Education GCSE is an academically rigorous, internationally recognised qualification awarded. A lot of these skills will be homed in class or coursework anyway Maths is one subject where it is easy to pass if you know how. Cookies make wikiHow. Coursework; Essay Advice; Exam Practice; Guidance; Personal Study; Revision Notes; Source Analysis; Ancient World: Pre-1000 Earliest periods, from ancient Egypt to. IGCSE and GCSE - What is the difference? (47. ever-increasing demands on pass rates for maths and English GCSE. There is coursework for the one English. 27 Responses to "AQA GCSE Mathematics Papers - Linear. Many thanks. I am helping with a community project to deliver maths revision over the Easter. Arrow_back Back to Home Free Maths Revision and Help for Students. Whatever your age or ability (I bet you are a lot better than you think you are), if you are. General Certificate of Secondary Education Mathematics: Coursework by Fiona Mapp starting at $2.59. General Certificate of Secondary Education Mathematics: Coursework.
Maths some boards do just exams now, so ask your teachers. altho some still do coursework !! history is 25% of your overall gcse grade (IMPORTANT!.
English GCSE Summary You also complete a range of Speaking and Listening tasks. These coursework components of the course will form 60 % of the overall GCSE.
GCSE CHEMISTRY RATES OF REACTION COURSEWORK. 11 Pages. GCSE CHEMISTRY RATES OF REACTION COURSEWORK. Uploaded by. Anjelina. Maths and Mechanics A Level;. Coursework Forms; Examination Results; Useful Examination Links;. GCSE Home Schooling. Apply Now Course Pricing. The New GCSE Maths series covers all exam boards and provides a fully integrated learning experience, whichever grade and syllabus you're following. For the first time 17-year-olds in England who gained a D grade in English or maths last year will have had to resit GCSEs in those subjects Published: 1:15 PM . GSCE statistics coursework relates to the gathering, analysis, and use of statistics. It's an important study in a range of areas like marketing, maths, demography. The General Certificate of Secondary Education (GCSE) is an academically rigorous, internationally recognised qualification (by Commonwealth countries with education. If you find your coursework to be too much An Online Quiz is a great way to test your GCSE Maths skills while you would highly benefit from using a Note to study. | 677.169 | 1 |
POLYNOMIAL MULTIPLICATION
AND THE ROLE OF VARIABLES
Joanna Taylor
MST Curriculum Project
December, 2005
1
ABSTRACT
"Very early in our mathematical education- in fact in junior high school or early in high
school itself- we are introduced to polynomials. For a seemingly endless amount of time
we are drilled, to the point of utter boredom, in factoring them, multiplying them,
dividing them, simplifying them. Facility in factoring a quadratic becomes confused with
genuine mathematical talent." (Herstein, 1975)
This sentiment is all too common in the study of polynomials. It is based on the view of
polynomials as a string of symbols and yet, this is only a small part of polynomials. The
following project presents an overview of polynomial multiplication, offering its
historical evolution as well as its progression from elementary school mathematics to
university level mathematics. It also traces the role of variables in the different contexts
in which polynomial multiplication is studied. By considering the differences in the
polynomial multiplication studied at different levels and in different areas of
mathematics, this paper presents a curriculum designed to help students understand the
processes of polynomial multiplication rather than just memorize how to perform the
computations. This paper also includes a description of the need for such a curriculum
and a justification for the design of this one. In addition, there are several possible
extension activities for the curriculum and a description of my experience using these
activities.
2
TABLE OF CONTENTS
A Brief History of Polynomials
Rhetorical and Syncopated Algebra
Symbolic Algebra
Overview of the History of Polynomial Multiplication
Polynomial Multiplication in Modern Mathematics
4
5
8
10
11
Mathematical Explorations Involving Polynomial Multiplication
Generating Functions
Recurrence Relations
Counting Problems
Ring Theory
A Specific Case
Summary
13
14
16
18
22
23
25
The Rationale for Developing a Curriculum on the Multiplication of Polynomials
26
The Rationale for Developing this Curriculum
The Multifaceted Role of Variables
How my Mathematical Explorations Informed the Design of This Curriculum
Other Factors that Informed the Design of This Curriculum
28
29
30
31
Curriculum Overview
33
Presentation of Curriculum
Note to Teachers
Activity 1: Collecting Like Terms
Activity 2: The Distributive Property
Activity 3: Polynomial multiplication
Activity 4: Anticipating Terms of Products
Activity 5: Culmination
36
37
39
47
57
64
75
Extensions
Extension 1: Rational Expressions
Extension 2: History
Extension 3: Ethnomathematics
Extension 4: Binomial Theorem
82
39
47
57
64
Solutions
123
My Experience with Curriculum
141
References
146
3
CHAPTER 1
A BRIEF HISTORY OF POLYNOMIALS
4
History
"In most sciences one generation tears down what another has built and what one has
established another undoes. In mathematics alone each generation builds a new story to
the old structure." (Hermann Hankel as quoted on p 207 of Burton, 2002). This evolution
can clearly be seen in the case of polynomials. Polynomials, as we often think of them
today, are a string of symbols: numbers, variables, exponents, relational operators, and
parentheses to name a few. Polynomials, however, were not originally born into this
form. It took generation after generation of mathematicians building on each other's
ideas to create the concept of polynomials as we know them today. In order to
understand this evolution, one needs to trace the development of algebra as well as the
development of the symbolism now intrinsic to polynomials.
Algebra was originally expressed verbally; this rhetorical algebra included detailed
instructions about how to obtain solutions to specific problems and geometric
justifications for the processes. Over time, the lengthy verbal method of communicating
algebra was shortened by abbreviating commonly used words, and using symbols for
commonly used quantities and operations (Joseph, 2000). This transitional form between
the rhetorical and symbolical algebra is known as syncopated algebra. Eventually, words
gave way to symbols that represented relational operators and unknowns. Introducing
symbols to algebra allowed mathematicians to reify many mathematical constructs that
had been around for centuries (Sfard, 1995) because they were able to manipulate
complicated ideas and expressions as objects. This revolutionized elementary algebra
and helped mathematicians discover new relationships inherent in the constructs.
Rhetorical and Syncopated Algebra
The first written records of algebra that historians have uncovered come from Egypt, and
date back to approximately 1550 BCE. Some of this work can be attributed to specific
individuals, while other works were discovered on Egyptian papyri, suggesting that many
people were exploring such ideas at that time (Smith, 1953). The rhetorical algebra of
the Egyptians is quite similar to the algebra we now use, but there is no reason to assume
the Egyptians used reasoning similar to the algebraic reasoning we use today (Joseph,
2000).
Historians believe that the Babylonians began exploring the ideas of algebra almost as
early as the Egyptians did (Katz, 1998). Joseph (2000) argues that the Babylonians were
able to develop more sophisticated numerical methods of solving for unknown quantities
than the Egyptians because they had an efficient number system that facilitated
computations. He adds that, although the Babylonians mainly explored rhetorical
algebra, they were forerunners in syncopated algebra because they used geometric terms
to denote unknown quantities. For example, they used the term for square to refer to the
square of an unknown quantity.
The ideas explored by the Egyptians and the Babylonians were further explored in
Alexandria, as students traveling between the regions shared the knowledge they acquired
during their journeys (Katz, 1998). Although this mathematics would not be considered
algebra by modern standards, the Greeks of the classical period could solve many present
5
History
day algebra problems (Smith, 1953). For instance, in approximately 300 BCE, Euclid
wrote The Elements, in which he showed how to find solutions to quadratic equations by
using geometry to represent polynomial multiplication. Over the next 600 years, the
Greeks revolutionized algebra by introducing analytic approaches to the study of algebra
(Smith, 1953). This was also when Diophantus introduced symbols, but symbol use did
not become widespread at this time (Sfard, 1995).
Historians are not sure when the Chinese first began studying algebra, but there are
records of Chinese work that historians believe date back to approximately the same time
as the Egyptian, Babylonian, and Greek works. Over the following centuries the Chinese
made many advances in algebra, including writing a book that offered rules for solving
algebra problems and deducing many algebraic identities in a fashion similar to how we
deduce them today (Smith, 1953).
Around 500 CE, the Hindus joined the fray by solving linear and quadratic equations
(Smith, 1953). Indian algebra was quite different from its predecessors in that it had
letters denoting unknown quantities and abbreviations representing mathematical
operations. This was the first systematic method for representing unknowns, and it
allowed the Indians to generalize in a way that other mathematicians of their time could
not. For instance, an algorithm similar to the quadratic formula first appeared in Indian
manuscripts of this time (Joseph, 2000).
Shortly after the Hindus began studying algebra, the Arabs and the Persians began the
study of algebra as well (Smith, 1953). Islamic inheritance laws were quite complicated,
so the Muslims needed an efficient method of solving for unknowns constrained by
complicated stipulations, such as those for dividing assets (Berggren, 1986). Explorations
for such a method coincided with the creation of a library in Baghdad filled with books
that arrived with intellectuals fleeing from persecution in their homelands. In creating
this library, the Muslims were able to integrate intellectual advances of diverse cultures
into cohesive collections by subject (Katz, 1998). By combining the geometric traditions
of the Greek Empire with the arithmetic traditions of Babylon, India, and China, Islamic
mathematicians were able to make huge contributions to already established algebraic
theories. This led to advances that cultures limited to one of these approaches were not
able to make (Joseph, 2000).
The very name algebra shows its Arab roots because algebra is the distorted Latin
transliteration of the title of al-Khowarizmi's book al-jabr w'al muqabalak (restoration
and opposition). This book was intended to be practical rather than theoretical (Katz,
1998), and, for the first time, systematically studied the ideas of algebra independent
from number theory (Berggren, 1986). In this book, Al-Khowarizmi separated equations
into six categories, and described algorithms for finding the solutions to each type. In
many of these cases he offered geometric justifications as well as numerical examples
(Joseph, 2000). This work transformed what had been a systematic approach to solving
equations into a science that both showed that the processes worked and explained why
they worked (Berggren, 1997).
6
History
The Muslim world at this time was also home to the first mathematician to define the
laws of exponents, which were deduced from well-known mathematical relationships and
definitions (Berggren, 1986). With our modern notation, the laws of exponents seem
quite obvious to us. Nevertheless, given the symbolism of that earlier era, deducing this
general relationship was quite a feat.
In the medieval era, when al-Khowarizmi's book was translated into Latin, it quickly
spread throughout Western Europe (Katz, 1998). Although many advances in algebra
were made during this era, Fibonacci is considered one of the greatest algebraists of the
Middle Ages because of his ingenuity in finding solutions to equations that were not
solved in al-Khowarizmi's book. Another highly influential algebraist of the Middle
Ages was the German Jordanus Nemorarius, whose book contained equations quite
similar to the ones presently found in algebra textbooks. Despite these advances, algebra
was not studied very deeply during this time period because mathematicians of this age
were more interested in applications of mathematics than in the study of mathematics for
its own sake (Smith, 1953).
Algebra was first treated as a topic that deserved serious study during the renaissance.
According to Smith (1953), in 1494, a book was written that roughly summarized the
existing knowledge about algebra. A crude symbolism was used in this book, and the
focus was on solving equations expressed with this new symbolism . Thirty years later,
another book on the big ideas of algebra was published. This book offered no advances
in algebraic theory, nevertheless, it advanced algebra because the improved symbolism
used in it enabled mathematicians to see existing relationships and expand well known
algebraic theories. As new editions of this book were released, this work continued to
influence the evolution of algebra. In 1545 the Ars Magna was published, whose study
of solving equations and introduction of complex numbers was incredibly influential in
advancing algebra.
By this time, the ideas of elementary algebra were fairly well developed, but an efficient
and consistent system of symbolization was needed (Smith, 1953). This elementary
algebra included mathematical operations on unknowns and solving for unknowns.
Polynomial multiplication is an example of operation involving unknowns, so it clearly
was one of the original ideas of algebra. Therefore, in theory, it too was largely perfected
by the end of the seventeenth century.
Nevertheless, since polynomial multiplication was originally treated verbally and
geometrically, polynomials of this era were different beasts than the symbolic ones with
which we currently interact. Therefore, this survey of the development of rhetorical and
syncopated algebra thus far still neglects many of the important pieces in the
development of polynomial multiplication. In order to understand the evolution of
polynomials to their present reified structure, we need to understand how the symbols
used to represent them came into being.
7
History
Symbolic Algebra
Historians widely argue that Viete revolutionized algebra by replacing geometric
methods with analytic ones. By using symbols to represent unknown quantities and
operations, he was able to reify many of the ideas that mathematicians had been working
with for years (Sfard, 1995). This drastically changed algebra as well as the future of
mathematics (Smith, 1953). This arithmetization of algebra faced much resistance:
generations of mathematicians had grown up with geometric proofs along side their
algebraic processes, and they were not eager to replace their traditional methods with this
new methodology based only in logic (Goldstein, 2000). Nevertheless, rhetorical
algebra, with its geometric proofs, eventually gave way to symbolic algebra and its
accompanying analytic proofs.
Although the complete transition to symbolical representations of algebra occurred
relatively recently, symbols were occasionally used for common ideas centuries before
their formal adoption. For instance, unknown quantities were a central focus in the study
of algebra from the very beginning. Therefore, there were many different names for this
idea throughout different eras and cultures (Joseph, 2000; Katz, 1998; Smith, 1953).
One of the first symbols introduced to mathematics, and to algebra, was a symbol to
represent the unknown. The symbol used, however, varied from region to region. In the
Middle Ages mathematicians began using letters to represent algebraic and geometric
quantities (Smith, 1953). This idea, however, was not widely adopted by mathematicians
(Sfard, 1995). In the sixteenth century, Viete began representing algebra symbolically,
and the Europeans adopted that symbolism to represent unknown quantities. It was not
until the seventeenth century, however, that a symbol structure was constructed to
represent more than one unknown within a given expression (Smith, 1953).
Although Europeans widely adopted symbol use in the seventeenth century, there was
still much resistance through the nineteenth century. This resistance stemmed from the
fact that symbols were originally introduced to represent a specific unknown or object, an
object that had a clear meaning in the physical universe. However, the reification of
algebra enabled mathematicians to perform operations that had no reasonable explanation
in an individual's physical reality (Pycior, 1982). This tension made many
mathematicians skeptical about the new mathematics being performed.
As symbols for unknowns entered algebra, a structure to symbolize what we now call
coefficients needed to be introduced. Although a few mathematicians created a word for
the idea of coefficients, through much of history no specific word existed. In 1250, the
Chinese used sticks to solve equations, and the sticks represented what we now call
coefficients (Smith, 1953). There was nothing representing the unknown in this system,
so these coefficients were still quite different from our present day coefficients. It was
around this time that al-Khowarizmi's book was translated into Latin. Both this book
and its translation expressed algebraic ideas rhetorically, so there was no reason for
people to have thought of the idea of coefficients. Both the word "coefficient" and its use
were relatively recent creations, being introduced by Viete when he introduced much of
the symbolism currently used in algebra (Smith, 1953).
8
History
The idea of exponents predates its modern symbolism as well. Originally exponents were
limited to small numbers because they represented geometric ideas such as area and
volume. This meant that mathematicians had no reason to consider the existence of a
power they could not represent physically (Sfard, 1995). These low powers were used so
often, that symbols for such powers of an unknown quantity were among the original
symbols introduced to mathematics and they were given specific names so that they
could be easily referenced (Smith, 1953).
The integral exponents we presently use are generally attributed to Descartes, although he
was not the first to use them. Before the invention of this symbolism, repeated
multiplication was often used to represent this idea. Even once the symbolism was
created, it was only used for powers of five or higher, and concatenation was used to
represent smaller powers (Smith, 1953). Eventually mathematicians adopted exponent
notation for all powers, including small ones.
Most of the symbols commonly used in mathematics were originally used in algebra, and
were later transferred to arithmetic and other areas of mathematics. Over time, many
different symbols were used to represent common operations, but the conventions widely
used today originated in Europe. For instance, the symbol we currently use for addition
comes from the abbreviation of the fourteenth and fifteenth century German word for
addition. Likewise, the minus sign evolved from the abbreviation used in German
syncopated algebra. These symbols were used to indicate increases and decreases far
before they were used to indicate operations. These symbols were in use this way for
close to 100 years before the Germans and the Dutch began to use them as operators as
well. These symbols slowly spread to England and then throughout the rest of Europe
and beyond (Smith, 1953).
The symbols for multiplication developed much more slowly. The lack of such a symbol
naturally led to the concatenation we presently use (Smith, 1953). A dot was eventually
introduced to separate numbers, and was adopted to symbolize multiplication in some
areas. The cross that is presently used to symbolize multiplication evolved from
diagrams used in seventeenth century England to demonstrate the process of multiplying
two digit numbers. A cross was employed to remind students to cross-multiply, examples
of this are shown below (taken from Smith, 1953 p. 404).
This symbol was not commonly used in arithmetic until the nineteenth century, and was
never widely adopted in algebra because it so closely resembles the letter "x", which was
most commonly used to represent unknowns in algebra (Smith, 1953).
9
History
Symbols of aggregation originated with the study of radicals. In syncopated algebra, the
letter "L" and a backward "L" were used to surround terms in an expression, when
describing the square root of that expression (Smith, 1953). This symbolism eventually
expanded to other mathematical ideas, and evolved into the parentheses and brackets we
use today.
Overview of the History of Polynomial Multiplication
Having sketched out the development of algebra and the symbols now used to represent
polynomials, we now consider polynomials multiplication itself. The idea of multiplying
polynomials can be traced throughout the history of algebra, and, therefore, their
evolution parallels that of algebra itself. For instance, polynomials, and the computations
involving them, were originally introduced rhetorically. Later, the Greeks represented
polynomial multiplication geometrically. One example of this is shown below, where
Euclid shows the square of a sum.
(adapted from Heath, 1953 pp 588-589). The Arabs adopted the Greek method of
representing relationships geometrically, and even al-Khowarizmi used these geometric
models in his derivation and explanations (Berggren, 1986; Joseph, 2000). The Hindus,
on the other hand, described all operations and processes, including those involving
polynomials, completely verbally (Smith, 1953).
The Chinese created an algorithm for finding the coefficients in the expansion of (a+b)n
before Euclid showed the expansion of (a+b)2. This was not known outside of China, and
Khayyam, a Persian mathematician, surprised the mathematics community in 1100 by
expanding (a+b)n for values of n that could not be represented geometrically. He claims
to have established this process on his own, but historians have not found any work that
shows how he came up with these ideas (Smith, 1953).
Although informal ideas about polynomial multiplication have been around for as long as
algebra, implicit discussion of these ideas was not limited to algebra. For example,
number systems that use place value are based on the principles that polynomials
represent. That said, cultures that used such number systems did not necessarily
explicitly connect the ideas of place value do the ideas of polynomials. For instance, our
decimal system is actually polynomials in powers of ten and the Babylonian system
involved polynomials in sixty. Calculations using such number systems require the
notion of collecting like terms when numbers are added and the process of multiplying
10
History
polynomials when numbers are multiplied. The algorithms used to facilitate these
processes allow people to overlook these connections, nevertheless, these ideas and
processes are highly related.
Polynomials formally came into existence as a construct of mathematical definitions. For
instance, Pacioli began his explorations with a few definitions and then he defined
operations involving monomials based on those. Polynomials were how he expressed
these operations involving an unknown and its powers. Operations involving
polynomials came up, by chance, in the process (Smith, 1953).
The history of polynomials is clearly embedded in the history of algebra. Therefore, the
knowledge of their properties progressed alongside every innovation in algebra. This can
be seen in the verbal method in which the multiplication of polynomials was originally
described as well as in the geometric proofs that were used in the original derivations of
their expansions. These processes were advanced by the Arabs, just as the rest of algebra
was, when algebra was turned into a science that described generalized relationships.
Finally, polynomials, and the operations on them, became reified with the introduction of
symbolic algebra, as did all other algebraic ideas. Therefore, the evolution of the
symbols used to represent polynomials is as important as the evolution of the ideas
themselves. This means that the form in which polynomials were represented changed
not only as algebra transformed from verbal expressions to symbolic expressions, but
also as each new symbol was created.
Polynomial Multiplication in Modern Mathematics
Mathematicians' interest in polynomial multiplication did not, however, end with the
invention of symbols and the reification of polynomials. As mathematics progressed,
polynomial multiplication continued to play a large role in higher mathematics. For
instance, identities involving the expansion and simplification of products of polynomials
are regularly used for substitutions in deductive proofs in areas such as advanced calculus
and number theory. Polynomial multiplication also plays a large role in discrete
mathematics, being used in the derivation of formulas, such as the formulas to quickly
compute number theoretic functions. The sciences also depend heavily on polynomial
multiplication, and the ideas inherent in it, for derivations as well as for computations.
Furthermore, according to Pycior (1981), the eventual acceptance of symbolic algebra
forced mathematicians to accept manipulations of ideas, regardless of whether the ideas
could be manifested in the physical world. At first, the reification of algebra brought its
legitimacy into question, but this transformation eventually inspired mathematicians to
consider the study of algebra as an intellectual pursuit that can be removed from its
utilitarian functions. This eventually led to the development, and to the acceptance of
modern algebra. Algebraists continued to explore the ideas inherent in polynomial
multiplication as they developed modern algebra.
For instance, algebraists study rings of polynomials. Although modern day algebraists
explore polynomials with integral, rational, and real coefficients, just as earlier
mathematicians did, the polynomials studied in modern algebra are not limited to these
11
History
polynomials. Since ring theory involves two operations, "multiplication" and "addition",
polynomial multiplication is a fundamental aspect of the study of polynomial rings
regardless of the coefficients involved.
As Gallian (1998) explains, in ring theory the variables do not necessarily represent
unknowns, instead they often function more as placeholders. In that sense these
polynomials could be considered sequences, where multiplication of these sequences is
defined in a similar fashion to polynomial multiplication. Nevertheless, these sequences
are referred to as polynomials and polynomial multiplication is a crucial component of
them.
Another concept in higher mathematics that utilizes the ideas inherent in polynomial
multiplication that was inspired by the reification of algebra is the concept of generating
functions. In this case, polynomials are used to represent the distribution of objects and
the coefficients in the product of those polynomials are used to determine the number of
ways to distribute the items. Well-established theorems about the products of
polynomials, such as the binomial theorem, are used to determine those coefficients.
Therefore, the development and use of generating functions relied heavily on
understanding the process of multiplying polynomials, and the well-known theorems
created from centuries of studying polynomials made it possible for combinatorists to
efficiently count such possibilities.
These are just a few of the many examples of the role polynomial multiplication
continues to have in mathematics. In the next chapter these ideas will be explored in
more detail. Furthermore, although it took many great minds centuries to develop
polynomial multiplication as we know it today, mathematics students and mathematicians
alike regularly use polynomial multiplication and often treat it as a fundamental concept.
Therefore, it is likely that polynomial multiplication will continue to have an important
role in the future of mathematics and the evolution of new mathematical ideas.
12
CHAPTER 2
MY MATHEMATICAL EXPLORATIONS INVOLVING
POLYNOMIAL MULTIPLICATION
13
Mathematical Explorations
I became inspired to study polynomial multiplication when I saw that many successful
university students could not readily see algebraic substitutions that needed to be made in
their proofs. These difficulties were often rooted in an inability to anticipate terms in
products of polynomial multiplication. Furthermore, many successful university students
know the binomial theorem, but did not understand from where it came. This also comes
from viewing polynomial multiplication as manipulation of symbols rather than
understanding how the terms arise in the product. Since the students in my mathematics
classes have been very successful in mathematics, I feared this problem could be worse in
less successful students. Therefore, I set out to create a curriculum that would help
students develop a better understanding of the process of polynomial multiplication.
In the situations where successful university students had trouble making appropriate
substitutions in their proofs, polynomial multiplication was not the explicit object of
study. In these cases, nevertheless, it was a necessary skill. In advanced mathematics,
however, there are situations where polynomial multiplication is a direct object of study.
These topics include generating functions, ring theory, and convolution of sequences. In
this chapter, I will explore these situations as well as whole number multiplication as
polynomials in powers of ten.
Generating Functions
A generating function is a polynomial or power series whose coefficients are the terms of
a sequence. In a sense, "a generating function is a clothesline on which we hang up a
sequence of numbers for display" (Wilf, 1994). The generating function of the sequence
a0, a1, a2,… is f(x) = a0 + a1x + a2x2 + … (Goodaire & Parmenter, 1998). Although it is
written in function notation, the purpose is not that it be evaluated at a specific number,
so to call it a function is something of a misnomer.
Nevertheless, generating functions help answer questions about the sequence that is used
of coefficients. Indeed, writing a sequence in this way reifies many properties of the
sequence. This provides mathematicians with another tool with which to answer
questions about the sequence. For example, generating functions can help
mathematicians create a simple formula for recurrence relations and can help
mathematicians count the number of ways to distribute a given number of objects.
Operations can be performed on generating functions just like they are performed on
polynomials. For instance,
given f(x) = a0+ a1x + a2x2 + … and g(x) = b0 + b1x + b2x2 + … ,
f(x) + g(x) = (a0 + b0) + (a1 + b1)x + (a2 + b2)x2 + … ,
f(x)g(x) = (a0b0) + (a1b0+ a0b1)x + (a2b0 + a1b1 + a0b2)x2 + …
+ (anb0 + an - 1b1 +…+ a0bn)xn + …
For a more concrete example,
consider f(x) = 1+ x + x2 + x3 + … and g(x) = 1 - x + x2 - x3 + …
The sum of these two polynomials is
f(x) + g(x) = (1 + 1) + (1 - 1)x + (1 + 1)x2 + (1 -1)x3 + ...
= (2) + (0)x + (2)x2 + (0)x3 + ...
= 2 + 2x2 + 2x4 + …
14
Mathematical Explorations
The product of these two polynomials is
f(x)g(x) = [(1)(1)] + [(1)(1) + [(1)(-1)]x + [(1)(1) + (1)(-1) + (1)(1)]x2 +[(1)(1) +
(1)(-1) + (1)(1) + (1)(-1)]x3 + …
= (1) + (1 - 1)x + (1 – 1 + 1)x2 + (1 – 1 + 1 –1)x3 + ...
= (1) + (0)x + (1)x2 + (0)x3 + ...
= 1+ x2 + x4 + …
Identities involving polynomial multiplication and power series are also an important
component of working with generating functions. For instance,
(1 – x)-1 = 1 + x + x2 + …, is a well known identity involving power series. From that, it
is clear that (1 – x)-1 is another way of representing the generating function of the
sequence of 1, 1, 1, …. Therefore, we can learn about the sequence 1, 1, 1,… by
examining the power series 1 + x + x2 + … or by examining the function (1 – x)-1.
Merris (1996) refers to the latter as the "freeze-dried version." He explains the benefits
of this form by asking "if you had to stuff f(x) into a backpack, which version would you
prefer?" He adds that "it is easy to reassemble (or generate) the sequence from [the
freeze-dried version]."
Using identities such as this, and properties of polynomial multiplication, freeze-dried
versions of many generating functions can be obtained. For instance,
(1 + x + x2 + …)2 = 1 + 2x + 3x2 + …
so [(1 – x)-1]2 = (1 – x)-2 = 1 + 2x + 3x2 + …,
which means that (1 – x)-2 represents the generating function for the natural numbers.
This works for more complicated sequences as well. For instance, consider the sequence
3, 4, 22, 46, 178,…One way to describe the structure of this sequence is to provide
information about its generating function. The generating function for this sequence is
3 + 4x + 22x2 + 46x3 + …
3 + 4x + 22x2 + 46x3 +… = (3 + x)(1 + x + 7x2 + …)
and
1 + x + 7x2 + … = (1 – 2x + 4x2 – …)(1 + 3x + 9x2 + …)
then 3 + 4x + 22x2 + 46x3 +… = (3 + x)(1 – 2x + 4x2 - …)(1 + 3x + 9x2 + …).
By using the identity above
(1 – x)-1 = 1 + x + x2 + …,
and noting that
1 – 2x + 4x2 - … = 1 + (-2x) + (-2x)2 + …
and
1 + 3x + 9x2 + … = 1 + (3x) + (3x)2 + …
we see that
3 + 4x + 22x2 + 46x3 +… = (3 + x)(1 + 2x)-1(1 – 3x)-1.
Thus, the generating function for the sequence above can also be rewritten as
3 + x ____ or as
1
2__ both of which are
+
2
1 – x – 6x
1 + 2x
1 – 3x
much simpler and more compact in form than the original expression.
These examples show some methods that can be used to freeze-dry sequences and their
corresponding generating functions. However, freeze-drying is not always the goal of
15
Mathematical Explorations
problems involving generating functions. For instance, often the goal is to determine the
coefficient of the nth term of the generating function. This could help answer several
questions such as what the nth term of a sequence is given a recurrence relation or how
many ways n objects can be distributed given specific constraints. Although the above
work can be used to find the coefficient, there are sometimes easier ways to find the
coefficients when given different pieces of information.
Some of the ways these coefficients can be determined are by decomposing complicated
generating functions into known ones, using the binomial theorem, and using Taylor's
theorem. Logic involving the structure of polynomial multiplication is also crucial. At
times one of these methods is enough, but at times they have to be used together. The
following sections will provide examples of each of these methods.
Generating Functions and Their Relation to Recurrence Relations
In this section, I use the method of decomposing an unknown generating function into
familiar generating functions to determine coefficients and solve some recurrences. When
given a recurrence relation with an initial condition, generating functions can help find an
explicit formula for the nth term of the sequence. This is done by finding the coefficient
of xn in the generating function. Sometimes it is easy to find such an expression without
using generating functions, but at other times it can be quite difficult.
As an example, consider the recurrence relation an = 3an-1 for n ≥ 0 given a0 = 1.
Although this expression is fairly simple and probably could have been easily found from
listing terms, this provides a good example of this principle. Furthermore, when the
relationship is more complicated, however, this is not always the case. Start by naming
the generating function of the sequence, for example
f(x) = a0+ a1x + a2x2 + a3x3 +…Given this function, clearly the nth term of the sequence
is the coefficient of xn. Given that each term in the relation is three times the previous
term, another way to write f(x) is a0 + 3a0x + 3a1x2 + 3a2x3 +… Notice that this is
3xf(x) + a0. In other words, f(x) = 3xf(x) + a0 or f(x) - 3xf(x) = a0. This equation can be
solved for f(x) yielding, f(x) = a0/(1-3x). Since a0 = 1, f(x) = 1/(1-3x). This is equivalent
to 1 + 3x + (3x)2 + (3x)3 + … In other words, f(x) = 1 + 3x + 32x2 + 33x3 + …From this
form of the expression, it is clear that the coefficient of xn is 3n, so the nth term of the
sequence is 3n. This answer makes sense given the problem.
Consider the recurrence relation an = 2an-1 - an-2 for n ≥ 2 given a0 = 3 and a1 = -2. Start
by naming the generating function of the sequence, for example f(x) = a0+ a1x + a2x2 +
a3x3 +…Once again, it is clear that the nth term of the sequence is the coefficient of xn.
Given that each term in the relation is dependent on the two preceding terms rather than
just one, this is more complicated than the last example. Given that each term in the
relation is two times the previous term less the term before that, another way to write f(x)
is a0 + a1x + (2a1 – a0)x2 + (2a2 – a1)x3 +… In the last problem, we used 3xf(x) since
each term depended on three times the preceding one. Similarly, in this case we need to
consider 2xf(x) and x2f(x) since each term is dependent on twice the preceding term as
well as the term before that one. Note that 2xf(x) = 2a0x + 2a1x2 + 2a2x3 + 2a3x4 +…
and x2f(x) = a0x2+ a1x3 + a2x4 + a3x5 +…
16
Mathematical Explorations
so f(x) = 2xf(x) - x2f(x) + a0 + (a1 – 2a0)x
and f(x) - 2xf(x) + x2f(x) = a0 + (a1 – 2a0)x.
Factoring out f(x) produces (1 – 2x + x2) f(x) = a0 + (a1 – 2a0)x
or (1 – x)2f(x) = a0 + (a1 – 2a0)x
By solving for f(x) and substituting in a0 = 3 and a1 = -2, we find that
f(x) = (3 – 8x)(1 - x)-2. You may recognize the second factor from above: it is the
generating function for the natural numbers. Therefore, we can replace (1 - x)-2 with
1 + 2x + 3x2 + … + (n + 1)xn + …, yielding
f(x) = (3 – 8x)( 1 + 2x + 3x2 + … + (n+1)xn + …). By distributing we find that
f(x) = 3(1 + 2x + 3x2 + … + (n+1)xn + …) – 8x(1 + 2x + 3x2 + … + (n+1)xn + …)
= ( 3 + 6x + 9x2 + … + 3(n+1)xn + …) + ( -8x - 16x2 - 24x3 - … - 8(n+1)nxn+1 - …)
= 3 + (6x -8x ) + (9x2 - 16x2) + … + (3(n + 1)xn – 8nxn) + …
= 3 + (6 – 8)x + (9 - 16)x2 + … + [3(n + 1) – 8n]xn + …
= 3 - 2x - 7x2 + … + (3 -5n)xn + …
From this it is clear that the coefficient of xn is 3 – 5n, so the nth term of the sequence is
3 – 5n. This answer is less obvious than the earlier one, but can easily be checked by
comparing the terms produced by this expression to the first few terms of the recurrence
relation.
Generating functions can help with non-homogenous recurrence relations as well. For
instance, consider the recurrence relation an = -an-1 +2n - 3 for n ≥ 1 given a0 = 1. Once
again, start by naming the generating function f(x) = a0 + a1x + a2x2 + a3x3 +…+ anxn +…,
where the nth term of the sequence is the coefficient of xn. Given that each term is
dependent on the preceding term as well as n, this problem is slightly more complicated
than the last two. Nevertheless, it is similar. By using the process used in the last two
problems, another way to write f(x) is a0 + (2 – a0 – 3)x + (2 – a1 – 3)x2 + (4 – a2 – 3)x3
+ … + (2n – an-1 – 3) xn + … Note that xf(x) = a0x + a1x2 + a2x3 + a3x4 + … + anxn+1 +
…,so f(x) = -xf(x) + a0 + (2 -3)x + (4 – 3)x2 + (6 – 3)x3 + … + (2n – 3) xn + …
and f(x) + xf(x) = a0 + (2 -3)x + (4 – 3)x2 + (6 – 3)x3 + … + (2n – 3) xn + …Factoring out
f(x) produces (1 + x) f(x) = a0 - x + x2 + 3x3 + … + (2n – 3)xn + …
By solving for f(x) and substituting in a0 = 1, we find that
f(x) = (1 - x + x2 + 3x3 + … + (2n – 3)xn + …)(1 + x)-1, which can be rewritten by
replacing (1 + x)-1 with 1 + (-x) + (-x)2 + … + (-x)n + …, yielding
f(x) = (1 - x + x2 - 3x3 + … + (2n – 3)xn + …)(1 - x + x2 - … +(-1)nxn + …) By
distributing we find that
f(x) = 1(1 - x + x2 - 3x3 + … + (2n – 3)xn + …) - x(1 - x + x2 - 3x3 + … + (2n – 3)xn
+ … ) + x2(1 - x + x2 - 3x3 + … + (2n – 3)xn + …) - … + (-1)n(1 - x + x2 - 3x3
+ … + (2n – 3)xn + …)xn + …
= 1 + (- x – x) + (x2 + x2 + x2 ) + (- 3x3 + 3x3 - 3x3 + 3x3)+ …
+ [(2n – 3) xn – (2n – 5) xn + .. + (-1)n-1(-1) xn + (-1)nxn] + …
= 1 + (- 1 – 1)x + (1 + 1 + 1)x2 + (-1 + 1 – 1 + 1)x3 + …
+ [(2n – 3) – (2n – 5) + .. + (-1)n-1(-1) + (-1)n]xn + …
= 1 - 2x + 3x2 + 0x3 - … + [(2n – 3) – (2n – 5) + ... + (-1)n-1(-1) + (-1)n]xn + …
= 1 - 2x + 3x2 + … + [(2n – 3) – (2n – 5) + ... + (-1)n-1(-1) + (-1)n]xn + …
From this, it can be obtained that the coefficient of xn is n + 1 if n is even and n – 3 if n is
odd. So the nth term of the sequence is n + 1 if n is even and n – 3 if n is odd. This
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Mathematical Explorations
answer is much less obvious than the earlier ones, but can be checked by listing the first
few terms of the recurrence relation.
These two examples above show how, through the use of generating functions, the
process of polynomial multiplication is used to solve recurrence relations. Recurrence
relations are a nontrivial and notoriously difficult mathematical topic, one that the simple
process of polynomial multiplication drastically simplifies. Therefore, understanding the
process of polynomial multiplication is an important tool for solving such problems.
Generating Functions and Counting Problems: Binomial Theorem and Taylor's Theorem
In addition to being a powerful tool for calculating terms in recurrence relations,
generating functions are also useful tools in counting problems. In combinatorics,
generating functions are used to determine the number of ways to "do something" with a
specific number of objects from a collection of objects (Tucker, 2002). In this situation,
we do not yet know the coefficients of the generating function; the goal is to find those
coefficients.
By examining the process of polynomial multiplication and the terms that arise from it, a
variety of mathematicians from different cultures noticed a pattern for determining the
coefficients of the expansion of (a + b)n. The formula for the coefficient of the kth term
of this expansion when written in descending order is the same as the number of ways to
select k objects out of n identical objects. This is because each term of the product is
going to involve one term from each the n factors. Therefore, all terms of this product
will be of the form ak bn-k, where k of the factors are a's and the remaining n-k are b's. In
order to get ak, k a's must be multiplied together. This means the number of ways to get
the term ak bn-k is the same as the number of ways to choose k a's from n factors. In other
words, the number of ways to choose k objects from n identical objects (often written as
(n, k)). This means the coefficient of the kth term of the expansion of (a + b)n when
written in descending order is the same as the number of ways to select k objects out of n
identical objects. This gives us (a + b)n = (n, n)an + (n, n - 1)an-1b + (n, n - 2)an-2b2 + …
+ (n, 1)abn-1 + (n, 0)bn, which is commonly referred to as the binomial theorem. By
using this relationship, mathematicians realized properties of polynomial multiplication
could help with more complicated counting problems as well.
To extend this reasoning beyond the binomial theorem, mathematicians create
polynomials or power series to represent the constraints of the problem. They then use
the properties of multiplication of polynomials and power series to rewrite the product as
a single generating function. The sequence "hanging" on the generating function is the
number of ways to do whatever it is you are modeling. That is, the coefficient of the term
ak is the same as the number of ways to "select" k objects within the constraints of the
problem.
In this process, the exponents and the coefficients are important. The variables act as
place holders, generally having no meaning in the problem itself. The variables provide
the formal structure necessary to algebraically manipulate the terms of the corresponding
sequence, allowing us to represent complicated problems in forms with which
18
Mathematical Explorations
mathematicians and mathematics students are quite comfortable. By doing this,
mathematicians and mathematics students can easily see how to use the properties of
polynomial multiplication to help them answer complicated counting questions.
Although generating functions can be used to count situations with and without
constraints, generating functions are particularly useful when there are constraints in the
problem. The use of generating functions to count the number of ways to select objects
relies heavily on properties of polynomial multiplication and identities derived from
those properties. This is because polynomials or power series that represent the
constraints of the problem are multiplied together to give us the generating function from
which the answer is obtained.
The first step in using generating functions to count is to create polynomials or power
series that represent the number of each object that can be distributed. For instance, if
you have fifteen different types of flowers and you want to count how many different
bouquets can be made using no more than one of each type of flower one would write
(x0 + x1)15 or (1 + x)15. The first term represents using zero of a given flower and the
second term represents using one of that type of flower. The polynomial is raised to the
fifteenth power because there are fifteen different types of flowers with identical
constraints. The expression of xk in the resulting expression would count the number of
such bouquets with k flowers.
Similarly, if the question had been how many different bouquets can be made using at
least two of each type of flower, but no more than three of each type of flower the
corresponding polynomial expression would be (x2 + x3)15. The exponent of the first term
is two because at least two flowers of each type need to be used, the exponent of the
second term is three because three flowers can be used, and there are no more terms
because no more than three flowers can be used. There are once again fifteen different
types of flowers with identical constraints so we need to multiply together fifteen such
polynomials.
If instead you only wanted to ensure that you had at least two of each type of flower, the
corresponding expression would differ depending on whether there was an infinite supply
of flowers or a finite supply of flowers. In the former case, the expression would involve
power series rather than polynomials, but would use the same general idea, using (x2 + x3
+ x4 + … )15. In the latter case, the representative polynomial expression would be (x2 +
x3 + … + xm)15, where m is the number of flowers available, if there were the same
number of each type of flower available. If there were different numbers of each type of
flower available, the representative polynomial expression would be (x2 + x3 + … +
xm1)(x2 + x3 + … + xm2)… (x2 + x3 + … + xm15), where m1 is the number of flowers of
type 1 available, m2 is the number of flowers of type 2 available, etc.
Finally, if you had different restraints for each type of flower, the polynomial
representing the possibility for each flower would be different. For instance, consider the
situation where you want to include no more than two sunflowers, no more than five
19
Mathematical Explorations
roses, at least two tulips but no more than five of them, and no more than three of any
other type of flower. In this case, the representative polynomial product would be
(1 + x + x2)( 1 + x + x2 + x3 + x4 + x5)(x2 + x3 + x4 + x5)(1 + x + x2 + x3)12.
After setting up polynomials whose exponents represent the constraints presented, the
next step is to manipulate the expressions in a way that it is easy to determine the
coefficient of the relevant term or terms. As with recurrence relations above, we will use
the technique of expressing the desired generating function in terms of more familiar
ones. In these problems, however, we often encounter the generating function of the
binomial theorem along the way. To proceed, we generally begin by factoring the
greatest common factor out of each factor of the product until each factor of the product
begins with a one. After simplifying each of the factors, algebraic identities such as
1 + x + x2 + … + xm = (1 – xm+1) / (1 – x) and 1 + x + x2 + … = 1 / (1 – x) are applied to
the factors. Note that only the latter was used in the discussion of recurrence relations.
That is because we were discussing infinite sequences. The latter identity is only relevant
when power series are involved. So in the case where we are discussing a finite number
of terms, or a polynomial, the former identity needs to be used.
Once the factors of the product have been rewritten in one of the simpler forms, they are
combined into as few terms as possible by using properties of exponents. After the
expression has been simplified, identities such as those for determining the coefficients of
the terms in the expansion of (1 + x)n, (1 - xm)n, and (1 - x)-n are used. Often this is
enough to manipulate the expression into a form from which the number of possibilities
can be counted. Occasionally, however, the result is the product of polynomials rather
than an expanded polynomial. In this case, the polynomials need to be multiplied, and
the ability to anticipate the terms in the product will help determine the appropriate
coefficient.
In the end, the expanded polynomial or power series will be of the form
a0 + a1x + a2x2 + … + anxn or of the form a0 + a1x + …. The number of ways to select k
objects, subject to the constraints used to create the original expression will be the
coefficient of xk. We can revisit some of the scenarios discussed at the beginning to see
this in action.
In the first example, where we wanted to count how many different bouquets could be
made using no more than one of each of fifteen different types of flowers, we began with
the expression (1 + x)15. From this, we immediately know that the coefficient of xk is
(15, k) or 15!/[k!(15-k)!]. Therefore, the number of ways to create a bouquet with k
flowers using no more than one of any type of flower is (n, k). If instead we had wanted
to know that number of such bouquets that could be created using at most k flowers, the
answer would be (15, 1) + (15, 2) + … + (15, k).
The second example we considered was where the bouquets had at least two of each type
of flower, but no more than three. We began this problem with the expression (x2 + x3)15.
The first step here is to factor out x2, leaving us with [x2(1 + x)] 15. This expression is
equivalent to x30(1 + x)15. By applying the binomial theorem, we get
20
Mathematical Explorations
x30[1 + (15, 1)x + (15, 2)x2 + (15, 3)x3 + … + (15, 15)x15]. By distributing the first term
we now have x30 + (15, 1)x31 + (15, 2)x32 + (15, 3)x33 + … + (15, 15)x45. The number of
ways to select a bouquet with k flowers is (15, k-30).
In the case where there are at least two of each flower, and an infinite supply of each type
of flower, we start with (x2 + x3 + x4 + … )15. The first step is to factor out the greatest
common factor, leaving us with [x2 (1 + x + x2 + … )] 15 or x30(1 + x + x2 + … )15. This
is equivalent to x30(1 - x)-15, which is equivalent to x30[1 + (15, 1)x + (16, 2)x2 + (17, 3)x3
+ … + (14 + r, r)xr +…]. By distributing, we see that this is equivalent to x30 + (15,
1)x31 + (16, 2)x32 + (17, 3)x33 + … + (14 + r, r)xr + 30 +… By using this expansion, it is
clear that the number of ways to create a bouquet with k flowers is (k - 16, k - 30).
These processes can be applied to the remaining problems discussed above as well as to
many other counting problems. However, the remaining problems can also be solved
with Taylor's theorem. If the terms of a sequence are given by a differentiable function,
then we can often use Taylor's theorem to deduce the form of the coefficients from the
"freeze-dried" form of the generating function. This is a nice alternate method for
extracting coefficients because it does not require factoring the generating function into
the small set of familiar ones.
Taylor's theorem states that "if f has a power series representation (expansion) at a, that
is if f(x) = ∑ cn(x – a)n and |x – a| < R, then cn = f(n)(a)/(n!)," (Stewart, 2001) where f(n)(a)
is the nth derivate of f evaluated at a and R is the radius of convergence. For our
purposes, we will consider a special case of Taylor's series, that is when a = 0. These
series are referred to as Maclaurin series. Since we are only using the power series
expansion as a formal technique to manipulate the coefficient sequences, it turns out that
conclusions drawn from Taylor's theorem regarding the coefficients are often valid, even
in the case where the convergence of the power series is not attained.
To see this in action, let us consider one of the examples from earlier. Consider the case
where we could use any number of flowers of each type, but there was a finite supply of
each type (the same number were available of each type). Recall that the generating
function for this situation was f(x) = (1 + x + x2 + x3 + … + xm)15. The derivate of this is
15(1 + x + x2 + x3 + … + xm)14(1 + x + x2 + x3 + … + xm), so f'(0)/1! = 15(1 + 0) 14(1 +
0) = 15(1)(1) = 15. This means there are fifteen different bouquets involving one flower.
This answer is easily confirmed using logic.
When using this method, there is often not an easy way to create a general answer for the
coefficient of xn, nevertheless, coefficients for a given term can be calculated fairly
easily. This can be done by taking repeated derivative or by using some sort of computer
algebra system. Similarly, the process could be used to determine the number of possible
bouquets in the other situations, too. Those derivatives, however, are slightly more
complicated as they require several applications of the product rule.
The point of this latter example is to show that the determination of the coefficients of
products of polynomials is not merely a mechanical routine whose importance is limited
21
Mathematical Explorations
to rudimentary mathematics. In fact, it is a prevalent theme in many area of modern
mathematics. These areas include the discrete world of combinatorics and the continuous
world, via Taylor's theorem, of calculus.
Generating functions helps mathematicians keep track of complicated restriction and
efficiently count possibilities involving such restrictions. Identities and processes that do
involve polynomials are essential to such problems. Although the situations themselves
do not involve polynomials, understanding the process of polynomial multiplication
allowed mathematicians to see that identities about polynomial multiplication could assist
them in efficiently counting complicated situations.
Ring Theory
In modern algebra, mathematicians study systems that have certain properties. Two of
these systems are groups and rings. The operations of polynomial arithmetic allow us to
construct a new group or ring from any other group or ring. In particular, the structure of
polynomial multiplication allows us to construct a new ring from a given ring.
Rings involve groups, so I will start by describing a group. A group involves a set and a
binary operation, it also meets certain specifications. These specifications are closure,
associativity, existence of an identity, and existence of inverses.
Note that commutativity is not a requirement for groups. However, such groups do exist
and are called abelian groups. Furthermore, there are always elements in a group that
commute with all other elements in the group; the subset comprised of all such elements
is called the center of the group.
Another system algebraists study is called a ring. A ring involves an abelian group and a
second binary operation. For simplicity's sake, the operation with which the set forms a
group is often referred to as addition and the other operation is often referred to as
multiplication. The second operation, multiplication, must also be associative and it must
distribute over the first operation on the left side as well as on the right side.
Furthermore, the set must also be closed with respect to the second operation. The
multiplication operation need not be commutative, but it can. In the case that it is, the
ring is called commutative.
Obviously the set of polynomials (with integer, rational, or real coefficients) is an
example of a group when the operation is addition and is an example of a ring when the
operations are addition and multiplication. There is clearly closure because the sum of
two polynomials is a polynomial and the product of two polynomials is a polynomial.
The zero polynomial is clearly the additive identity, and the additive inverse of a
polynomial is clearly a polynomial. Furthermore, in the case of polynomials,
multiplication is distributive. Finally, since addition of polynomials is commutative, the
set of polynomials together with addition is clearly an abelian group.
"To a certain degree, the notion of a ring was invented in an attempt to put the algebraic
properties of the integers into an abstract setting" (Gallian, 1998). Therefore, because the
22
Mathematical Explorations
set of polynomials together with ordinary addition and multiplication is a ring as well,
many of the similarities between operations on polynomials and operations on integers
are obvious. Likewise, many of these similarities also exist between operations on
polynomials and operations on real numbers, which also form a ring.
The obviousness of many of the proofs listed above depends on our understanding of the
real numbers and the properties polynomials share with the real numbers. Therefore,
these ideas can be extended to polynomials with coefficients from any ring. In other
words, polynomials whose coefficients come from a ring will form a ring with the
operations defined on the ring. In this situation, however, the procedure with which
polynomials are multiplied needs to be combined with the "addition" and
"multiplication" operations from the ring from which the coefficients come.
In all of these descriptions, I have assumed that the variables used in the polynomials
represent unknown elements from the coefficient ring, just as they do in the mathematics
leading up to abstract algebra. However, this need not be the case. The variables can
also merely serve as place holders to define the multiplication of sequences. This is
sometimes referred to as convolution of sequences.
To illustrate convolution of sequences, consider the two sequences (a0, a1, a2,…, an) and
(b0, b1, b2,…, bm). For simplicity assume n equals m. By using convolution, the product
of the two sequences is (a0b0, a0b1 + a1b0,…, anbm). To see where these terms came from,
consider the two polynomials a0 + a1x + a2x2 + … + anxn and b0 + b1x + b2x2 +… +
bmxm, also assuming that n equals m. The product of these two polynomial is
(a0b0) + (a0b1 + a1b0)x + (a0b2 + a1b1 + a2b0)x2 + … + (anbm)xm + n. Thus the
convolution of two sequences is a sequence whose terms are the coefficients of the
product of the corresponding generating functions.
By considering two polynomial of different degree, this example can be extended to show
the sum and product of two sequences of different lengths. To keep track of the terms in
the convolution of two sequences, it helps to rewrite the sequences as polynomials.
However, it is not essential to do so. Ignoring the variables, however, requires a strong
understanding of how to arrive at the coefficients in polynomial multiplication.
When the elements of a ring are sequences and the multiplicative operation is
convolution, the rings are called polynomial rings regardless of whether or not the
variables are written. Furthermore, when the elements are written as polynomials, the
variable does not need to represent an unknown.
Polynomial Multiplication- A Specific Case
In addition to exploring where polynomial multiplication is an explicit object of study in
advanced mathematics, I also explored how polynomial multiplication is implicitly
studied in elementary mathematics. I did this by rewriting whole numbers as
polynomials in powers of ten and then exploring the process of arriving at the product by
using the format generally used for multiplying polynomials. Several interesting things
came out of this experience.
23
Mathematical Explorations
Writing multi-digit whole numbers as polynomials in powers of ten helped me notice that
ten is a very easy value at which to evaluate a polynomial, especially when the
coefficients are positive. This was an interesting revelation that was uncovered during a
conversation with Sean Larsen (personal communication, February 17, 2005). People
usually evaluate polynomials at zero, one, and negative one assuming they are the
simplest values, but ten is often a simple value as well. Although this is an interesting
fact about polynomials, it is not directly relevant to the study of polynomial
multiplication. So I did not spend much time on this fact.
The next thing I noticed was that replacing the variable with ten changes the procedure
for multiplying polynomials. This is because of the similarity between the coefficients
and the "variables." This blurred distinction creates a slight different in the process of
multiplication. This is illustrated below.
Consider the multiplication problem 53 x 24. Writing the numbers as polynomials in ten
produces the expression (5 x 10 + 3)( 2 x 10 + 4). Applying the standard procedure for
polynomial multiplication yields (10 x 102) + (20 x 10) + (6 x 10) + (12). From here the
blurring between the coefficients and the variable is obvious. Consider the first term of
the expanded polynomial. In this case, the coefficient is identical to the "variable".
Several processes can be used to simplify this expression. These simplification methods,
however, do not work when a variable that actually varies is used. One such
simplification process would be to begin by collecting like terms, producing the
expression (10 x 102) + (26 x 10) + (12). If this is done, the next step could be to rewrite
the expression as (1 x 103) + [(20 + 6) x 10)] + (10 + 2). This expression could then be
rewritten as (1 x 103) + (20 x 10) + (6 x 10) + (1 x 10) + (2) and then as (1 x 103) + (2 x
102) + (6 x 10) + (1 x 10) + (2). Collecting like terms produces (1 x 103) + (2 x 102) + (7
x 10) + (2). The last three steps could not be performed when true variables are used.
The blurry line between the coefficients and the variables makes whole number
arithmetic slightly different from polynomial arithmetic.
On further reflection, this blurry line directly relates to ring theory. In ring theory it is
assumed that the variable is transcendental over the coefficients, that is that there is no
polynomial satisfied by x. By substituting 10 for x, we are not only evaluating the
polynomial as a function, we are replacing it with an algebraic element from the
coefficient ring.
In fact, evaluating polynomials with integer coefficients induces a function mapping the
polynomials to the integers (similarly if the coefficients were rational or real, the function
would map the polynomials to the rationals or the reals respectively). This function is a
homomorphism, not an isomorphism. That explains why some of the properties transfer
from system to system while others do not. In particular, facts true of polynomial
arithmetic are true of base ten arithmetic, whereas the opposite is not always true.
24
Mathematical Explorations
This blurry line also encouraged me to look at long division involving polynomials as
well as synthetic division. I realized that the connection between the coefficients and
variables with base ten numbers allows us to work only with positive numbers, creating
the long division algorithm commonly used in the United States. This difference forces
us to use signed numbers when attempting to use the same procedure for long division
involving polynomials. This same difference is inherent in synthetic division, but other
complications arise in that situation as well.
Summary
Reflecting on these explorations, I became aware of the differences in the role of the
variable in each of these situations. The role varies from completely irrelevant to so
relevant that it changes the algorithm. In modern algebra the variable can serve as a
place holder, or it can serve the crucial role of representing all or some numbers. In
generating functions the variable has no meaning, serving only as a place holder. This
impacts the identities we can use to rewrite the expressions and the statements we can
make about the product of power series (infinite polynomials). Convolution of sequences
does not even include the variable at all. In whole number multiplication, the variable
plays such an important role that it changes the algorithm.
The wide spectrum of the role played by variables in processes involving polynomial
multiplication ranges from the role it plays in whole number multiplication to the role it
plays in convolution of sequences. Exploring the role of the variable in these topics,
helps to explain the role that polynomial multiplication plays in higher math.
Furthermore, it shows that polynomial multiplication exists in more mathematics than
first meets the eye.
25
CHAPTER 3
RATIONALE FOR DEVELOPING A CURRICULUM ON THE
MULTIPLICATION OF POLYNOMIALS
26
Rationale
As Bob Moses said in his book Radical Equations, "algebra, once solely in place as the
gatekeeper for higher math and the priesthood who gained access to it, now is the
gatekeeper for citizenship; and people who don't have it are like the people who couldn't
read and write in the industrial age." This quote accurately describes the role algebra
currently plays in our society and the weight many people put on it. Therefore, algebra
education plays an important role in promoting social justice. The gate-keeping role of
algebra makes it increasingly important that all people have the opportunity to develop
algebraic reasoning and algebraic thinking. Therefore, it is important to make all
concepts in algebra, even the most abstract ones, accessible to all.
Furthermore, being able to manipulate symbols and arrive at an answer is not the same
thing as being adept at algebra. Students need to understand algebraic concepts rather
than merely memorize algorithms for symbolic manipulations. This means that in order
to create equal access for all people, more lessons need to be created that promote student
thought about algebraic concepts. By creating such lessons, teachers will have
alternatives for teaching topics that are often difficult for students (Herscovics, 1989).
Without access to this type of lesson, teachers are often dependent on the "drill and kill"
approach they experienced in their schooling.
Polynomials are a core component of beginning algebra; therefore, it is crucial that
students have a solid understanding of their properties (Eisenberg & Dreyfus, 1988).
Many teachers of algebra, however, tend to emphasize procedure rather than structure
when teaching polynomials (Coopersmith, 1984). This type of learning makes it difficult
for students to apply what they are learning. Therefore, it is crucial that new lessons are
created that promote a conceptual understanding of algebraic structures (referred to from
here on out as structural understanding) of polynomial operations.
Building a structural understanding of polynomials is beneficial in many ways. This
deeper understanding could help students anticipate the form of the products of
polynomials. This will help students in elementary algebra with topics such as factoring
and working with rational expressions. Skills involving polynomials are not only
important in beginning algebra; they are also important in almost all math classes after
algebra. As was discussed earlier, the structure involved in polynomial multiplication
becomes more important in higher mathematics, where the variables are often eliminated
from the conversation. Therefore, this structural understanding will help students directly
and indirectly if they continue on to higher math classes.
In addition, as mentioned in the last chapter, polynomial operations, including
polynomial multiplication, is an object of direct study in higher math classes such as
abstract algebra and combinatorics. Furthermore, the study and use of polynomials is not
limited to math classes. For instance, polynomials are also a foundational topic in many
science classes, so a structural understanding of polynomials will also help students later
on in their academic careers by enabling them to understand derivations of existing
formulas and possibly derive new ones. The effects of a solid understanding of
polynomials directly and indirectly affect many aspects of peoples' mathematical lives.
Therefore, curricula that promote such an understanding are essential in this day and age.
27
CHAPTER 4
RATIONALE FOR DEVELOPING THIS CURRICULUM
28
Rationale
For the reasons I mentioned earlier, such as the importance of a structural understanding
of algebra, the role of algebra in higher mathematics, and the superficial understanding of
polynomial multiplication some successful university students seemed to have, I felt that
it was important to design a curriculum on multiplying polynomials. Therefore, I set out
to explore mathematical topics that would help me design a curriculum on polynomial
multiplication. I originally thought learning about the historical evolution of polynomial
multiplication would help me develop a curriculum.
As time went on, however, I learned algebra was originally developed rhetorically. This
meant a process for polynomial multiplication was determined before the polynomials we
use today were. Furthermore, I learned that when variables were first introduced, there
was a lot of reluctance to accept them, even by the greatest mathematical minds of the
time. This meant that the ideas of rhetorical algebra do not easily transfer to symbolic
algebra. So I decided that the historical evolution of polynomials would not help design a
curriculum involving polynomial multiplication as it is manifested in symbolic algebra.
Therefore, I explored several mathematical topics, in search of a topic that would uncover
aspects of polynomial multiplication that would help me develop a curriculum. None of
the topics alone did this, but looking at the topics together did. As I mentioned earlier,
when I reflected on the different topics I explored, I realized that the role of the variable
was quite different in the different topics. I thought this could help me create a lesson
involving polynomial multiplication.
The Multifaceted Role of Variables
On one extreme there was convolution of sequences, which did not use variables at all. It
adopted the process of polynomial multiplication, but variables were not used, not even
as placeholders. This is similar to the role of variables in polynomials as they are studied
in abstract algebra. However, the variable is sometimes used in abstract algebra. When
the variable is used in the expression it can be used as an object that can replaced by
another value, but this is not necessary.
In generating functions variables are used only as placeholders. In order to help count,
generating functions rely on properties of polynomial multiplication such as identities
about products and quotients of polynomials. Other than helping reference those
properties, however, the variables serve no function.
This is in contrast to the role of variables in the polynomials we see in elementary
algebra. In this case, the role of the variable ranges from representing a specific number
to representing any number. For instance, when polynomials are being multiplied
together in expressions, the variable can represent any number. Whereas when variables
are used in equations they can represent all numbers, some numbers, or even no number
at all.
On the other end of this spectrum are whole numbers in our base ten system, or in
systems using other bases. In this case the "variable" is a specific number. Having this
specific number distorts the algorithm, because the coefficients merge with the
29
Rationale
"variable", allowing terms to be combined that are not combined when the "variable" is
not a known constant.
Although all of these topics use the same basic concept, the topics differed because of the
role the variable plays. After noticing this spectrum, I realized that it was the role of the
variable that kept me from using history as a starting point for my curriculum. Therefore,
I realized that the role of the variable had been a key component of all of my
explorations, and, therefore, needed to be a key component of my curriculum.
How My Mathematical Explorations Informed this Design on this Curriculum
I began to think about the differences between the roles of variables in the topics I
explored, and how that could help me teach students polynomial multiplication. I also
thought about my exploration into ring theory, which reminded me about the similarities
between operations involving polynomials and operations involving integers. I decided
to start by thinking about whole numbers, the polynomials with which students are most
familiar. Then I thought about how to highlight the differences between whole numbers
and polynomials as they appear in elementary algebra.
I realized the important difference between whole numbers and polynomials was the
blurred distinction between the coefficients and variables in whole numbers. This
impacted the process of collecting like terms in whole number addition. So I began to
think about how to get students to think about whole number addition as collecting like
terms and how to transition students from that view of whole number addition to
collecting like terms in the polynomial world.
The addition algorithm used for whole number addition is based on collecting like terms,
so asking students to explain the process seemed like a good starting point. Then asking
students to apply the algorithm to time allows them to realize that bases other than ten
work as well. This extends students ideas of like terms, but is still limited because the
distinction between variables and coefficients is still fuzzy. Therefore, I decided to
expand the idea of time to include months, something that does not have a clear exchange
rate with any smaller unit. This could help students begin to address the lack of closure
that exists when true variables are used. After expanding the idea of like terms to address
the lack of closure in algebra, it seemed appropriate to introduce variables.
After students have begun to think about whole numbers and time as polynomials, it is
time to introduce multiplication. This is done by first having students look at
multiplication involving a single digit number and then a multi-digit number. This allows
students to develop the idea of the distributive property. This also calls on students'
methods for mentally performing calculations. It seemed important for students to look
at the distributive property in the context of whole numbers, time, and variables. That
way, they could eventually extend their ideas to reinvent the algorithm for polynomial
multiplication.
This idea easily extends itself to rational expressions. This seems like a logical extension
because thinking about the process of operations involving fractions clearly extends to
30
Rationale
rational expressions in the same way operations with whole numbers extend to
polynomials. Furthermore, students often memorize the algorithm for operations
involving fractions and do not understand the processes. Therefore, encouraging students
to think deeply about the processes involved in operating on fractions would be beneficial
for students.
Understanding the process of polynomial multiplication and anticipating terms in
products also helps students understand the binomial theorem, which in turn helps
students better understand polynomial multiplication. Therefore, I also decided to extend
this curriculum to include the binomial theorem. This activity was inspired by a
technique used by Kieran and Saldanha (forthcoming) and allows students to use
technology to further their understanding of polynomial multiplication and the binomial
theorem.
Furthermore, the idea of thinking of whole numbers as polynomials in base ten enables
students to better understand number systems using other bases. This is valuable in this
technological age since computers are based on binary code. This also helps students
understand the differences between number systems used by different cultures at different
time. Therefore, I also extended this activity to help students understand these ideas.
Finally, I found learning the history of polynomial multiplication very interesting and it
supported my belief that geometric models are not the best way to teach polynomial
multiplication. Therefore, I also decided to extend this activity to teach students about
the history of algebra and the obstacles encountered by considering only the geometric
basis of polynomial multiplication. Hopefully this will help students interested in history
become more interested in math and students interested in math become more interested
in history.
Other Factors that Informed the Design of this Curriculum
This curriculum was not only influenced by my mathematical explorations, it was also
influenced by pedagogical considerations. For instance, it was based on many suggestion
made by organizations such as the National Council of Teacher's of Mathematics
(NCTM). It was based on ideas expressed by other researchers as well.
In the 1991 document Professional Standards for Teaching Mathematics, NCTM argued
that "tasks that require students to reason and to communicate mathematically are more
likely to promote their ability to solve problems and to make connections." Since a goal
of this curriculum is to foster an understanding of polynomial multiplication that
facilitates extension of the concepts, this curriculum was designed with the goal of asking
students to reason and communicate about the concepts they are studying. The design of
the problems presented in this curriculum is also consistent with the NCTM stance
because the problems were developed "to create opportunities for students to develop this
kind of mathematical understanding, competence, interests, and dispositions."
Another goal for students of mathematics presented in the NCTM 2000 document
Principles and Standards for School Mathematics "is to develop increased abilities in
31
Rationale
justifying claims, proving conjectures, and using symbols in reasoning." In designing
this curriculum, this idea was a central focus because it enables students to make sense of
new concepts they confront and because it empowers students to construct mathematical
concepts even when they are not formally presented to them. This is important because it
helps students assess the reasonableness of their assumptions and conjectures.
A third important component of mathematics instruction described by the NCTM (1991,
2000) is mathematical discourse. They argue that by "making conjectures, proposing
approaches, and solutions to problems, and arguing about the validity of particular
claims, [students] should learn to verify revise, and discard claims on the basis of
mathematical evidence." The NCTM further argues that teachers need to "pos[e]
questions…that elicit…students' thinking [by] asking students to clarify and justify their
ideas." The problems presented in this curriculum are posed for exactly this purpose.
Furthermore, they are "based on sound and significant mathematics" in order to "develop
students' mathematical understanding and to stimulate students to make connections and
develop a coherent framework for mathematical ideas." NCTM argues, and I agree, that
by asking students "what makes something true or reasonable in mathematics?" students
investigate "fundamental issues about knowledge." Also, "over time, [students] learn
new ways of thinking from their peers." Learning to learn from others will help students
learn in other classrooms as well.
In addition to designing this curriculum with the beliefs of the NCTM in mind,
Freudenthal's (1991) belief that "students should be given the opportunity to reinvent
mathematics" was influential as well. The problems in this curriculum were chosen to
encourage students to reinvent ideas in algebra by explicitly reflecting on the
mathematics they already do. The goal is for students to generalize the mathematics they
do in concrete situations to the more abstract concepts of algebra. Although this aspect of
the curriculum design was not based directly on NCTM documents, the idea of asking
students to extend their prior knowledge is consistent with the beliefs of the NCTM.
Furthermore, ideas in mathematics are interrelated; the ideas in arithmetic are abstracted
into the ideas of elementary algebra, just as the ideas of elementary algebra are abstracted
into the ideas of calculus and modern algebra. Therefore, this curriculum is designed to
help students see those connections. This could help students build the new and abstract
ideas of algebra on their prior knowledge of arithmetic as well as be more prepared to
extend the ideas of elementary algebra to modern algebra and linear algebra. It was also
designed with the goal that students would construct ideas about polynomial
multiplication in a way that they could easily extend to the more abstract ideas they will
see in higher mathematics rather than in a way that will cause confusion in later classes.
Finally, this curriculum is designed to help students learn to use the mathematics they
know to understand the mathematics they do not yet know. As Yogi Berra once said,
"you can see a lot just by looking." If we teach students to look, who knows what they
will see.
32
CHAPTER 5
CURRICULUM OVERVIEW
33
Overview
The goal of this curriculum is for students to construct a process for multiplying
polynomials. It is designed to be used in an algebra or pre-algebra class in which
students are learning the ideas of collecting like terms, the distributive property, and
polynomial multiplication. It can be used equally well in a junior high class, high school
class, or college class. The only skills it is assumed students have are an understanding
of the basic arithmetic operations (involving whole numbers, decimals, fractions, and
time) and exponents. It is not assumed that students know rules for working with
exponents, but it would not be a problem if the did either.
In order to do this, students will examine the role of the "variable" in different types of
"polynomials." They start by looking at the structure of arithmetic with multi-digit whole
numbers and time. It is hoped that students will, if they do not already, understand the
relationship between the base ten system and the way we write numbers. By doing this,
students will start thinking of multi-digit whole numbers as polynomials in powers of ten.
This idea should help students think of measurements of time as linear combinations of
its components as well. This is important because it will help students generalize the
mathematics implicit in the arithmetic they perform in familiar situations to operations on
polynomials. By making informal understandings of arithmetic in familiar situations
more explicit, students will develop informal understandings of polynomials, which will
help them create their own algorithms for multiplying polynomials.
This curriculum is divided into four broad activities, each of which addresses a particular
aspect of working with polynomials. These four aspects are collecting like terms, the
distributive property and working with exponents, multiplying polynomials, and
anticipating the form of products of polynomials.
An overview of these four activities is listed below.
Activity 1: Collecting Like Terms
Part A: Students think critically about addition and subtraction of multi-digit
whole numbers.
Part B: Students think critically about addition and subtraction involving time and
reflect on the relationship between the questions in Part A and those in Part B. In
this section , students confront the lack of closure in algebra for the first time.
Part C: Students extend the ideas they constructed in Parts A and B to variables
and symbolic algebra.
Activity 2: The Distributive Property
Part A: Students think critically about multiplication of a multi-digit whole
number by a single digit whole number.
Part B: Students think critically about multiplication involving time and reflect on
the relationship between the questions in Part A and those in Part B.
Part C: Students extend the ideas they constructed in Parts A and B to variables
and symbolic algebra (only involving the first power of variables).
Part D: Students extend the ideas they constructed in Part C to problems
involving powers greater than one of variables. In doing so, they construct some
of the laws of exponents as well.
34
Overview
Activity 3: Polynomial Multiplication
Part A: Students think critically about multiplication involving time and reflect on
the relationship between the questions in Part A and those in Part B.
Part B: Students extend the ideas they constructed in Part A to variables and
symbolic algebra.
Activity 4: Anticipating terms of products
Part A: Students observe the "special products" of polynomial multiplication such
as the difference of two squares and perfect square trinomials.
Part B: Students think critically about terms that arise in products of polynomials.
Part C: Students extend the ideas they constructed in Parts A and B to factoring
polynomials.
Activity 5: Culmination
Part A: Students factor more difficult polynomials.
Part B: Students factor by grouping.
Part C: Students individually assess their ideas from all five activities.
Each activity is comprised of several thought provoking problems, each of which
involves private think time, small group discussions, and whole class discussions. The
whole class discussions are a time for students to compare their answers with those from
other groups. They are also a time for the teacher to encourage reflection on the problem
as well as reflection on the similarities and differences between the different problems
students have explored during the preceding activities.
Each activity has a homework assignment related to the topics covered, which should be
completed by students and discussed in both small and large groups before the next
activity begins. There is a group assessment after every other activity, which should be
completed by students after the homework assignments have been completed and
discussed. After the small groups have completed the quizzes, the whole group should
discuss their responses to the quizzes as a class before the class moves on to the next
activity. At the end of the curriculum, there is a review of the material covered in all four
activities as well as an individual assessment.
During these activities, the teacher serves as a facilitator. It is the students' responsibility
to answer the questions and to defend their ideas. It is the teacher's role to ask questions
that direct students in the appropriate direction and to help students realize how and when
to prove or disprove ideas. Student interaction and discussion are a crucial part of these
activities.
35
CHAPTER 6
PRESENTATION OF THE CURRICULUM
36
Curriculum
A NOTE TO TEACHERS:
I should mention a few things about this curriculum before you begin exploring it. First,
I have included a description of all problems, assignments, and activities in the appendix.
This should help you recognize the goals intended in each activity. It should also help
you figure out what questions to ask your students to facilitate thinking about topics at
the heart of each activity.
Second, during small and large group discussions, refer back to the concrete examples to
help students prove or disprove examples. Encourage students to do the same on their
own too. This will help students learn to make convincing arguments. It will also
encourage students to explicitly reflect on the previous problems and to explicitly abstract
from them.
Third, if students need more problems for discussion than are listed, have students repeat
the protocol with a similar problem. Be sure to choose a problem that is different enough
from the other problems that students still have to think. If the problem is too similar,
students may begin to memorize a procedure to solve the problem rather than thinking
about the problem. It is important to remember that thinking critically is the goal of these
activities, not the computations themselves. If students are not actively reflecting on and
abstracting from the problems, they will not develop the understandings they are intended
to be developing. Therefore, supplementary problems need to be chosen carefully.
Fourth, it is important to give students adequate private think time so they can create their
own understandings. It is also important to encourage multiple ways of solving the
problems, so students feel comfortable construct their own procedures rather than
performing someone else's. Furthermore, it is important to ensure that all students have
an opportunity to share their ideas in small groups. It is also important that you listen to
small group discussions so you know what ideas are being expressed in each group. This
way, you can ensure that ideas are shared in whole class discussions. This is important
because it allows for both the sharing of good ideas with all students and addressing any
misconceptions they have.
Fifth, the length and content of this curriculum can easily be modified to suit your needs.
It is important, however, that you do not rush your students; give them enough time to
explore each question so they can fully develop the ideas they are constructing. That
said, I have included a few suggested modifications that could be made. These are
intended as ideas to start your thinking about how to adapt this curriculum to your needs,
and not as an exhaustive list of possible extensions.
For instance, this curriculum can be shortened if you do not want to cover all of the
topics. In a pre-algebra course that covers collecting like terms and the distributive
property, but not the multiplication of polynomials, you could stop after the first two
activities and the first group assessment. If the intended course of study is algebra,
however, rational expressions would be a reasonable topic with which to follow these
activities. An activity on rational expressions is included in chapter six. An example of a
37
Curriculum
concept from a course after algebra 1 that relies heavily on this idea of anticipating the
products of polynomials is the binomial theorem. An activity that derives the binomial
theorem by using this skill is also included in chapter six. Chapter six includes activities
that make historic, geometric, and ethnomathematic connections to polynomial
multiplication as well. This curriculum could also be extended to linear algebra. The
ideas about collecting like terms could easily be extended to matrix addition and
subtraction. It would be a bit more work to extend the ideas to matrix multiplication, but
that too could be done.
Finally, it is up to the individual teacher to decide how to use these activities to best meet
the needs of given students in a given course. These activities are intended not only to
cover the specific topics presented, but also to inspire ideas that would be beneficial for a
specific class. I hope this curriculum facilitates student learning of polynomial
multiplication and inspires teacher adaptation and modification of the ideas presented
here.
38
Curriculum
Activity 1: Collecting like terms
In this activity students will discuss the structure of multi-digit whole numbers. They
will do this by examining addition and subtraction of multi-digit whole numbers. By
discussing the structure of the base ten system in the context of addition and subtraction,
students will develop a deeper understanding of both multi-digit whole numbers and
addition. In doing so, they will develop an informal idea of collecting like terms, which
they will apply to unfamiliar situations to construct a formal idea of collecting like terms.
39
Curriculum
CLASSWORK - Part A
Ask students how to add 428 + 173
Ask students to think privately about how they solve this problem and why each action is
performed. After students have had adequate time to think, ask them to discuss their
ideas in pairs. Once the small groups have thoroughly discussed the issues at the heart of
multi-digit whole number addition, have the groups share their ideas in a whole class
discussion.
The goal of this question is for students to realize they are adding the ones, then tens, and
finally the 100's. They should see this when they discuss why one "carries" (i.e. that
carrying is combining the tens with the tens, the hundreds with the hundreds, etc, in other
words, collecting like terms) . In the process they should begin viewing 428 as 400 + 20
+ 8 and 173 as 100 + 70 + 3. It would be even better if students thought of these
numbers as 4x102 + 2x10 + 8, though, it is not essential that they do so. If students are
not ready for this concept the foundation for this idea can still be established (that is that
carrying is combining powers of ten). If students experience difficulties getting started
analyzing this problem, either have them explain their procedures for performing this
computation in their head and why they separate the numbers into the parts they choose
or suggest that students line the numbers up to perform the standard algorithm and to
explain why the numbers line up the way they are and what the process of carrying is.
Make sure they explain that they are combining powers of ten (it is not crucial that they
think of them as powers of ten, however, combining ones, tens, hundreds, etc is a good
start in that direction).
40
Curriculum
Next have students think individually about the process of subtracting 638 from 1037.
Once they have had enough individual time, have them discuss their ideas in dyads.
Follow up the small group discussions with a large group discussion of small group
findings.
The goal of this problem is to check for student understanding of the previous problem by
seeing whether students can extend the concepts to subtraction. That is, to see if students
can extend the idea of carrying to the ideas of borrowing. Once again, realizing that
"like terms" (powers of ten) are being collected. If students cannot do so, it is important
to assess whether this is because they did not understand the first problem (i.e. they are
still thinking of addition and subtraction in terms of algorithms they memorized) or if it is
because they do not see the connections between addition and subtraction (i.e. that
addition is the opposite of subtraction). Both of these skills are essential for a structural
understanding of algebra. Therefore, if students are not successful with this problem it is
important to have them discuss more related examples. If students are successful, they
are ready to move on to the examples using time.
41
Curriculum
CLASSWORK – Part B
Next have students think individually about the process of solving the following problem:
You have a layover at the airport of 6 hours and 12 minutes and you have
already been waiting for 4 hours and 37 minutes, how much longer do you
have to wait?
Encourage students to find multiple ways to solve this problem. Once they have had
enough individual time, have them discuss their ideas in small groups. Follow up the
small group discussion by a large group discussion of the findings from the small groups.
In the large group discussion ask students how this discussion is related to the previous
one.
The goal for this problem is that students will once again consider which terms are
combined when solving this problem. Encourage students to solve this problem by
setting up the subtraction problem in the standard form and to consider what it means to
carry within these constraints. The aim is to help students understand that not all
systems work in base 10, which will in turn help them begin to generalize the processes of
addition and subtraction. The goal of the reflective question in the large group
discussion is to encourage students to explicitly address the idea of like terms. To
address the question, students should be asked to take some private think time to think of
how they would answer the question, then students should take turns sharing their
responses to the question and discuss the ideas shared. Then the groups should share
their ideas with the class and the class can comment and ask clarifying questions about
the points made by other groups. Hopefully students will mention that in each question
similar units are being combined (i.e. minutes with minutes and tens with tens etc).
42
Curriculum
Next have students add 4 months, 2 weeks, 6 days, and 7 minutes to 3 months, 1 week, 3
days, and 4 hours. Once students have taken private think time, have them work with
their small groups before moving to a whole class discussion. During small group
discussions and the whole class discussions, make sure that students debate what a month
is (i.e. is a month four weeks, thirty days, or thirty-one days). Once again, in the large
group discussion, encourage students to explicitly state the ideas they are developing
about like terms.
One of the goals for this problem is to have students really think, once again, about
which terms can be combined. This is an occasion to uncover and address student
misconceptions of this issue. It is also a chance for students to address the lack of
closure that they might struggle with in algebra. By addressing this issue in a specific
example, students can use calendar arithmetic to confront the lack of closure they will
encounter in algebraic equations. This question addresses lack of closure because there
is no clear conversion between months and weeks or months and days since a month is
sometimes thought to be four weeks, while other times it is thought to be thirty days etc.
This way students can deal with this concept on its own rather than tackling it at the
same time that they confront the issues related to variables. Try to ensure that all
students have explicitly searched for connection between the problems before moving on
to the next problems, by once gain asking them to think about the similarities and
differences between the problem and the other problems. It is important that all students
have formally searched for meaning in the problems they are addressing so they are
ready to address the more abstract problems toward which they are moving.
43
Curriculum
CLASSWORK – Part C
Now ask students to think about how they would solve a + 3a + 4a + 5b + 2b and to
justify their procedures. Have students take private think time, then move to small group
discussions, and finally to a whole class discussion. In the whole class discussion, ask
students how this problem relates to the previous ones. Also ask students to discuss the
differences between the answer to this problem and the answers to the earlier problems.
The goal of this problem is for students to explicitly formalize their perception of like
terms, an idea they have been constructing. Addressing expressions with no conversion
rate between the terms at all will force students to address any misconceptions they have
as well as reinforce correct ideas they have been constructing. This is also an
opportunity for students to address lack of closure in algebra. Make sure students
explicitly state that, unlike arithmetic, the sum of algebraic expressions is not always a
single term let alone a number. Asking students how this answer is different from
answers in arithmetic is one way to get to that point.
44
Curriculum
Next, ask students to find the following sums
1)
2)
3)
3a + 2b + 1 + 7a + 3 + 4 + b
3a + 2c + 3c + 4
2b + 4a + 5c
Once students have worked on the problems on their own, have them compare their
answers with their group. Encourage them to justify their answers, particularly when
students have different answers for a given problem. Once all groups have reached
consensus, have the groups compare their responses. In the whole class discussion, just
as in the small group discussions, urge students to justify their reasoning. As always,
encourage students to refer back to the concrete examples discussed earlier to support
their claims.
The goal here is for students to test the ideas they have developed about like terms and to
formally address the "lack of closure" that can haunt beginning algebra students. In
these problems, students should address constants as well as problems that cannot be
simplified. Another goal of this problem, however, is for students to get accustomed to
defending their answers and questioning answers offered to them. This encourages
students to question their own ideas too.
45
Curriculum
Next, have students work individually on
1)
2)
3)
4)
5)
7a – 2a + 4b – 3b + 3a
6a + 2b – 3a – b
5a – 4b + 3a
3a – 4b + 3b – 2a
5a + 2b – a – 2b + 4
Follow the routine in the problem above.
The goal of this problem is for students to see how these problems change when
subtraction is introduced. That is, that they do not change. However, students do need to
realize that the negative sign is attached to the number immediately following it. In
doing this problem, students can refine their ideas about collecting like terms and about
terms involving negative coefficients.
46
Curriculum
Finally, once again using the routine from the last two problems, have students work on
1)
2)
3)
3m + 4n + 2mn + 2m + n
3x + 4x + 3x2 + 2 + 6
2x + 4y + 5x2 + 6y2 + 1
The goal of this problem is for students to address the issues that arise when terms share
a variable, but are not like terms. Students often want to collect terms that share a
variable, even when they are not like terms. Addressing this will help students cement
their ideas about like terms as well as confront any misconceptions they have. During
these problems students are likely to face some conflict because these problems require
some deep thought. Understanding that terms that share a variable are not necessarily
like terms will require some accommodation for students, but they will have a more
complete idea of like terms after they address these conflicts.
47
Curriculum
HOMEWORK
1) Explain to a young child how to subtract 1,074 from 24,153. Make sure to explain
both what you do and why you do it.
2) a) Explain to a visitor from another country, who uses a different currency than we do,
precisely which coins he will receive as change if he pays for something that
costs $1.33 with 1 dollar bill and 2 quarters. Assume he is given the fewest
possible number of coins.
b) Explain how this problem is related to our class discussions.
3) Find the following sums
a) 5q + 4r + 3r + 2q + 6r + 7q
b) 1 + 2a + 3b –ab + b + 3 + 4a
c) 2x + 3x3 + 6x2 – x + x3
d) 4a + 5b + 1
e) 3m + 2m – n + 6m + n
Extra credit: Describe addition and subtraction of decimals in the same way we
described addition and subtraction of whole numbers in class. Be sure to use examples.
Give students an opportunity to compare their answers with the other students in small
groups before turning in their homework.
The goal of this is to give students the opportunity to apply the processes they have
constructed. It is important for students to discuss their theories with others so they have
the opportunity to explicitly defend and question their theories.
48
Curriculum
Activity 2: The distributive property
In this activity students will examine both the structure and the process of multiplication
of multi-digit whole numbers. By discussing the structure of multiplication in the base ten
number system, I aim for them to develop a deeper understanding of the process of
multiplication (that is to think about the processes involved in solving multiplication
problems and understand why they work rather than merely performing them). In doing
so, it is hoped that students will develop an informal idea of distributing, which they will
apply to unfamiliar situations to construct a formal idea of the distributive property.
49
Curriculum
CLASSWORK – Part A
Ask students to think about what happens when you multiply 3 x 123. Once students
have taken some time to think on their own, have them discuss their ideas in small groups
and then discuss their observations with the whole class. Ask students how this
discussion relates to the earlier discussion of addition and subtraction.
The goal of this problem is for students to realize that each of the digits is multiplied by
three. It is important that they think of 123 as 100 + 20 + 3 so they realize that each of
the terms is multiplied by 3. It is even better if they think of it as 1x100 + 2x10 + 3x1
1x102 + 2x10 + 3 and realize that they can think of it as only the coefficients that are
multiplied by 3. The goal of asking students to explicitly address the connections to the
earlier problems is so they are actively thinking about collecting like terms because that
is an important part of the distributive property.
50
Curriculum
CLASSWORK – Part B
Next have students think on their own about what happens when you multiply 2 hours
and 12 minutes by 4. Then, in small groups, have students discuss their answers.
Finally, have a whole class discussion. Ask students to explicitly address the relationship
between these two problems.
The goal of this problem is for students to assess their ideas. That is, for students to
realize each unit of time is multiplied by four. Another motivation of this problem is for
the teacher to assess student ideas in order to address student misconceptions. This can
be done by seeing if students correctly extend the ideas from the previous problem to this
problem.
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CLASSWORK – Part C
Following the procedure in the previous problem, ask students to consider 3(a + 2b + 3c).
Also ask students how this problem relates to the earlier activities. Discuss the
differences between the answer to a multiplication problem with whole numbers and the
answer to multiplication problems with algebraic expressions.
The goal of this problem is to have students extend their experience with arithmetic to
construct the distributive property in abstract expressions. Hopefully they realize that
each term in the expression needs to be multiplied by three just as each "unit" was
multiplied in the earlier problems. This problem also requires coordination of ideas,
because students have to remember that the terms in the product cannot be combined as
they are not like terms. Students may still be struggling with their ideas about collecting
like terms, so they may need reminding that those issues are still relevant.
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Now ask students to work on the following problems on their own
1)
2)
3)
4)
5)
6)
3(2a + 4b + 5c)
2(2p -3q + 4r)
4( 2m + n + m + 3n +1)
-2(x + 2y -3z)
-(2a – b + 2c)
2(p + 2q – r) + 2p + q
Once students have worked on the problem on their own, have them share their answers
and the processes by which they arrived at them with their small group. Have students
debate their answers until a consensus is reached. Once all groups have reached
consensus, discuss and debate answers as a class. Have students share their solution
methods in addition to their solution. Discuss different ways to reach the same answer.
There are several goals in this activity. The first one is for students to assess their
algorithms for distributing. The second goal is for students to revisit their ideas of
operations with signed numbers. The third goal is for students to combine their ideas of
the distributive property with their ideas of collecting like terms. The fourth goal is for
students to establish an order of operations for combining the two processes.
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CLASSWORK – Part D
Now ask students to take some private think time on the following problems.
1)
2)
3)
4)
5)
6)
a(2a + 4b + 5c)
2p(2p -3q + 4r)
4m2( 2m + n + m2 + 3n2 +1)
-2xy(x + 2y -3z)
-ab(2ab – a2 b + 2ab2)
2p(p + 2q – r) + 2p + q
After students have worked on these problems by themselves, have them discuss the
problems with their groups. Once all groups have reached consensus, discuss and debate
answers as a class. Have students share their solution methods and discuss different ways
to reach the same answer. Make sure to discuss the properties students discover about
working with exponents. It is assumed students are already familiar with exponents and
what they represent. If this is not the case, take some time to explain the notation to
students.
The goal of this activity is for students to revisit their ideas about collecting like terms as
well as their ideas about the distributive property. Another goal is for students to
construct rules for working with exponents. Students may need to be encouraged to think
about what the exponents mean in order to construct operation involving them. To do
this, it could be suggested that students rewrite the terms with exponents without
exponents and then to rewrite their answers with exponents. If students do this, they
should be encouraged to reflect back on the relationship between exponents in the factors
and the exponents in their product. Furthermore, this problem gives students an
opportunity to assess the order of operations the developed for combining the two
processes.
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HOMEWORK
1) Simplify and explain how you arrived at your solution
a) 2x(3xy)
b) 2x2(4xy)
c) (2xy)2
d) (3xy)2(4xy)
2) Simplify
a) 2(x + y) + 3(y + z)
b) 4a(a + b) + 2b(a + 2b)
c) 3x(x – y + z) – 2(x + 2y + 3z)
d) m2(m – n) + n2(n – m)
Give students an opportunity to compare their answers with other students before turning
in their homework.
The goal of this is to give students the opportunity to build theories about how to apply
their processes for distributing, collecting like terms, and working with exponents.
Students should discuss their theories with others so they have the opportunity to assess
their ideas about collecting like terms, distributing, and exponents.
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GROUP QUIZ #1
Simplify
1) (2x)2 + 3x(x + 3) +2(x + 1)
2) x2(x + 1) + (3x) 2 + x(2y + 1) – y(x – 2)
The square root of a number, symbolized by √, is what you multiply by itself to get the
original number
For example √ 9 = 3, because 3x3 = 9
Since 3 is a whole number, we can write √9 or we can write 3,
√5 is not a whole number, so we simply write √5.
Note that √9 x √9 = 3 x 3 = 9 and √5 x √5 = 5.
Also note that 2√4 = 2 x 2 = 4, which is not the same as √(2 x 4) because 4 is not √8.
3) (√3)2 + 3(y + 2)
4) (√3)(y – 2) + 3 + √3
5) y((√5) + y) + (√5)(y + 1)
After each group has completed the quiz, discuss the answers as a group
The goal of this quiz is to have students apply their concepts of collecting like terms,
distributing, and operations with exponents to both new and old situations. This will
show the teacher any misconceptions students have that need to be addressed (for
instance, students may have a hard time figuring out what like terms are in the context of
radicals). Address any misconceptions (such as confusions that rise in collecting like
terms involving radicals) before moving on to the next topic.
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Activity 3: Multiplying polynomials
In this activity students will discuss the process of multiplication with multi-digit whole
numbers. By discussing the structure of multiplication in the context of the base ten
number system, students will develop a deeper understanding multiplication. In doing so,
they will develop an informal idea of polynomial multiplication, which they will apply to
unfamiliar situations to construct a formal process for polynomial multiplication.
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CLASSWORK – Part A
Ask students to think about what happens when they multiply 12 x 43. Once students
have taken their private think time, have them discuss their ideas in small groups, then
share their observations with the whole class.
For this activity it is important that students think of 12 as 10 + 2 and 43 as 40 + 3,
which they should realize to do after the last two activities. The goal of this problem is to
have students use this realize that multiplying 43 by 12 is the same as multiplying 40 by
10, 3 by 10, 40 by 2, and 3 by 2 and adding the products. Thinking of the partial
products in whole number multiplication should help students develop ideas about how to
multiply polynomials. The multiplication algorithm is based on this idea, so asking
students to explain the algorithm is one way to get at this. However, students are likely
to have methods for mentally multiplying numbers that rely heavily on these ideas as
well. So asking students to explain their mental processes is another way to get at the
ideas. Having students share both methods will help students think about the process of
multiplication from multiple perspectives.
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CLASSWORK – Part B
Next have students think on their own about what happens when they multiply 2 hours
and 2 minutes by 14. Then have students discuss their ideas in small groups. Then have
their groups share their ideas with the whole class. Make sure you ask students to
explicitly address the relationship between these two problems.
The goal of this problem is for students to test their ideas about multiplication. This is
also a good opportunity for the teacher to assess student understanding of the ideas of
partial products in order to address any misconceptions students have developed in
regards to the topic (such as forgetting to multiply each unit by each number and
collecting like terms as appropriate).
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CLASSWORK – Part C
Following the procedure above, ask students to consider the product of (a + b)(3a + 2b).
Once they have worked on the problem on their own, have them discuss their ideas in
small groups. During the whole class discussion, ask students how this problem relates to
the earlier activities
The goal of this problem is to have students extend their informal ideas about arithmetic
to polynomial multiplication by thinking of the partial products they uncovered in the
earlier problems. Students may still be struggling with their ideas about collecting like
terms, distributing, and working with exponents, so it is important to help students
solidify those ideas as they develop new ideas about new processes.
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Now ask students to work on the following problems on their own.
1)
2)
3)
4)
5)
(x + y)(x + y +z)
(a – b)(2a – 3b)
(p + q + r + s)(p + q + r + s + t)
(2m + 3n) (4m + 5n + 6)
(x2 + x + 1)(2x2 + 3x + 4)
Then have students share their answers and the processes by which they arrived at them
with a small group. Have students debate their answers until a consensus is reached.
Once all groups have reached consensus, discuss and debate answers in the large group.
Have students share their solution methods and discuss different ways to reach the same
answers.
The goal of these problems is for students to extend and assess their ideas of polynomial
multiplication. By asking students to extend their ideas of polynomial multiplication to
polynomials with more than two terms, students are being asked to realize that all terms
in one polynomial still need to be multiplied by all terms in the other polynomial. This is
also an opportunity for the teacher to discover student misconceptions (such as only
multiplying some of the terms and confusion about like terms) and to address them
directly.
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Now ask students to take some private think time on the following
1)
2)
3)
4)
5)
6)
(x + 3)(x + 4)(x + 5)
(2x2 + 5)(3x + 4)
(2x2 + 5) (3x2 + 4x) + x + 5
(a2 + 2) (3a2 + 4a) + (a + 3)(2a + 5)
(a + ab + b)(2a + 3b + b2)
(2p + 3q)(4r + 5s)
Once students have worked on all six problem on their own, have them discuss their
answers in their small groups. Once small groups have agreed on answers, compare and
discuss answers with the whole class. Make sure students explain how they arrived at
their answers. Also acknowledge and encourage multiple methods of arriving at correct
answers.
The goal of these problems is, once again, to have students extend and assess their ideas
of polynomial multiplication. In this problem terms involving exponents and multiple
variables are being introduced for the first time. This requires students to readdress laws
of exponents and like terms in more complicated situations than the earlier problems.
This is also another opportunity for the teacher to discover student misconceptions and to
address them.
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HOMEWORK
1) Simplify
a) (a + b)2
b) (p + q)(p – q)
c) (2m – 3n)2
d) (2x – 1)(2x + 1)
e) (3a – 2b) 2
f) (2c + 2d) 2
g) Use your procedure to find 132 byt thinking of 13 as 10 + 3
2) Describe at least 3 patterns you notice in number 1.
3) Simplify
a) 2(x + 3)(x – 5)
b) 3(x2 + 4)(x + 4)
c) –(x – 3)(3 – x)
d) 2x(x2 + x + 1)2
Give students an opportunity to compare their answers with the other students in their
group before turning it in.
The goal of this activity is for students the opportunity to build theories about how to
apply their processes for collecting like terms, working with exponents, distributing, and
multiplying polynomials. It is also for students to begin recognizing patterns such as the
difference of two squares and the square of a binomial. Students should discuss their
theories with others so they have the opportunity to assess if their theories are correct or
not. This way students can be sure they noticed the patterns they were intended to notice
in the first part of the homework and to be sure they correctly multiplied the polynomials
and simplified the terms. Also note that this is the first time students are thinking about
squaring a polynomial. It is important they realize they need to multiply each term in the
polynomial by all of the other terms, and not just square each term. If students have
trouble with this, make sure you remind them what exponents mean and to rewrite the
expression without exponents. This may help them see how to square a polynomial.
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Activity 4: Anticipating terms of products
In this activity students will predict the terms of products of polynomials without actually
multiplying them. By predicting the structure of the products of polynomials, they will
develop a deeper understanding of both the process and the product of polynomial
multiplication. In doing so, they will develop an informal idea of factoring which will
help them to construct a process for doing so.
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Curriculum
CLASSWORK – Part A
Ask students to match the expressions on the left with the term(s) on the right that could
appear in their expansions without actually multiplying out the expressions. Also ask
them to explain why each term will appear in the product. Do not worry about the
coefficients.
For example: if the problem was r(p + q + 1),
the answer would be (d)
because r(p + q + 1) = pr + qr + r,
so r is the only term listed below that appears in the product
Note that in this example there is only one answer choice that works, for the following
problems, however, you might want to select more than one choice.
1) (p + q)(r + s)
2) (p + q)(r + 1)
3) (p + 1)(q + 1)
4) (1 + p)(q + 1)
5) (p + q) + (r + s)
a) 1
b) p
c) q
d) r
e) s to have students think about what terms arise in products of
polynomials. This will help students anticipate products of polynomials and will help
students factor because they will be more aware of how and why terms arise in the
product of polynomials.
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Curriculum
Ask students to match the expressions on the left with the term(s) on the right that could
appear in their expansions without actually multiplying out the expressions. Also ask
them to explain why each term will appear in the product. Once again, without worrying
about the coefficients.
1) (x + 1)2
2) (x2 + 1) 2
3) (x2 + x + 1) 2
4) (x2 + 1)(x + 1)
5) ( once again to have students think about what terms arise in
products of polynomials. This will help them extend their ideas from the earlier problem
to a situation where only one variable is used and where polynomials are being squared.
Because of those changes, these problems are slightly more difficult than the earlier
ones. This to should help students anticipate products of polynomials and will help
students factor.
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Ask students to match the expressions on the left with the term(s) on the right that could
appear in their expansions without actually multiplying out the expressions. Also ask
them to explain why each term would appear in the product. Once again, do not worry
about the coefficients ask them to multiply out the expressions to
check if they are right and then discuss the results as a class.
The goal of this activity is to have students continue thinking about what terms arise in
products. These problems are more difficult than the earlier problems because they
involve three factors rather than just two. This means students have to coordinate more
processes and revise their previous strategies for answering these questions to including
a third term. This will help reinforce student ideas about polynomial multiplication by
making students think about which terms get multiplied together to create the terms in the
product.
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Curriculum
Ask students to match the expressions on the left with the term(s) on the right that could
appear in their expansions without actually multiplying out the expressions. Once again
ask them to explain why they would. This time ask students to take the coefficients into
account, not just the variables.
1) (x + 1)2
2) (x2 + 1) 2
3)(x2 + x + 1) 2
4) (x2 + 1)(x + 1)
5) ( arise
in products. The difference between this activity and the earlier activities is that in these
activities students are not only thinking about what terms arise, they are also thinking
about how many of each term will arise. This too will help students anticipate products
of polynomials and will help students factor by helping students understand how terms
arise in the product of polynomials.
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Ask students to match the expressions on the left with the term(s) on the right that could
appear in their expansions without actually multiplying out the expressions. Once again
ask them to explain why. Once again, taking the coefficients into account, not just the
variables and
how many of them arise in the products polynomials. These problems are more difficult
than the problems on the previous page because they involve three polynomials rather
than just two. Once again, these means students have to coordinate more factors at one
time and think about how three polynomials are multiplied together. This will help
students anticipate products of polynomials and will help students factor.
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CLASSWORK – Part B
Ask students to fill in the blanks to make the following expressions true
1)
2)
3)
4)
5)
(x + 2)(x + __) = x2 + 5x + 6
(x + 1)(__ + 3) = 4x2 + 7x + 3
(x + __)2 = x2 + 6x + __
(2x + __)(x + 2) = 2x2 + 7x + __
(x + __) (x + __)= x2 + 5x + 4
Encourage students to multiply out the expressions to check if they are right before
discussing their answers with their groups. Once groups have arrived at consensus,
discuss the results as a class.
The goal of this activity is also to have students further their thinking about what terms
arise in products. These problems do this by having students think about how terms arise
in a product. These questions are different from the earlier problems in that they ask
students to think about the terms in the factors rather than just the terms in the products.
They also ask students to coordinate the terms of the factors with the terms on the
products. This will help students' understanding of polynomial multiplication because it
asks them to look at the process from a different perspective. Furthermore, this will help
students anticipate products of polynomials and will help students develop informal rules
for factoring polynomials by asking them to think deeply about the process of
multiplication as well as how factors and products are related.
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Again, ask students to fill in the blanks to make the following expressions true.
1)
2)
3)
4)
5)
(x2 + x + 2)(x + __) = x3 + __x2 +3x + 2
(x2 + x + 1)(__ + 3) = 2x3 + 5x2 + __x + 3
(x2 + x + __)2 = x4 + 2x3 + 3x2 + 2x + __
(x2 + 2x + __)(x3 + __ + 2) = x5 + 4x4 + 7x3 + 8x2 + 4x + __
(x2 + 2x + __)(2x2 + x + __) = 2x4 + 5x3 + 12x2 + 14x + 12
Encourage students to multiply out the expressions to check if they are right before
discussing their answers with their groups. Once groups have arrived at a consensus,
discuss the results as a class.
The goal of this activity is, once again, for students to further their thinking about what
terms arise in products and to coordinate the terms in the factors with the terms in the
products. These problems are more difficult than the ones on the previous apge because
they involve trinomials as well as binomials. This will help students anticipate products
of polynomials and will help students factor.
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HOMEWORK
1) Describe what the terms in the following products will look like WITHOUT actually
multiplying. Explain why the terms will look like you describe. (Note: explanations
matter more than answers, you will get full credit if even if your answers on parts a-d are
wrong, as long as you carefully explain your reasoning on parts a -f. You will not get full
credit if all are correct but your explanations are not thorough.)
a) (a + 1)(a + 2)(a + 3)
b) (a + 1)(b + 1)(c + 1) (d + 1)
c) (x3 + 1)( x2 + 1)(x + 1)
d) (2x2 + 1)( 2x2 – 1)(x + 1)
e) Multiply the polynomials to see if your predictions were correct
f) Explain why any incorrect predictions were incorrect
2) Fill in the blanks (Note: no partial credit will be given, so make sure your answer
works!)
a) 2x(x + 3)(x + __) = 2x3 + 8x2 + 6x
b) (x + 5)(x - __) = x2 + 2x + __
c) (2x + 3)(3x - __) = 6x2 + 11x + __
d) (__ + __)(x2 + 6) = 3x3 + 2x2 + 18x + __
e) (__ + __)(2x – 3) = 6x2 – 9x
f) (3x + 4)(3x + __) = 9x2 – 16
g) (2x2 + __)(x2 + 1) + __ = 2x4 + 5x2 + 1
3) Find polynomials whose product is (Note: no partial credit will be given here either.)
a) 16 - 49x2
b) 4x6 - 12x3 + 9
c) x2 + 5x + 6
d) x2 - 5x + 6
e) x2 + 5x - 6
f) x2 - 5x - 6
Students can check all of these problems on their own, so they do not need time to discuss
their answers with their classmates, but students need to be warned of this ahead of time
if you decide not to give students time discuss their answers. That way students will know
that they need to check their answers on their own carefully because they will not be able
to compare with their classmates.
The goal of this assignment is to have students think about what terms arise when
polynomials are multiplied and to test their theories about polynomial multiplication
(questions 1 and 2). It is also an opportunity for students to make their informal ideas
about factoring explicit (question 3)t. Furthermore, it is a chance for students to realize
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they can tell if their answers are correct on their own and to make students accountable
for checking their work whenever possible.
Be sure to discuss with students that what they are doing in question #3 is called
factoring. They may be able to guess this term because it is a term they are familiar with
from whole number multiplication. Allow them to guess the word before telling it to
them; you may need to make the connection to whole number multiplication explicit to
help them guess the word. To do this, you could remind students about the relationship
between whole numbers and polynomials we have been using throughout the curriculum.
Then, you could ask students for an example of an analogous problem to number three
using whole numbers rather than polynomials. They may answers this by asking for the
factorization of a number or by showing a factor tree. Either way, the word factor could
easily arise.
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GROUP QUIZ #2
Simplify
1) 2a2(a2 + 1)(a + 1) – a2 (a + 2) - (a - 1)
2) (a + b)2 + (a – b)2 – a(a + b) –b (a – b)
3) (m + 2)(m2 - 2m + 4)
4) (m - 3)(m2 + 3m + 9)
5) (m2 - 2)(m4 + 2m2 + 4)
6) (m3 + 3)(m6 - 4m3 + 9)
7) Describe any patterns you see in problems 3-6
After each group has completed the quiz, discuss the answers as a class
The goal of this quiz is to have students apply their concepts of collecting like terms,
distributing, multiplying polynomials, and operations with exponents to both new and old
situations. This will show the teacher any misconceptions students have that need to be
addressed. Another goal of this problem is for students to notice the identities for the
sum of cubes and the difference of cubes. Address any problems before moving on (such
as mistakes in multiplying, distributing, and collecting like terms and students not
observing the identities in questions three through six.
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Activity 5: Review
In this activity students will review the concepts they learned over the course of the past
four activities. They will test the processes they have constructed in new situations. This
will encourage students to reflect on the topics the covered during these activities and
generalize the processes and ideas they developed. Therefore, it is important that all
assignment and assessments have been discussed before moving on to this point in the
curriculum.
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Curriculum
CLASSWORK – Part A
Ask students to completely factor (if possible) the following expressions.
1) x6 - 1
2) x6 - 64
3) 8x6 - 1
4) 9x6 - 1
5) 10x6 – 1
6) 8x6 – 64
7) 9x6 – 64
8) 10x6 – 64, have students share their different solution strategies. Discuss
the fact that some strategies more easily lead to the answer than other strategies. Be sure
to discuss the fact that there are several equivalent answers.
The goal of this activity is, once again, for students to further their thinking about what
terms arise in products. It is also to help students begin to factor. These problems
involving using the difference of squares and the sum and difference of cubes identities.
These problems involve factoring multiple times, so they show students that factoring
can take multiple steps. Students may realize this as they use different strategies (for
instance the difference of squares versus the difference of cubes) and realize that the
answers are equivalent by fully factoring the expressions. Furthermore, these problems
helps students develop symbol sense (Arcavi, 1994) because it will help them learn to
determine which of several equivalent expressions is best for a given situation. They
learn this from these activities because some factored forms of the expressions make it
easier to factor completely than others. For instance, if the first problem is though of as
the difference of squares, it immediately become the sum and difference of cubes, both of
which are easy to factor. Whereas, if the students start with the difference of cubes, it
takes more work to factor the resulting trinomial.
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Ask students to factor (if possible) the following expressions.
1) x2 + 5x + 6
2) x2 - 5x + 6
3) x2 + 5x - 6
4) x2 - 5x - 6
5) x2 + 8x + 16
6) x2 - 8x + 16
7) x2 + 8x - 16
8) x2 - 8x - 16
9) x2 + 4x + 12
10) x2 - 4x + 12
11) x2 + 4x - 12
12) x2 - 4x - 12
Encourage students to multiply out their answers The first four problems on this page get at the heart of factoring
because different factors create similar looking products. By factoring these, students
are forced to think about when both factors involve addition and when both factors
involve subtraction as well as which factor has a positive term and which a negative term
when the operations are different. The last eight problems encourage students think
critically about how terms arise while they search for factors of the polynomials. This is
because some of the problems are not factorable, which means students have to think
about how they know if a polynomial is not factorable. All of these pieces of these
questions will help students formalize the process of polynomial multiplication by using
the concepts inherent in it to factor.
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Ask students to factor (if possible) the following expressions.
1) 3x2 - x + 4
2) 2x2 + 5x + 2
3) 35x2 - 57x - 44
4) 3x2 - 4x + 1
5) 6x2 + 13x + 6
6) 2x2 + x – 1
7) 2x2 - 5x + 12
8) 9x4 + 18x2 + 8
9) 18x2 + 3xy - 10y2
10) 12x2 - 31xy + 20 ask students to extend the processes they
constructed for factoring in the earlier problems to polynomials with leading coefficients
other than one. Furthermore, these problems ask student to formalize their ideas about
what terms arise in products of polynomials by using those ideas to factor.
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Curriculum
Ask students to factor (if possible) the following expressions.
1) 20x2 + 100x + 125
2) x3 - x2 – 6x
3) 12x2 + 28x - 24
4) 18x3 - 21x2 – 9x
5) 5x2 + 10x + 30
6) x2 + 8x
7) 36x - 49 x3
8) 81x2 - 64x
9) x5 - 6x3 + 5x
10) 5x2 + 8xy - 10 require students to factor out a common factor.
For some of these problems that is all that is required, while others require students to
factor further after extracting the common factor. This will remind students that
factoring out monomials is part of factoring. It will also show students that doing so is
an important first step in factoring. It is also for students to formalize this skill by using
it to factor.
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CLASSWORK – Part B
Ask students to factor (if possible) the following expressions.
1) x3 + 3x2 + 2x + 6
2) 3x3 + 2x2 + 3x + 2
3) 18x3 - 21x2 + 30x - 35
4) ax - bx + ay - by
5) 6x2 - 3x + 2xy - y
6) x13 + x7 + 2x6 + 2
7) 4x5 + 6x4 + 6x3 + 9x2
8) x6 – x4 – x2 + 1
9) 14x3 + 18x2 - 21x + 27
10) x2 + 2xy + y2 - 1
Students may need a lot of coaching when attempting these problems. Encourage
students to look for common factors in pairs, and then help them notice the like terms.
Also, encourage students to multiply out their solutions to check if they are correct answers are equivalent, even though they are from different
strategies and in different formats.
The goal of this activity is, once again, for students to further their thinking about what
terms arise in products. In these problems students confront factoring by grouping. If
students have difficulty constructing this idea, encourage them to group the terms into
pairs and to find the greatest common factor in each pair. Then point out to students that
they are left with like terms, which can be collected. Then help students realize that that
result is the product of two polynomials, so it is a factored form of the original
polynomial.
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INDIVIDUAL ASSESSMENT
1) Add
a) 7x3 + 6x2 + 4x + 1 - 7x3 + 6x2 - 4x + 5
b) 3x4 + -5x2 - 6x + 5 + -4x3 + 6x2 + 7x – 1
c) (7x4 – 5x + 6) – (3x2 + 8x – 12) + (8x2 – 10x + 3)
d) (-xy4 – 7y3 + xy2) + (-2xy4 + 5y – 2) – (-6y3 + xy2)
2) Multiply and simplify as appropriate
a) 2y(4y – 6)
b) 4x2(3x + 6)
c) 2x(3x2 + 4x – 3)
d) 10x(-y5 – xy3 + 12x)
e) (5x – 6)(x + 2)
f) (4x4 + x4)(x2 + x)
g) (2x + 3y)(2x – 3y)
h) (x + 2)3
i) (x2 + x + 1)( 3x2 + 2x + 1)
3) Factor (if possible)
a) x2 + 49 + 14x
b) x3 – 18x2 + 81x
c) 2x2 - 128
d) 3x3y – 2x2y2 + 3x4y – 2x3y2
e) –x4 + 7x2 +18
f) x8 - 1
g) x3 + 4x2 + x + 4
h) x4 + 9
81
CHAPTER 7
EXTENSIONS
82
Extensions
Extension 1
This lesson is intended to extend the ideas of polynomials students have constructed to
rational expressions. This is a good next activity for an algebra class because it builds on
the skills students just developed. Students are once again asked to deeply explore the
properties of arithmetic with which they are familiar in order to construct strategies for
working with rational expressions. In the first activity, students are asked to think about
what it means to reduce fractions. Immediately after that, they discuss multiplying and
dividing fractions. In the second activity, students deal with addition and subtraction of
fractions by using the ideas they developed in the earlier activity.
83
Extensions
CLASSWORK
Ask students to think privately about how to reduce12/30. reducing fractions, have the
groups share their ideas in a whole class discussion. In the whole class discussion, be
sure that students realize that they are looking at the numerator and denominator as the
product of numbers and not as the sum of numbers.
The goal of this problem is for students to think deeply about the fractions with which
they are familiar. This should help students extend their ideas about fractions to rational
expressions involving unknowns. This example will also serve as a concrete example to
which students can refer while they are constructing formal ideas about rational
expressions. This will be helpful because it will offer students an example to help them
justify their ideas.
84
Extensions
Now ask students how to reduce (x + 4)(x + 1) and x2 + 5x + 6.
(x + 1)(x + 2)
x2 – 9
After students have had adequate private think time, have them discuss their ideas with
small groups. Encourage students to think of a way to assess if their answers are correct.
Also encourage students to refer back to the concrete example they just worked with to
justify their claims.
Once the small groups have reached a consensus, have the small groups share their
answers with the other groups in a whole class discussion. Ask the groups to share the
different methods they found for assessing if their answers were correct.
The goal of this problem is for students to extend their ideas of reducing fractions to
reducing rational expressions. It is also to encourage students to think of polynomials as
products of other polynomials. This will help students more efficiently find common
denominators for rational expressions. Another goal of this problem is to help students
learn to assess if expressions are equivalent on their own. If students try to cancel terms
that are being added, point them back to the numerical example and ask what happens if
that process were used there.
85
Extensions
Now ask students to multiply the following rational expressions.
1) 3 . 2
4 9
2) (x – 4)(x – 2) . (x + 3)(x + 2)
(x + 3)(x – 2) (x – 4)(x + 5)
3) x2 - 6x + 9 . x2 – 5x + 6
x2 – x – 6 x2 - 9
4) x3 - 5x2 – 6x . x2 – 5x + 6
x2 – x – 6
x4 - 16
2
5) x - 6x + 9
. x2 – 4x + 5 . x2 – 5x + 4_
x2 – 7x + 12
x2 - 1
x2 – 7x + 12
2
2
6) x - 2x - 3 . x + 2x - 3
x2 – 1
x2 - 9
7) x2 – 5x – 6 . x2 – 5x + 6
x2 + 5x – 6 x2 + 5x + 6
8)
x2 - 4x + 9
. x2 – 6x + 5
3
2
x - 5x + 6x - 30 x2 – x + 3 to arrive at their answers. Make sure students realize that there are multiple
ways of expressing the correct answer.
The goal of this activity is for students to learn how to multiply rational expressions.
This will help students when they need to work with rational expressions later on and it
will reinforce students' factoring skills. This will also further students' understanding of
fractions and operations with polynomials by asking them to look at each of these
perspective from a different angel.
86
Extensions
Now ask students to divide the following rational expressions.
1) 3 ÷ 12
4
7
2) x2 - 6x + 9 ÷
x2 – x – 6
3) x2 + 2x - 8 ÷
x2 + x – 12
4) x2 + x + 6 ÷
x2 + x – 6
5)
x2 - 6x
3
x + x2 + x
x2 – 5x + 6
x2 - 9
x2 – 7x + 12
x2 – 2x - 8
x2 – x + 6
x2 – x - 6
2
2
÷ x – 1 ÷ x – 5x – 6
3
x -1
x2 + 5x – 6 as a class. Make sure students realize that there are multiple ways of
expressing the correct answer.
The goal of this activity is for students to learn how to divide rational expressions. This
will help them when they need to work with rational expressions later on and it will
reinforce their skills with factoring. This will also further students' understanding of
fractions and operations with polynomials by having students treat those as basic skills
they use while constructing new ideas.
87
Extensions
HOMEWORK
Compute the following given
a = 2, b = x – 1, c = x2 + 3x – 4, d = x2 + 5x + 6, e =
x + 4 , f = x2 + 4x + 1
x2 + 2
x2 + 3x - 4
x2 + 5x + 6
x2 + 5x + 6
1) a ∙ b
2) a ÷ b
3) b ÷ a
4) c ∙ a
5) c ÷ a
6) b ÷ c
7) c ∙ d
8) e ∙ d
9) e ÷ f
10) b ∙ c ÷ f
88
Extensions
CLASSWORK
Ask students to think privately about how to add ½ and ¾ . adding fractions, have the
groups share their ideas in a whole class discussion.
The goal of this problem is for students to think deeply about addition of fractions. In
doing so, encourage them to explicitly state that finding a common denominator creates
"like terms", which is why it is essential to find a common denominator before adding or
subtracting (you can do this by asking students how this problem relates to collecting like
terms). This will help students with operations involving fractions by having them think
more deeply about the process from a different perspective. This should help students
extend their ideas about fractions to rational expressions involving unknowns. This
example will also serve as a concrete example that students can refer to while they are
constructing formal ideas about the addition of rational expressions.
89
Extensions
Now ask students to add the following rational expressions.
1) 3 + 2
4 9
2) x2 + 5x - 6 + x2 + 9x + 18
x–6
x-6
3) (x – 5)(x – 2) + (x - 3)(x + 2)
(x + 3)(x – 2)
(x – 2)(x + 5)
4)
x – 6 __ + x2 – 5x + 6
x3 – x2 + x – 1
x4 - 16
2
2
2
5) x + 5x + 6 +
x –1
+ x + 3x
x2 – 9
x2 – x - 2
x2 + x
Encourage students to check their answers before comparing their answers with their
neighbors. Have students share their methods for solving and checking in addition to
sharing their answers. After all small groups have reached a consensus, discuss the
different answers and solution methods as a class. Make sure students realize that there
are multiple ways of expressing the correct answers.
The goal of this activity is for students to learn how to add rational expressions. This
will help them when they need to work with rational expressions later on and it will
reinforce their skills of working with fractions, collecting like terms, and factoring. This
will also further students' understanding of fractions and operations with polynomials.
Furthermore, this activity will help students develop a list of strategies to simplify
complicated problems.
90
Extensions
Now ask students to subtract the following rational expressions.
1)
1
_
x + 4__
x2 – 16 x2 – 3x - 4
2)
6
_
3 _
9 - x2
12 + 4x
3) x + 1 _ x - 1
x-1
x+1
4) x – 3 _ x + 3 _ 5x2 + 27
2x + 6
3x - 9
6x2 - 54
5)
x+2
_ 2+x + 2–x_
2
x + 5x + 6
4 - x2 x2 + x - 6
Encourage students to check their answers before comparing with their neighbors. Have
students share their methods for solving and checking in addition to sharing their
answers. After all small groups have reached a consensus, discuss the different answers
and solution strategies as a class. Make sure students realize that there are multiple ways
of expressing the correct answers.
The goal of this activity is for students to learn how to subtract rational expressions.
This will help them when they need to work with rational expressions later on and it will
reinforce their skills of collecting like terms, distributing, and factoring. This will also
further students' understanding of fractions and operations with polynomials.
91
Extensions
Now ask students to perform the following computations.
1) 2x3 + 4x2 – 6x . 2x3 – 2x_ _
4x2 – 1
2x3 + 5x – 3x
2)
6
3 _
+
2x2 – 5x - 3
2x3 + x2 – 2x - 1
3)
1
_
1
__
2x2 – x – 1
2x3 - 8x2 + 6x
4) 6x2 + x – 1 ÷
9x2 – 1 __
2
2
4x + 10x + 4
2x + 12x + 16
3
2
5) 8x + 4x – 4x ÷
8x2 – 2 __
2
2
x + 3x
2x + 7x + 3
6)
x
_
2
_
2x2 + x – 6
4x2 – 4x - 3
7)
2
1
_
+
2x2 + 8x + 6
2x3 + 6x2 + x + 3
8) 4x3 - 4x2 + 4x - 4 . 3x3 + 3x2 – 27x_- 27
4x2 – 1
2x4 – 2
Encourage students to check their answers before comparing with their neighbors. Have
students share their methods for solving and checking in addition to sharing their
answers. After all small groups have reached a consensus, discuss the different answers
and methods of getting there as a class. Make sure students realize that there are multiple
ways of expressing the correct answers.
The goal of this activity is for students to determine which process to use when solving
each problem. This will reinforce the ideas they have constructed about operations with
rationale expressions and it will reinforce their skills of collecting like terms,
distributing, and factoring. This will also further students' understanding of fractions and
operations with polynomials.
92
Extensions
HOMEWORK
Perform the indicated operation, reduce all answers
1)
x__ ___ ÷ y _
3x2 + 3xy – x + y
x–y
2) xy – 3x ÷ y - 3_
y
x
3) ( 1 + 1 ) ÷ ( 1 _ 1 )
m n
m n
4) ( 1 – 1 ) ÷ (1 1 )
x2
x
2
5) x
x _
÷
x2 - y2
y+x
6) ( x + y ) ÷ ( 2 _ 3 )
2 3
x y
These problems are more difficult that the earlier problems because they require
students to coordinate the different processes they constructed in the earlier activities
and to think deeply about order of operations. Reflecting on the processes
constructed in class will help students further their ideas about those processes.
93
Extensions
GROUP QUIZ
Find the value for ? that makes the following equations true
1)
4x
x2 – 1
+
? =
2x__
x-1
2) x2 – 5x – 6 _ ? = x2 – 5x + 6
x2 + 5x – 6
x2 + 5x + 6
3) x2 - y2
xy2
÷
? = x-y
y3
4) x2 – 5x – 6 . ? = x2 – 5x + 6
x2 + 5x – 6
x2 + 5x + 6
5) x2 - x - 20
x2 + 7x + 12
6)
÷
? =
1 _
x–5
x–5
- ? =
x2 – x – 26 __
3
x - 4x – 5
x + x2 - 25x - 25
2
7) 2x2 - x + 6
5x2 + 6x + 1
8) 5x + 10
x+4
+
? = 5x2 + x + 1
5x + 1
. ? = x2 + 5x + 6
x2 + 5x + 4
This quiz asks students to think deeply about the processes involved in operations
involving rational expressions because the problems ask students to think about how the
answers arise. Reflecting on where answers come from helps students solidify their ideas
about the processes themselves.
94
Extensions
Extension 2
In this lesson students learn a little about the historical developments of the multiplication
of polynomials. This activity will help students understand the development of algebra.
It will also help visual students see a geometric approach to the multiplication of
polynomials. Furthermore, it will help students make connections between algebra and
geometry. This lesson could be extended into a geometry lesson when students find areas
and volumes of unconventional shapes by breaking them in to simpler shapes.
95
Extensions
CLASSWORK
Ask students to find the area of a square with side length a.
Encourage students to draw a picture in addition to writing the answer. This problem
might be difficult for students who have not yet confronted the lack of closure in algebra
because they might be expecting a numerical answer. Make sure that all students are
comfortable with the answer to this question before moving on to ensure that all students
are comfortable with what is being asked of them.
The goal of this problem is to remind students about geometry and geometric models and
for students to review area. It is also for students to confront lack of closure the face in
algebra through a geometric model. These skills will all be essential in the later
problems.
96
Extensions
Once students are comfortable with that problem, ask students to find the area of a
rectangle with length a and width b.
This problem should not be difficult for students given the last problem, but it is an
important question to ask students to ensure that students are comfortable with the lack of
closure they will face in their answers to future problems.
The goal of this problem is to ensure that students are comfortable with the processes
they are using (how to use a geometric model) as well as with what is being asked of
them (to draw a picture and to write an expression) before moving on to more complex
problems.
97
Extensions
Now ask students to find the area of a rectangle with length a and width a + b.
Encourage students to draw a picture to help them find a solution. Discuss the different
answers students find as a class. Hopefully some students will have written a(a + b) and
other students will have written a2 + ab. Discuss why both answers are correct and why
that means they are equivalent. Also discuss how this problem relates to what students
have learned in the earlier activities.
The goal of this problem is for students to think about how to divide large areas into
smaller areas of which they already know the area. This will help students see a visual
model of the distributive property.
98
Extensions
Next have students find as many expressions as they can think of that express the area of
a square with side length a + b.
Ask students to explain how they found each expression and why it makes sense. Have
students discuss the answers they found individually in small groups, then discuss the
expressions and accompanying representations as a class. Once again, be sure to ask
students how these ideas relate to the ideas they constructed in class.
Use this opportunity to teach the students about the historical development of
polynomials. At this point, explain to them that these properties were originally
developed geometrically. Be sure to tell them that these processes are different than
those processes because they use variables. Variable use allows us to reify the algebraic
representation of the area model, which allows us to analyze the geometric processes
from an analytic stance. Although this is not true to the original discovery of these ideas,
it allows students to gain some insight into the evolution. If students have a strong
background in geometry, show them the original derivations, such as the ones found in
The Elements.
The goal of this problem is for students to make connections between the distributive
property, collecting like terms, polynomial multiplication, and the geometric models.
This will help students reinforce the algebraic ideas they have constructed as well as the
geometric ideas with which they are working.
99
Extensions
Once students are comfortable with the various expressions and how they are related, ask
students to find the following products. Encourage them to use any method they like to
expand the expressions, and to use a different method to check if their answers are
correct.
1)
2)
3)
4)
5)
6)
7)
8)
(a + 1)(b + 1)
a(a – b)
(a + b)(a – b)
(a + 1)(a + b)
a(a + b + c)
(a + b)(a + b +c)
(a + b)(a + b +c)
(a + b + c)2
Once students are confident in their answers, have them discuss their answers and
methods of finding them within a small group. Once the small groups have reached a
consensus, have the small groups share their results as a group. Encourage the class to
find as many different ways of expressing the products as possible. Discuss which
methods students liked best for which problems and why.
The goal of these questions is for students to reinforce the algebraic ideas they have
constructed and to have them use the geometric models to do so.
100
Extensions
HOMEWORK
Find the following products using any method you like, then check your answers by
using a different method and correcting whichever method was incorrectly used if the
answers differ.
1)
2)
3)
4)
5)
6)
7)
8)
2(x + y)
(x + 2)(x + 3)
(x + 2)(x + y)
(2x + 3)(x + 1)
(x + 2)(x – 2)
2(x + 3)(x + 2)
2x(x – 3)
(2x + 1)(2x – 1)
Students should have all of these problems correct if they checked the problems. It
would, however, be useful for students to discuss which processes they used and why.
The goal of this assignment is for students to reflect on the processes they are using and
to reinforce their algebraic ideas with geometric ones and vice versa. This will help
visual students with symbolic manipulation and help less visual students understand
geometric models.
101
Extensions
CLASSWORK
Provide students with materials such as legos, clay, and construction paper that they can
use to build three-dimensional models.
Ask students to determine the volume of a cube with sides of length a.
Encourage them to draw or build a model. Have students start on their own, and then
discuss their ideas in small groups.
Discuss the models students used as well as how they found the volume as a class. Make
sure students explain why a formula works if they cite a formula as their process.
The goal of this problem is for students to gain familiarity with a three-dimensional
model and to review volume. These skills will be essential as students explore the
following problems. Students often have difficulty visualizing things in three dimensions
from two dimensional pictures, so it is important to give them material with which they
can build model three dimensional models.
102
Extensions
Once students are comfortable with their models and answers, ask students to determine
the volume of a rectangular prism with length a, width b, and depth a + b. Have students
compare the various expressions and models students made first in small groups, and then
with the whole class. Once again, make sure to discuss the equivalence of the various
expressions, as well as how this relates to the ideas constructed in the earlier activities.
The goal of this problem is for students to become more familiar with the threedimensional model. It is also for them to begin making connections between the
algebraic ideas about polynomial multiplication they constructed earlier and the
geometric ideas they are exploring.
103
Extensions
Next, have students determine the volume of a cube with sides of length a + b. Have
students compare the various expressions and models used in small groups before
discussing with the whole class. Also have students discuss the connections between
these answers and the algebraic ideas they constructed in earlier activities.
The goal of this question is for students to more deeply explore the geometric
representations of the algebraic ideas they are constructing. This could help students
reinforce both concepts by providing visual students a visual representation of the ideas
and helping abstract students learn to visualize.
104
Extensions
Next have students find the expansion of (a + 1)3. Have students work on this problem
on their own before working in small groups. Discuss the answers and methods used in a
class discussion once all small groups have found at least one method for solving the
problem. Be sure to discuss the relationship between the various answers and methods
used.
The goal of this problem is for students to gain confidence in working with the models
and to reinforce the connections students are seeing between the different
representations.
105
Extensions
Next have students find the expansion of (a + 1)4. Have students work on this problem
on their own before working in small groups.
Discuss the answers and methods used in a class discussion. Discuss which methods
worked in this situation and which ones did not. Also discuss situations where each of the
methods would work and would not work. At this point spend some time discussing how
this relates to the historical evolution of polynomial multiplication. In doing so, explain
to students that since these ideas were originally explored geometrically, algebraists did
not have any reason to consider exponents that could not be expressed geometrically
because they did not realize such powers existed. Therefore, many of the ideas did not
exist until algebraic expressions were reified.
The goal of this problem is for students to see the limitation of geometric representations
for these problems (that only whole number powers of three or less can be used), and to
help them see the strength of symbolic algebra and its accompanying analytical methods
(that all exponents can be used). Another goal is to help students deeply understand the
impact of symbols on the historical evolution of algebra, and mathematics in general, as
well as the evolution of polynomials (they allowed mathematicians to study concepts that
could not be represented in our physical reality).
106
Extensions
HOMEWORK
Find the following products using any method you chose. Check your answers using a
different method and reconcile any differences that arise.
1) 2(x + y + 3)
2) 2(x + 1)(x + 2)
3) x(x + 1)(x + 2)
4) 2x(x + 1)(x + 2)
5) x(x + 1)(x – 1)
6) x2(2x + 1)
7) x2(2x + 1)(x + 2)
8) (x + 1)(x + 2)(x + 3)
9) (x + 1)(x + 2)(x + 3)(x + 4)
10) (x + 1)(x + 2)(x + 3)(x + 4)(x + 5)
Since students may not be able to check all of these products using alternate methods,
allow students an opportunity to compare their answers with their peers. Also have
students discuss which method they like better for each type of problem and why.
The goal of this activity is for students to practice working on problems and to have an
opportunity to choose which methods they prefer (geometric or analytic) in different
situations. It is also for students to determine methods for checking their answers.
Another goal of this problem is for students to strengthen their appreciation of the power
and elegance of symbol use.
107
Extensions
CLASSWORK
Next ask students to find the area of the following shapes. Encourage students to write as
many different expressions as they can think of to express the given relationships.
x
a
x
3
a
b
c
c
d
b
x
x
y
x
z
First have students work on their own, and then share their answers in small groups.
Once all suggested answers have been discussed in small groups, share the answers as a
class.
The goal of this task is for students to further their ideas of working with quadratic
expressions and geometric figures. This will help students with geometric processes as
well as the algebraic ones.
108
Extensions
Next have students simplify the following expressions.
1) (x + 3)(x + 2) + (x + 1)(x – 1)
2) (x + y)(x + 1) – (x – y)(x + 1)
3) (x + y)2 + x2
4) (x + y)2 + (x – y)2
5) 3(x + 1)(x + 3) + (x + 1)(x – 1) – 2(x + 1)
Encourage students to find as many different ways of finding the answer as possible, and
to use these to check their answers. Once students have worked on the problems on their
own, have students share their answers and approaches in small groups, and then as a
class.
The goal of this task is for students to deepen their understanding of algebraic
expressions by using geometric models. In turn, it is for students to deepen their
understanding of area by using the models to help develop algebraic ideas.
109
Extensions
HOMEWORK
Simplify the following expressions using any method you would like. Then use a
different method to check your answers. If your answers differ, correct your errors that
led to incorrect answers.
1)
2)
3)
4)
5)
(x + 1)(x + 2)(x + 3) + (x + 1)(x)(2)
(x + 1)3 + x(x + 1)(x – 1)
(x + y)2x + x2(x + y)
(x + y)3 – x3
(x + y)3 + x2y + xy2
The goal of these problems is for students to practice using all of the processes they have
learned and to select which methods (geometric or analytic) they like best in given
situations.
110
Extensions
Extension 3
In this activity students will use their understanding of the base ten system to work with
numbers in other bases. This will help students understand systems used in our society,
such as binary code, as well as the number systems used in other cultures. This will help
students gain an understanding of how different number systems work. It will also help
students become aware of differences and similarities between different cultures. This
activity involves a few historical connections as well. Understanding other number
systems with different bases will help students better understand operations on the base
ten system as well as on polynomials.
111
Extensions
CLASSWORK
Begin by discussing the idea of binary systems with students. For instance, ask them if
they have heard of binary before and, if so, what it is. If students do not know, ask them
to think about the prefix bi. Remind students that we have a decimal system, and ask
them to think about the prefix dec. Use the prefixes to have students make connections
between the use of a base ten and the name decimal system. Have them use that idea to
predict the base of a binary system. In this discussion, be sure to discuss the fact that the
metric system is another example of a base ten system. Also make sure that students are
aware that the binary system is the framework upon which digital computing is based.
The goal of this activity is for students to develop ideas about the existence of other
number systems. It is also for them to begin making connections between the decimal
system and the binary system. Furthermore, students will begin developing informal
ideas about how the binary system works. This is also an opportunity for students to see
connections between the math they are learning and life outside of the classroom. This
activity will do this because computers rely heavily on the binary system.
112
Extensions
Ask students to think privately about how we write 2, 5, 12, and 128 in the base ten
system. Ask them to express these numbers in terms of powers of ten. Once students
have taken adequate private think time, ask them to compare their answers with their
groups.
Once all groups have reached an agreement, discuss the answers as a class. Make sure
students discuss writing the numbers using powers of ten written with exponent notation,
not just with expanded powers of 10. During this discussion ask students why we used
powers of ten, and ask them what numbers we would use powers of in the binary system.
The goal of this activity is for students to start thinking about what it means to work with
a base ten system and to begin formalizing their ideas about the binary system.
113
Extensions
Ask students to think privately about how to express 2, 5, 12, and 128 as powers of two.
Ask them to predict how we would write these numbers in a base two system. Once
students have taken adequate private think time, ask them to compare their answers with
their114
Extensions
Now tell the students that the Babylonians used a sexigesimal system. Ask students to
think privately about how to express 2, 5, and 128 as powers of sixty. Ask them to
predict how we would write these numbers in a base sixty system. Once students have
taken adequate private think time, ask them to compare their answers with their115
Extensions
All of the numbers above required one or two places to express the numbers, ask students
to try to find a number that would take exactly three places to express and one that would
take more than three. You may wish to give examples of what this means using the
decimal system.
After students have had adequate private think time, have them compare their answers as
a group. Also have them compare how they found their answers.
Once individual groups have had enough time for discussion, have the small groups
discuss their answers and their solutions for finding them. During this discussion, ask
students how these number systems relate to polynomials.
The goal of this activity is for students to think critically about how numbers are
expressed. It is also for students to gain a deeper understanding of number systems with
different bases. It is hoped that this understanding will help students better understand
operations involving polynomials.
116
Extensions
HOMEWORK
Explore different cultures from present day or the past. Find a culture other than the
Babylonians that used a system other than the three we have discussed. Give a brief
description of the culture, then describe the number system they use(d). Explain how to
work with the number system to someone who has not studied number systems using
different bases. Also, express at least five different numbers using the number system
you are studying (they should be non-trivial examples). Finally, try to find other cultures,
past or present, or subjects that use the same number system.
117
Extensions
Extension 4
In this activity students will use their skills in anticipating the terms in the products of
polynomials to generate the binomial theorem. This activity is designed for students who
know the formula for combinations, but can be modified for students who do not.
Students will begin by looking at several concrete examples, and will use the patterns
they recognize to generate the general formula. The binomial theorem is a useful
theorem in mathematics, one that many students merely memorize. The ability to
anticipate terms in the product of polynomials, which students have developed over the
course of this curriculum, will help students derive this formula. This activity is designed
to incorporate computer algebra systems (CAS) into the classroom, if such systems are
available. CAS is not, however, required for this activity.
118
Extensions
CLASSWORK
Ask students to complete the following chart on their own.
Complete the following chart by working across each row. Be sure to complete the
ENTIRE row for a given expression before moving on to the next expression.
Predict the coefficient of the given variable Check your predictions by expanding
for the given expression.
the expression. If your answer is
different, explain why that is.
1
x
x2
x3
x4
x5
x6
(x + 1)2
(x + 1)3
(x + 1)4
(x + 1)5
(x + 1)6
It is important that students complete the chart one row at a time so they reflect on the
solutions before moving on to the next expression. This approach follows an idea
elaborated in Kieran & Saldanha (forthcoming). Encourage students to generate a
general formula for the expansion of (x + 1)n before they discuss their answers with their
group members. Once students have completed the chart and made their predictions,
have students compare their answers with their group. Ask the small groups to discuss
how they predicted their answers in addition to comparing their answers. Once the small
groups have shared their strategies, discuss the different strategies and reasonings used in
a whole class discussion. Be sure that students explicitly state the patterns they have
observed. If students know the formula for counting combinations be sure they use it. If
they do not know the formula, teach it to them, or have them create an expression that
expresses the same relationship.
Note that if you have access to CAS, it would be a good idea to have students use that to
check their products, otherwise checking the products could take a long time! If you do
not have access to CAS, you might want to skip the final column and have students
carefully explain their reasoning, only checking the expansions for small values of n and
when there is a disagreement.
119
Extensions
The goal of this activity is for students to revisit their ideas about when and why certain
terms arise in products. It is also for students to begin constructing the binomial
theorem. If CAS is used, another goal is for students to learn to use technology to help
them check their answers and make generalizations.
120
Extensions
Ask students to use the pattern they observed above to predict the expansion of the
following expressions
Complete the following chart by working across each row. Be sure to complete the
ENTIRE row for a given expression before moving on to the next expression.
Predict the expansion of the given
Check your predictions and explain any
expression.
discrepancies in your predictions.
(a + b)2
(a + b)3
(a + b)4
(a + b)5
(a + b)6
Once students have completed the chart, have them compare their answers with their
group. Ask the small groups to discuss how they predicted their answers in addition to
comparing their answers. Once the small groups have shared their strategies, discuss the
different strategies and reasonings in a whole class discussion. Once again, make sure
that students explicitly state the pattern they have observed, and encourage students to
generate a general formula for the expansion of (a + b)n.
Once again it would be a good idea to have students use CAS to check their products, if
they have access to it, otherwise you may want to modify when and how students check
their answers.
Once again the goals of this activity are for students to revisit their ideas about which
terms arise in products and why, to formally construct the binomial theorem, and to learn
to use technology to check answers and make generalizations.
121
Extensions
HOMEWORK
Using the logic we used in deriving the formula for the binomial theorem, can you predict
a formula for the following expressions?
1) (a + b + 1)n
2) (a + b + c)n
3) (x2 + x + 1)n
These problems are challenge problems; they may be quite difficult for many students!
122
Solutions
CHAPTER 8
SOLUTIONS AND RESPONSES TO THE ACTIVITIES
123
In the following section there are possible solutions to all of the activities in this
curriculum. Many of the questions have several possible equivalent solutions, so the
absence of an answer does not necessarily mean the answer is incorrect. Furthermore,
since many of the activities are discussion based, the "solutions" contain many of the
important issues that hopefully will arise in class conversations. That does not mean that
all possible conversations are included. Teachers need to decide which strands of
conversation to follow and which ones to leave behind. Finally, although these solutions
are intended to work as an answer key, they are also intended to help people using this
curriculum understand the goal of a given task if the goal is not clear to them from the
prompt. The solutions are organized in order they are presented, and each individual
prompt is specified by the page number on which it appears.
124
Activity 1
Pg 40:
I began by thinking of the number 428 as 400 + 20 + 8 and 173 as 100 + 70 + 3.
Then I started to actually add the numbers by looking at the ones. 8 + 3 is 11,
which is 10 + 1. Therefore, I knew there was a 1 in the ones place, and I added
the ten I just mentioned to the 2 tens and 7 tens in the problems. That gave me 10
tens, or 1 hundred. Then I added that hundred to the other hundreds in the
problem (4 and 1): 4 + 1 + 1 = 6, so there are 6 hundreds. So in the end, we have
6 hundreds, no other tens, and 1 one, making the answer 601.
Pg 41:
After breaking 1037 into 1000 + 30 + 7 and 638 into 600 + 30 + 8, I started with
the ones. However, because 8 is bigger than 7, I had to look to the tens place as
well so I had more than 8. Therefore, I instead broke 1037 into 1000 + 20 + 17.
After taking 8 ones away from 17 ones, I was left with 9 ones. Then I attempted
to do the same with the tens, however, I once again did not have enough tens to
complete the process, so I had to redivide the 1020 I had left into 900 + 100 + 20
or 900 + 120. The 120 meant I had 12 tens, so I could now easily take 3 tens
away, leaving me with 9 tens. Finally, I took the 6 hundreds in 638 away from 9
hundreds I still had left from 1037. That left me with 3 hundreds. This means
that I ended up with 3 hundreds, 9 tens, and 9 ones, or 399.
Pg 42:
I started with the hours. When I had been waiting for 4 hours, I had 2 hours and
12 minutes to wait still. Then, I had to deal with the 37 minutes. After 12 more
minutes of waiting I would only have two more hours to wait. Since 37 is 25 +
12, I still had 25 minutes to wait. To figure out how much I had to wait still, I
broke one of the 2 hours I still had to wait into 60 minutes. 60 – 25 is 35, so after
waiting for another 25 minutes, I would have 1 hour and 35 minutes left to wait
still.
Pg 43:
I started this problem in a similar manner to the last one. 4 months plus 3 months
makes 7 months. Then I got to the weeks, and, since 2 + 1 is 3, I got 3 weeks. 3
days plus 6 days is 9 days, and 1 week is 7 days, so I broke the 9 days into 1 week
and 2 days. I combined this with the 7 months and 3 weeks from before. This
gave me 7 months, 4 weeks, and 2 days. The 4 weeks got me thinking, I often
think of 4 weeks as a month. If we are talking about work days that is fine, but if
we are trying to think about how much medicine to take on a trip, we need to
think of months as 30 or 31 days, depending on the month. Because I did not
know what to do next, I waited until it was time to discuss our answers. The
people in my small group all assumed 4 weeks was a month, but when I brought
up the issue I was having, they realized it was an issue too. So we called the
teacher over, and she smiled at us. We then had a class discussion, and I
discovered that other people were having similar issues to the ones I was having,
and we discussed how we needed to know more information in order to solve the
problem. In the conversation, the teacher told us that was the point of the
problem. She explained that in the earlier problems we knew the exchange rate,
so we could simplify the problems into simple and neat solutions, but here, we
125
could not combine months, days, and weeks without knowing more information.
She likened this to the proverbial combining apples and oranges.
Pg 44:
I solved this problem by combining the a's and the b's. I had 1 a, 3 a's, and 4 a's.
Since 1 + 3 + 4 is 8, I knew I had 8 a's. Next I combined the b's, giving me 7 b's.
Therfore, my answer was 8a + 7b. Although it was hard to leave my answer
looking like that, I thought about the earlier problem and how we could not
combine some of our objects because we did not know how they related to each
other, and since I did not know how a and b related to each other, I left my answer
broken up into a's and b's.
Pg 45:
Using the reasoning described above, I got
1) 10a + 3b + 8
2) 3a + 4c + 4
3) 4a + 2b +5c.
Pg 46:
1) 8a + b
2) 3a + b
3) 8a – 4b
4) a – b
5) 4a + 4
Pg 47:
1) 5m + 5n + 2mn
2) 7x + 3x2 + 8
3) 2x + 5y + 5x2 + 6y2 + 1
Pg 48:
1) First think of 1,074 as 1000 + 70 + 4 and 24,153 as 20,000 + 4,000 + 100 + 50
+ 3. Then Since 4 is bigger than 3, regroup 24,153 into 20,000 + 4,000 + 100 +
40 + 13. Since 13 is bigger than 4, you can easily take 4 from 13, giving you 9
ones. Next look at the tens, since 40 is smaller than 70, you run into the same
problem as before, so regroup again, giving you 20,000 + 4,000 + 140. 70 from
140 is 70, so for now we are left with 70 + 9 or 79. Now we move to the
hundreds, but since neither number has any hundreds, we move to the thousands.
We have to take 1 thousand from 4 thousands, giving us 3 thousands. Since the
smaller number ends at thousands, we are done. We now have 20, 000 + 3,000 +
70 + 9 or 23,079.
2) a) First, since what you bought costs more than a dollar, obviously you will not
get the bill back. The question becomes how 33 cents relates to 2 quarters. 2
quarters is 50 cents, and 50 cents take away 33 cents is 17 cents. The biggest coin
smaller than 17 cents is a dime, so you will get a dime back, leaving you with 7
cents. The biggest coin less than 7 cents is a nickel, so you also get a nickel,
leaving you with 2 cents. The only way to get 2 cents is 2 pennies. Therefore,
you get a dime, a nickel, and 2 pennies.
b) First of all, it is subtraction, and we talked about subtraction today. Secondly,
we had to trade the quarter into cents so we could relate the 33 cents to the 2
quarters. Thirdly, we had to rebreak the 17 cents we were left with into coins in
126
order to make the change. This is like what we were doing with ones, tens, and
hundreds as well as with hours, months, days, and weeks as well as with the
letters.
3) a) 17q + 13r
b) 6a + 4b –ab + 4
c) x + 6x2 + 2x3
d) 4a + 5b + 1
e) 11m
EC) We have to add like pieces, just like with whole numbers. With decimals we
have to add tenths to tenths, hundredths to hundredths etc. Also, When we find
ourselves with more than 10 hundredths (for example), we have to break them
into tenths and hundredths. Likewise with other place values. Also, we have to
regroup when we are subtracting just like with whole numbers.
127
Activity 2
Pg 50:
When I multiply, I start by thinking of 123 as 100 + 20 + 3, so 3 x 123 is 3 x 100
+ 3x20 + 3x3, or 300 + 60 + 9 =369.
Pg 51:
Similarly, I get 4 x 2 hours and 4 x 12 minutes, which is 8 hours and 48 minutes.
Pg 52:
Given how I did the last problems I guess I would do this by 3xa + 3x2b + 3x3c
which is 3a + 6b + 9c.
Pg 53:
1) 6a + 12b + 15c
2) 4p – 6q + 8r
3) 12m + 16n + 4
4) -2x -4y + 6z
5) -2a + b -2c
6) 4p + 5q -2r
Pg 54:
1) 2a2 + 4ab + 5ac
2) 4p2 – 6pq + 8pr
3) 8m3 + 4m2n + 4m4 + 12m2n2 + 4m2
4) -2x2y – 4xy2 + 6xyz
5) -2a2b2 + a3b2 -2a2b3
6) 2p2 +4pq-2pr + 2p + q
Pg 55:
1) a) 6x2y
b) 8x3y
c) 4x2y2
d) 36x3y3
2) a) 2x + 5y + 3z
b) 4a2 + 6ab + 4b2
c) 3x2 +-3xy + 3xz -2x -4y -6z
d) m3 - m2n - n2m + n3
Pg 56:
1) 7x2 + 11x + 2
2) x3 + 10x2 + x + xy + 2
3) 5 + 3y
4) 3y - 3 + 3
5) 25y + y2 + 5
128
Activity 3
Pg 58:
I think of 12 as 10 + 2, so the problem becomes 10x43 + 2x43. The question then
becomes how to do those problems. Since 43 is 40 + 3, I can solve those by
thinking of it as (10x40 + 10x3) + (2x40 + 2x3). That gives me (400 +30) + (80
+6) which is 400 + (30 +80) + 6 or 400 + 110 + 6 which is 400 + 100 + 10 + 6
which simplifies to 516.
Pg 59:
14(2 hours + 14 minutes) is 10(2 hours + 14 minutes) + 4(2 hours + 14 minutes)
= (20 hours + 140 minutes) + (8 hours + 56 minutes)
= (20 hours + 2 hours + 20 minutes) + (8 hours + 56 minutes)
= 30 hours + 76 minutes
= 30 hours + 1 hour + 16 minutes
= 31 hours and 16 minutes
Pg 60:
Following the process from the previous problems, I guess I would break (a + b)
into a + b, giving me a(3a + 2b) + b(3a + 2b). That equals 3a2 +2ab +3ab + 2b2.
That simplifies to 3a2 + 5ab + 2b2.
Pg 61:
1) x2 + 2xy + y2 + xz + yz
2) 2a2 -5ab + 3b2
3) p2 + 2pq + 2pr + 2ps + pt + q2 + 2qr + 2qs + qt + r2 + 2rs + rt + s2 + st
4) 8m2 + 22mn + 15n2 + 12m + 18n
5) 2x4 + 5x3 + 6x2 + 7x + 4
Pg 62:
1) x3 + 12x2 + 47 x + 60
2) 6x3 + 8x2 + 15x + 20
3) 6x4 +8x3 + 15x2 + 21x + 5
4) 3a4 + 4a3 + 8a2 + 19a + 15
5) 2a2 + 5ab + 3b2 + 2a2b + 4ab2 + b3
6) 8pr + 12 qr + 10ps + 15qs
Pg 63:
1) a) a2 + 2ab + b2
b) p2 - q2
c) 4m2 – 12mn + 9n2
d) 4x2 - 1
e) 9a2 -12ab + 4b2
f) 4c2 + 8cd + 4d2
g) 169
2) When the two sets of parentheses have the same terms, but one is a sum and the
other is a difference, you get the difference of each of the terms squared. When
you square a binomial your answers has the square of each of the terms and twice
the product of the two terms.
3) a) 2x2 – 4x -30
b) 3x3 + 12x2 + 12x + 48
c) 9 -18x + x2
129
d) 2x5 + 4x4 + 6x3 + 4x2 + 2x
130
Activity 4
Pg 65:
1) none
2) b & c
3) b & c
4) b & c
5) b, c, d, e
Pg 66:
1) a, b, c
2) a, c, e
3) a, b, c, d, e
4) a, b, c, d
5) b, c, d
Pg 67:
1) a, b, c, d
2) a, b, c, d, e
3) a, b, c, d, e
4) a, b, c, d, e
5) c, d, e
Pg 68:
1) b & c
2) d
3) b & e
4) a & c
5) a & d
Pg 69:
1) e
2) a & d
3) none
4) b & d
5) c
Pg 70:
1) 3
2) 4x
3) 3, 9
4) 3, 6
5) 1, 4 or 4, 1
Pg 71:
1) 1, 2
2) 2x, 5
3) 1, 1
4) 3, 2x2, 6
5) 2, 6
Pg 72:
1) a) You will get an a3 because there are 3 a's, 3 + 2 + 1 a2's because each pair
of a's will then be multiplied by the number in the other set of parentheses, you
131
will get 3x2 + 2x1 + 3x1 a's because each time you use just one a you will have to
use the constant in each of the other factors, and your constant will be 3x2x1
b) You will have abcd, abc, abd, acd, bcd, ab, ac, ad, bc, bd, cd, and 1 because
you will either use none of the 1'a, 1 on the 1's, 2 of the ones, 3 of the ones, or all
of the ones
c) If you use all of the ones, you will get a 1, , if you use none of the ones you will
get x6, if you use one of the ones you will get x5, x4, or x3 and if you use two ones
you will get x3, x2, or x.
d) Using the same reasoning you will get: -1, 4x5, 4x4, 3x2, -3x2, -2x2, -2x, and 2x2
2) a)1
b) 3, -15
c) -1, 3
d) 3x, 2
e) 3x (or 2x + x or 4x – x etc)
f) -4
g) 3, -2 or 1, 2x2
3) a) (4 – 7x)(4 + 7x)
b) (2x – 3) 2
c) (x + 3)(x + 2)
d) (x - 3)(x - 2)
e) (x + 6)(x – 1)
f) (x – 6)(x + 1)
Pg 74:
1) 2a5 + 2a4 - 2a2 – a + 1
2) a2 – 2ab + 3b2
3) m3 + 8
4) m3 - 27
5) m6 - 8
6) m9 + 27
7) You end up with two terms in all of the expressions.
132
Activity 5
Pg 76:
1) (x – 1)(x + 1)(x2 + x + 1)(x2 – x + 1)
2) (x8 + 8)(x – 1)(x7 + x6 + x5 + x4 + x3 + x2 + 1)
3) (2x3 + 1)(2x2 -1)
4) (3x3 + 1)(3x3 – 1)
5) Fully Factored
6) 8(x2 -2)(2x4 + 2x2 + 4)
7) (3x3 + 8)(3x3 – 8)
8) 2(5x2 – 32)
Pg 77:
1) (x + 3)(x + 2)
2) (x – 3)(x – 2)
3) (x + 6)(x – 1)
4) (x – 6)(x + 1)
5) (x + 4) 2
6) (x – 4) 2
7) Not possible
8) Not possible
9) Not possible
10) Not possible
11) (x + 6)(x – 2)
12) (x – 6)(x + 2)
Pg 78:
1) (3x -4 )(x + 1)
2) (2x + 1)(x + 2)
3) (5x – 11)(7x + 4)
4) (3x – 1)(x – 1)
5) (3x + 2)(2x + 3)
6) (2x – 1)(x + 1)
7) (2x + 3)(x – 4)
8) (3x2 + 4)(3x2 + 2)
9) (6x + 5y)(3x – 2y)
10) (4x – 5y)(3x – 4y)
Pg 79:
1) 5(2x + 5) 2
2) x(x – 3)(x + 2)
3) 4(3x2 + 7x – 6)
4) 3x(3x + 1)(2x – 3)
5) 5(x2 + 2x + 6)
6) x(x + 8)
7) x(6 – 7x)(6 + 7x)
8) x(81x – 64)
9) x(x2 – 5)(x + 1)(x – 1)
10) Not possible
133
Pg 80:
1) (x2 + 2)(x + 3)
2) (3x + 2)(x2 + 1)
3) (6x – 7)(3x2 + 5)
4) (a –b)(x + y)
5) (2x – 1)(3x – y)
6) (x7 + 2)(x + 1)(x – 1)(x2 + x + 1)(x2 – x + 1)
7) (2x + 3)(2x2 + 3)
8) (x + 1) 2(x – 1)2(x2 + 1)
9) (7x + 9)(2x2 – 3)
10) Not possible
Pg 81:
1) a) 12x2 + 6
b) 3x2 – 4x2 + x2 + x + 4
c) 7x2 + 5x - 23x + 21
d) -3xy4 – y3 + 5y - 2
2) a) 8y2 – 12y
b) 12x3 + 24x2
c) 6x3 + 8x2 – 6x
d) -10xy5 – 10x2y2 + 120x2
e) 5x2 + 4x - 12
f) 5x6 + 5x5
g) 4x2 – 9y2
h) x3 + 6x2 +12x + 8
i) 3x4 + 5x3 + 6x2 + 3x + 1
3) a) (x + 7)2
b) x(x – 9)2
c) 2(x + 8)(x – 8)
d) x2y(3x – 2y)(1 + x)
e) –(x2 + 2)(x + 3)(x – 3)
f) (x4 + 1)(x2 + 1)(x + 1)(x – 1)
g) (x2 + 1)(x + 4)
h) Not possible
134
Extension 1
Pg 84:
12 is 4 groups pf 3 and 30 is 10 group of 3, so 12/30 is the same as 4/10.
However, 4 is 2 groups of 2 and 10 is 5 groups of 2, so 4/10 is the same as 2/5.
Pg 85:
The numerator has (x + 4) groups of (x + 1) and the denominator has (x + 2)
groups of (x + 1), so it reduces to (x + 4)/(x + 2). Similarly, we have to factor the
numerator and denominator of the second fraction so we can see what common
size pieces they can each be grouped in to. The numerator is (x + 3)(x + 2) and
the denominator is (x + 3)(x – 3), so the numerator is (x + 2) groups of (x + 3) and
the denominator is (x – 3) groups of (x + 3), so it is equivalent to (x + 2)/(x – 3)
Pg 86:
1) 1/6
2) (x + 2)/(x + 5)
3) [(x – 3)(x – 2)]/[(x + 3)(x + 2)]
4) [x(x – 6)(x + 1)]/[(x2 + 4)(x + 2)2]
5) (x – 5)/(x – 4)
6) 1
7) already reduced
8) [(x2 – 4x + 9)(x – 1)]/[(x2 – x + 3)(x2 + 6)]
Pg 87:
1) 7/16
2) [(x – 3)(x + 3)]/[(x + 2)(x - 2)]
3) [(x – 2)(x + 2)]/[(x - 3)2]
4) already reduced
5) [(x – 6)(x + 6)(x – 1)]/[(x + 1)(x – 3)(x – 2)]
Pg 88:
1) 2x - 2
2) 2/(x – 1)
3) (x – 1)/2
4) [2(x + 4)(x – 1)]/(x2 + 2)
5) [(x + 4)(x – 1)]/(2x2 + 4)
6) (x2 + 2)/(x + 4)
7) [(x + 3)(x + 2)]/(x2 + 2)
8) 1/(x – 1)
9) (x + 4)/(x2 + 4x + 1)
10) [(x – 1)(x + 4)(x – 1)(x + 3)(x + 2)]/[(x2 + 2)(x2 + 4x + 1)
Pg 89:
To add ½ and ¾ I first need to find a common denominator so I am talking about
the same size pieces (just like when we were talking about adding whole numbers
and collecting like terms, we needed to be talking about the same unit of
measurement to combine terms). 1/2 is the same as 2/4, which can be found by
breaking each of the halves into 2 pieces. 2/4 plus 3/4 is 5/4, so the answers is
5/4.
135
Pg 90:
1) 35/36
2) [2(x + 6)(x + 1)]/(x – 6)
3) (x2 – x -31)/(x2 + 3x -10)
4) (2x4 – 9x3 – 6x2 – 25x -18)/(x6 + 4x4 - x2 – 4)
5) (3x3 – 4x2 – 14x + 17)/(x3 – 4x2 +x + 6)
Pg 91:
1) (-x2 – 7x -15)/(x3 + x2 – 16x – 16)
2) (15 + 3x)/(36 – 4x2)
3) 4x/(x2 – 1)
4) (5x2 + 12x + 27)/(54 – 6x2)
5) (x – 3)/(x2 + x – 6)
Pg 92:
1) (2x4 + 4x3 + 2x2 + 6x)/(4x2 – 1)
2) (6x2 + 3x – 15)/(2x4 – 5x3 – 5x2 + 5x + 3)
3) (2x2 – 8x – 1)/(2x4 – 14x3 – 2x2 + 6x)
4) (x + 4)/(3x + 1)
5) (2x2 + 2x – 2)(2x – 1)
6) (2x2 – x - 4)/(4x3 + 4x2 – 11x – 6)
7) (2x2 + x + 2)/(2x4 + 8x3 + 7x2 + 4x + 3)
8) [6(x2 + 1)(x + 3)(x – 3)]/[(2x + 1)(2x – 1)]
Pg 93:
1) x/(3xy – y)
2) x2/y
3) (n + m)/(n – m)
4) (x + 1)/(x2 – x)
5) x/(x – y)
6) (3x2y + 2xy2)/(12y – 18x)
Pg 94:
1)(2x – 2)/(x + 1)
2) -120x/(x4 + 10x3 + 25x2 – 36)
3) y(x + y)/x
4) (x4 -25x2 + + 60x – 36)/(x4 – 25x2 – 60x – 36)
5) (x – 5)2/(x + 3)
6) 1/(x2 – 25)
7) (5x3 + 4x2 + 3x – 5)/(5x2 + 6x + 1)
8) (x + 3)/(5x + 5)
136
Extension 2
Pg 96:
The area of a square with side length a is a2.
Pg 97:
The area is ab
Pg 98:
I broke the rectangle into 2 parts, one with area a2 and one with area ab, since the
area of the entire rectangle is covered by both of those areas, the entire area is a2 +
ab.
Pg 99:
I could do it by looking at two rectangle, that would give me a(a + b) + b(a + b) or
(a + b)a + (a + b)b. Instead I could break it into 4 rectangles, giving me a2 + ab +
ab + b2.
Pg 100:
1) ab + a + b + 1
2) a2 - ab
3) a2 - b2
4) a2 + a + ab + b
5) a2 + ab + ac
6) a2 + 2ab + ac + b2 + bc
7) a2 + 2ab + ac + b2 + bc
8) a2 + 2ab + 2ac + b2 + 2bc + c2
Pg 101:
1) 2x + 2y
2) x2 + 5x + 6
3)x2 + 2x + xy + 2y
4) 2x2 + 5x + 3
5) x2 - 4
6) 2x2 + 10x + 12
7) 2x2 – 6x
8) 4x2 – 1
Pg 103:
a2b + ab2
Pg 104:
a3 + 3a2b + 3ab2
Pg 105:
a3 + 3a2 + 3a
Pg 106:
I tried to build a model, but I couldn't.
a4 + 4a3 + 6a2 + 4a + 1
Pg 107:
1) 2x + 2y + 6
2) 2x2 + 6x + 4
3) x3 + 3x2 + 2
4) 2x3 + 6x2 + 4
5) x3 - x
137
6) 2x3 + x2
7) 2x4 + 5x3 + x2
8) x3 + 6x2 + 11x + 6
9) x4 + 10x3 + 35x2 + 50x + 24
10) x5 + 15x4 + 85x3 + 225x2 + 274x 120
Pg 108:
x2 + 3x + ax, 2ab + 2cd + ad, 2xz + 2xy
There are many different ways to write all of these expressions, these are just the
most simplified versions of them
Pg 109:
1) 2x2 + 5x + 5
2) 2xy + y
3) 2x2 + 2xy + y2
4) 2x2 + 2y2
5) 4x2 + 10x + 7
Pg 110:
1) x3 + 8x2 + 13x + 6
2) 2x3 + 3x2 + 2x + 1
3) 2x3 + 3x2y + y2x
4) 3x2y + 3xy2 + y
5) 2x3 +4x2y + 4xy2 + y
138
Extension 3
Pg 113:
2x 100, 5x100, 1x10 + 2x100, 1x102 + 2x10 + 8
Pg 114:
1x2, 1x22 + 1x20, 1x23 + 1x22 + 1x20, 1x27
10, 101, 1101, 10000000
Pg 115:
2x600, 5x600, 2x60 + 8
2, 5, 28
Pg 116:
For binary, the base ten numbers 4-7 work. For base sixty, the base ten numbers
216,000-12,959,999 work.
139
Extension 4
Pg 119:
Predict the coefficient of the given variable
for the given expression.
1
x
x2
x3
x4
x5
x6
(x + 1)2
1
2
1
0
0
0
0
(x + 1)3
1
3
3
1
0
0
0
(x + 1)4
1
4
6
4
1
0
0
(x + 1)5
1
5
10
10
5
1
0
(x + 1)6
1
6
15
20
15
6
1
Pg 121:
Predict the expansion of the given expression.
(a + b)2
a2 + 2ab + b2
(a + b)3
a3 + 3a2b + 3ab2 + b3
(a + b)4
a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4+ b5
(a + b)6
a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
140
CHAPTER 9
MY EXPERIENCE WITH THE CURRICULUM
141
Analysis
I taught this curriculum, or at least parts of it, in different forms in two different
situations. I had very different experiences in each situation, but I enjoyed both of them.
My first experience working with this curriculum was with my thirteen year old neighbor
who struggles with math. We worked on the entire curriculum with additional practice
on the topics implicit and explicit in this curriculum, but we did not do all of the
extensions. My next experience with this curriculum was with an algebra class at a large
university. In this situation we covered the same material I covered with my neighbor.
In the class, however, several students showed interest in various aspects that were
addressed in the extensions, so I invited them to work on those extensions with me in
small groups. I will now describe these experiences in more detail.
Experiences with an Individual Student
When my neighbor approached me about working with her son on his math over the
summer, this curriculum seemed like the perfect activity for us to do together. This was
because it would give him a chance to look critically at the math he struggled with,
helping him build a deeper understanding of that mathematics. He could then use that
understanding to help him build an understanding for algebraic ideas, which could help
him have more success when he formally studies algebra in school.
When we began the activity, he did not have trouble completing the first addition
problem. He did, however, have trouble explaining the process, which provided us with
the perfect opportunity to begin the discussion about the addition algorithm. We then
moved on to a multi digit subtraction problem to see how deeply he had internalized our
conversation. He did not solve the subtraction problem as quickly as he had the addition
problem and he struggled at explaining the process. When I reminded him of our earlier
conversation and encouraged him to think of the problem in terms of it he was eventually
able to talk through the algorithm, although it took a fair amount of coaching.
We next moved on to considering arithmetic involving time. He was able to answer the
question without using the algorithm, but also had difficulty explaining his ideas. He
tried completing the problems using the addition algorithm, to see if that helped. When
doing this, he got a different answer than he had mentally. This did not bother him.
When I pointed out this paradox, he had a hard time extending the ideas we had explored
with whole number arithmetic to time. After a lot of discussion, he eventually was able
to explain what it meant to carry in this context and correctly use the algorithm.
When we moved on to the problem where there was no clear unit to use for carrying, he
did not notice the dilemma. He immediately assumed a month had four weeks and was
ready to go on. After we discussed how many days there were in different months, he did
not immediately see a connection to the problem. After some discussion, he asked me if
the appropriate conversion for this problem was in fact four weeks. This nicely led us to
a conversation as to whether or not you could always carry. I felt that we were now
ready to move on to variables.
My neighbor did not immediately collect like terms correctly, but after some discussion
he was successful at the problems where the coefficients were positive. He had more
142
Analysis
difficulty, however, when the problem contained different powers of the same variables
and when there were signed numbers involved. So we spent a while practicing with
signed numbers.
Having become more comfortable with signed numbers, we moved on to the issue of
different powers of the same variable. I directed him back to the multi digit numbers, but
he had hard time seeing them as linear combinations of powers of ten. He could see the
numbers as their partial sums, but he experienced difficulties seeing the exponents
inherent in the problem. He eventually saw that the different powers of variables could
not be collected, but it was through substitution rather than from the intended analogy.
He did not come up with the idea of using substitution on his own, and it took a while for
him to figure out why and how substitution could help him.
He successfully completed much of the homework assignment, but eraser marks
suggested that he still had some difficulty with the problems involving the signed
numbers. He had difficulty with the extra credit, and it became clear it was because he
did not really know what the decimals places represented.
Having successfully completed the addition and subtraction problems, we moved on to
multiplication problems. My neighbor completed the problems that were building up to
the distributive properly very easily. In fact, he completed many of the problems on the
distributive property themselves quite easily as well. Although he forgot about the idea
of like terms at first, combining his answers in to a single term. He was a little confused
about what to do with the problems involving the rules of exponents, but with a little
prodding he quickly derived the rules and completed those problems as well.
My neighbor performed quite well on the assessment at the end of these two units. He
was able to extend this idea of like terms to radicals that were already simplified, but he
had a hard time simplifying the radicals. I decided that that aspect was outside of the
goal of the problem, so we did not spend much time on it. Furthermore, this was
intended to be a group quiz, meaning that it was probably particularly difficult for him to
complete on his own. Therefore, I skipped most of the problems involving radicals.
We next moved on to multi digit multiplication, which caused us some difficulty. I
believe this is because my neighbor had memorized, rather than understood the algorithm
he was taught in school. We spent some time talking about how we do mental arithmetic
in our head, which helped a bit. I decided the discussion of mental arithmetic got to the
heart of the issue as much as the algorithm did, so we moved on.
He did not readily extend the idea of multiplication to polynomials. He still had some
difficulties coordinating all of the ideas he had constructed during the process of the
curriculum. That said, in short order he had all of the processes coordinated and was
successfully completing the problems. He did not immediately notice the patterns I
hoped he would notice in the special products, but once he began to notice patterns he
noticed many of them.
143
Analysis
Furthermore, he was successfully able to anticipate the products of polynomials. He was
also quickly able to complete the fill in the blank problems. In addition he had relatively
little difficulties factoring polynomials with leading coefficients of one. He did make
more mistakes with problems involving signed numbers than those involving all positive
numbers. Nevertheless, he was beginning to make fewer mistakes with signed numbers.
He did not, however, successfully develop an efficient process for factoring polynomials
in general.
I tried to move on to the rational expressions extension, however, his difficulties with
factoring made it difficult for him to successfully complete many of the problems. We
did discuss fractions in the same way we had discussed whole numbers and decimals. He
did not immediately extend his ideas about decimals to fractions; nevertheless, with some
prodding he did connect the two ideas. He did not extend the ideas about fractions to
rational expressions as quickly as I would have thought; nonetheless, he eventually
successfully completed some of the problems.
I think my neighbor benefited from doing these activities because it gave him a different
perspective from which to view arithmetic. It also provided him with an opportunity to
practice the types of problems with which he needed help. Furthermore, it showed him
where the ideas were going and that he needed to master them because he would continue
to use them. That said, some of the weaknesses in his background made it difficult for
him to complete many of these activities. Also, many of these activities were designed to
be done with partners or in small groups, so it was difficult for him to complete these
activities on his own. At times I was not able to help him develop ideas without prodding
him a lot, meaning that he was not constructing these ideas as much as I would have
liked.
Experiences of a Group of Students
My algebra class at the university proved to be just as interesting, but in very different
ways. My students seemed to appreciate starting out their first class with a discussion of
arithmetic. I believe this is because many of them have math phobia, and it made the
class less threatening. Nevertheless, they did not know quite what to make of me or the
class at first. Also, they were looking for what answer I wanted them to give me. I think
it was helpful to start with this problem because the students were able to learn the
structure and style of the class before we moved on to more difficult math.
Many different answers were offered, and eventually a foreign student explained that it
was because we were using the decimal system (he had studied the idea of number
systems with different bases in high school). As he explained what he meant by that,
many students were interested in learning more. This immediately made me think of my
ethnomathematic extension I had already written, and the next day I handed out the
assignment as extra credit. However, no one turned it in.
As we moved on to the first problem involving time, the first student to offer a solution in
the class discussion converted the hours into minutes and used exactly the algorithm we
had used before. Although this was a perfectly valid solution, it defeated the purpose of
144
Analysis
the problem. Luckily, a few people eventually offered solutions that used the idea of like
terms.
When I presented the second problem involving time, a student raised her hand and asked
how many days she should assume are in a month. This immediately thwarted the
discussions that were supposed to happen in the dyads, however, it presented us with the
opportunity for a class discussion. The class quickly began using the phrase like terms,
which I assume is because this was not their first experience with the idea. Nevertheless,
I was impressed that they made the connection that we were working towards so quickly
on their own.
The students successfully completed the questions on completing like terms, something
that I cannot necessarily attribute to the activity since this was not the first time many
students had seen this idea. I decided to introduce the idea of radicals at this point, to
challenge the students a bit. At first they were afraid of the radicals, but after a little
discussion most groups were able to add radicals that were already simplified. Some
students had intuitions that some could be simplified, but no one figured out on his or her
own how to do so. With a class discussion and some coaching from me they eventually
came up with an idea, however, many students still had trouble with the idea.
On the homework, very few of the students attempted the extra credit problem asking
them to look at decimals in this light. Of those that did, very few went into detail, with
most just showing the problem or stating what they did. This showed that we still needed
a little more time on the ideas of explaining and exploring.
As we moved on to the distributive property, the students flew through it. A few students
were still grappling with the idea of explaining where the product came from rather than
just stating the answer, while other students easily described multiplication as partial
products. The students also quickly did the problems involving the distributive property,
including those involving rules of exponents. Once again, however, this could be based
on their previous experience with the concept.
I then extended the idea to include radicals. Some students immediately used intuition to
correctly simplify the problems, some were stumped, and yet others used their calculators
to assess if answers were correct. The idea of using calculators quickly caught on around
the class, which lead the students to want to give approximate answers. I reminded them
about the difference between rounded answers and approximate answers and placed the
ideas of radicals in a historical context. Some of my students seemed interested in
learning more about the history, so I invited all interested students to do the historical
extension. Nevertheless, none of the students took me up on the offer.
When we moved towards polynomial multiplication, most students described the
multiplication of multi digit whole numbers in terms of partial products. They did this
through a variety of ways including describing their mental processes and describing the
reasons behind the algorithm many of them performed. When students simply performed
145
Analysis
the algorithm, their classmates pushed for more, which told me they had been
enculturated into the activities.
As we moved to using variables, several students reverted to the process of FOIL they
had learned at other points in their careers, which worried me quite a bit. These students,
however, all got the problems involving trinomials wrong. This showed them that they
needed to have a deeper understanding of the process rather than merely memorizing an
algorithm. Although I was disappointed that my students had gotten problems wrong, I
was glad that it motivated them to understand rather than to memorize.
As we moved on toward factoring students were beginning to anticipate terms in the
products. When I first asked students to do so, they were confused by the directions.
Once they got the hang of it, however, they did quite well with the problems. They were
also able to successfully fill in the blanks in multiplication problems quite quickly.
Furthermore, they were able to successfully factor problems involving leading
coefficients of one. This was not, however surprising since students often factor those
with ease. It took them a while to extend their ideas to other polynomials. Nevertheless,
they eventually did.
Many students were using their calculators that can factor to help them find answers, and
then worked backwards from the answers their calculators gave them. Since this was
consistent with the extension I created on the binomial theorem, it seemed like a great
time to mention that activity. The students generally seemed excited to be invited to use
their calculators and to learn how to use their calculators to help figure out processes.
Ten of the students turned in the assignment, but none of them completed the entire thing.
By the time we began working on the rational expressions extension, students were quite
used to the format of the activity. Many of them had difficulty with the concepts in
fractions, however, as they worked together to remember how fractions worked they
described the concepts to each other in the same terms we had been describing the other
number systems. Furthermore, they quickly extended their ideas to those of rational
expressions.
In general, I enjoyed teaching this curriculum. As I asked students to think about the
questions, I immediately realized some of the discussions were better suited for pairs than
for small groups. At other times, even though I thought the conversations might not
warrant more than two people, many students seemed to need more support. So I
switched between dyads and small groups depending on the air in the classroom. At
times I let the students decide on their own how many people they wanted in their groups
as well. It was my first time using discourse in this way in a class, but these activities
seemed to successfully help my students learn the intended concepts. Furthermore, my
students seemed to enjoy it, as they regularly came up to me after class to tell me how
much fun they were having. I look forward to adapting these activities over time and I
would love to hear any modifications others have to offer.
146
REFERENCES
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Learning of Mathematics, 14(3), 24-35.
Berggren, J. L. (1986). Episodes in the mathematics of medieval Islam. New York, NY:
Springer-Verlag.
Berggren, J. L. (1997). Mathematics in medieval Islam. Historia Mathematica, 24, 407440.
Burton, D. M. (2002). Elementary Number Theory: Fifth Edition. Boston, MA:
McGraw-Hill.
Coopersmith, A. (1984). Factoring trinomials: Trial and error? Hardly ever!.
Mathematics Teacher, 77, 194-195.
Durbin, J. R. (2005). Modern algebra: An introduction. United States of America: John
Wiley & Sons Inc.
Eisenberg, T. & Dreyfys, T. (1988). Polynomials in the school curriculum. In A.
Coxford & A. Shulte (Eds.), The Ideas of Algebra, K-12 (122-118). Reston, VA:
National Council of Teachers of Mathematics.
Euclid (1956). The Elements, Book II (T. L. Heath, trans). New York, NY: Dover
Publications Inc.
Freudenthal, H. (1991). Revisiting mathematics education. Dordrecht, The Netherlands:
Kluwer.
Gallian, J. A. (1998). Contemporary abstract algebra. Boston, MA: Houghton Mifflin.
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16th, 17th, and 18th century English editions of Euclid's Elements books I through
VI (1561-1795). Historia Mathemtica, 27, 36-53.
147
Goodaire, E. G & Parmenter, M. P. (1998). Discrete mathematics with graph theory.
Upper Saddle River, NJ: Prentice Hall.
Herscovics, N. (1989). Cognitive obstacles in the learning of algebra. In S. Wagner & C.
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Herstein, I. N. (1975). Topics in algebra. United States of America: Wiley Inc.
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Katz, V. J. (1998). A history of mathematics: An introduction. Reading, MA: Addison
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Kieran, C. & Saldanha, L. (forthcoming). Designing tasks for the co-development of
conceptual and technical knowledge in CAS activity: An example from factoring.
In K. Heid & G. Bloome (Eds.), Research on Technology in the Teaching and
Learning of Mathematics, V. 2
Merris, R. (1996). Combinatorics. Boston, MA: PWS Publishing Company.
Moses, R. P. (2001). Radical equations: Civil rights from Mississippi to the algebra
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National Council of Teacher's of Mathematics. (1991). Professional standards for
teaching mathematics. Reston, VA: Author
National Council of Teacher's of Mathematics. (2000). Principles and standards for
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Pycior, H. M. (1981). Peacock and British origins of symbolical algebra. Historia
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148
Sfard, A. (1995). The development of algebra: Confronting historical and psychological
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Wilf, H. S. (1994). Generatingfunctionology. Retreived Novermber 17, 2005 from
149 | 677.169 | 1 |
This Advanced Course from the "28 SAT Math Lessons"
This book is a Common Core course for students taking 8th grade math. It can be used prior or during the 8th grade to help students excel. It teaches students the material that will be covered in the classrooms that follow Common Core curriculum. | 677.169 | 1 |
however. Typical in this respect is the
treatment of factoring in
many
text-books
In this book
all
methods which are of
and which are applied in advanced work are given. short-cuts that solve only examples
real value.
and ingenuity
while the cultivation of the student's reasoning power is neglected. in order to make every example a
social
case of a memorized method. chief
:
among
These
which are the following
1. etc. and conse-
."
this book.
owing
has certain distinctive features.
specially
2.
Such a large number of methods.PREFACE
IN
this
book the attempt
while
still
is
made
to shorten the usual course
in algebra.
omissions serve not only practical but distinctly pedagogic " cases " ends..
manufactured for this purpose.
All unnecessary methods
and "cases" are
omitted. not only taxes a student's memory unduly but in variably leads to mechanical modes of study.
All
practical
teachers
know how few
students understand and
appreciate the more difficult parts of the theory. Until recently the tendency was to multiply as far as possible.
giving to the student complete familiarity
with
all
the essentials of the subject. The entire study of algebra becomes a mechanical application of memorized
rules. but "cases" that are taught only on account of tradition.
Elementary Algebra.
"
While
in
many
respects
similar to the author's
to its peculiar aim. All parts of the theory whicJi are beyond the comprehension
of
the student or wliicli are logically
unsound are
omitted. are omitted.
" The book is designed to meet the requirements for admis-
sion to our best universities
and
colleges. are placed early in the course.
may
be used to supplement the other. there has been placed at the end of the book a collection of exercises which contains an abundance
of
more
difficult
work. all elementary proofs theorem for fractional exponents. in particular the
requirements of the College Entrance Examination Board. and it is hoped that this treatment will materially diminish the difficulty of this topic for young students.
In regard
to
some other features of the book. Moreover.
TJie exercises are slightly simpler than in the larger look. as quadratic equations and
graphs. The presenwill be found to be tation of problems as given in Chapter
V
quite a departure from the customary way of treating the subject.
Topics of practical importance. all proofs for the sign age
of the product of
of the binomial
3.
two negative numbers.
differ
With very few
from those
exceptions
all
the exer
cises in this
book
in the
"Elementary Alge-
bra". etc. especially problems and factoring.
The best way to introduce a beginner to a new topic is to offer Lim a large number of simple exercises. e. This made it necessary to introduce the theory of proportions
. hence either book
4. the following
may
be quoted from the author's "Elementary Algebra":
which
"Particular care has been bestowed upon those chapters in the customary courses offer the greatest difficulties to
the beginner. For the more ambitious student.vi
PREFACE
quently hardly ever emphasize the theoretical aspect of alge bra. a great deal of the theory offered in the avertext-book is logically unsound .
enable students
who can devote only a minimum
This arrangement will of time to
algebra to study those subjects which are of such importance for further work. however.g.
while in the usual course proportions
are studied a long time after their principal application. and commercial are numerous.
viz.'
This topic has been preit is
sented in a simple. are
frequently arranged in sets that are algebraically uniform. the student will be able to utilize this knowledge where it is most
needed.
By studying proportions during the first year's work."
Applications taken from geometry. Moreover. elementary way. but the true study of algebra has not been sacrificed in order to make an impressive display of sham
life
applications. and
of the
hoped that some
modes of representation given
will be considered im-
provements upon the prevailing methods. in
"
geometry
.
and they usually involve difficult numerical calculations. " Graphical methods have not only a great practical value. an innovation which seems to mark a distinct gain from the pedagogical point
of view. and hence the student is more easily led to do the work by rote
than when the arrangement
braic aspect of the problem. physics.
but they unquestionably furnish a very good antidote against 'the tendency of school algebra to degenerate into a mechanical application of
memorized
rules.
of the Mississippi or the height of Mt. based upon statistical abstracts.
McKinley
than one that gives him the number of Henry's marbles.PREFACE
vii
and graphical methods into the first year's work.
to solve a
It is
undoubtedly more interesting for a student
problem that results in the height of Mt.
nobody would find the length Etna by such a method. such examples. The entire work in graphical methods has been so arranged that teachers who wish
a shorter course
may omit
these chapters. But on the other hand very few of such applied examples are
genuine applications of algebra.
is
based principally upon the alge-
.
desires to acknowledge his indebtedness to Mr.
William P.
NEW
YORK.viii
PREFACE
problems relating to physics often
offer
It is true that
a field
for genuine applications of algebra. Manguse for the careful reading of the proofs and
many
valuable suggestions. however.
edge of physics.
genuine applications of elementary algebra work seems to have certain limi-
but within these limits the author has attempted to
give as
many
The author
for
simple applied examples as possible. 1910. is such problems involves as a rule the teaching of physics by the
teacher of algebra.
.
pupil's knowlso small that an extensive use of
The average
Hence the
field of
suitable for secondary school
tations.
April.
ARTHUR SCHULTZE.
23.
Read the expressions
of Exs. a = 4. 6 = 5. Six times a plus 4 times
32.
and
If the three sides of a triangle contain respectively c feet (or other units of length).6 -f c) (6
a
+ c).
35. = 3.
sible to state
Ex.
12 cr6
-f-
6 a6 2
6s. and the area of the
is
triangle
S
square feet (or squares of other units selected). 2-6 of the exercise.
Six
2
.
The quantity a
6
2 by the quantity a
minus
36.
24. 6 = 2.
6. 6 = 5.
a. a = 4.
Express in algebraic symbols 31.
w
cube plus three times the quantity a minus
plus 6 multiplied
6.
30. 10-14
The
representation of numbers by letters makes it posvery briefly and accurately some of the principles of arithmetic. if
:
a = 2.
:
6. physics. 25. and other sciences.
a =3.
.6 . a =4.
6=2.
28. 6 = 6.
Twice a3 diminished by 5 times the square root of the quantity a minus 6 square.
a = 3. of this exercise?
What kind of expressions are Exs.c) (a .
then
8 = \ V(a + 6 + c) (a 4.
27. a
a=3. 38.
37. geometry. 33. 6 = 3. a = 3. 6 = 1.
26. 6.
6 = 4.
34. 6 = 6. a = 2.
29.
Six times the square of a minus three times the cube of Eight x cube minus four x square plus y square. 6 = 7.12
17
&
*
ELEMENTS OF ALGEBRA
18
'
8
Find the numerical value of 8 a3
21. 30.
22.
g. 15 therefore feet. A carrier pigeon in 10 minutes.
(b) 5. if v . if v = 30 miles per hour.
the area of the triangle equals
feet.
=
(a)
How
far does
a body fall from a state of rest in 2
seconds ?
(b)
*
stone dropped from the top of a tree reached the ground in 2-J.
By
using the formula
find the area of a triangle
whose
sides are respectively
(a) 3. A body falling from a state of rest passes in t seconds 2 over a space S (This formula does not take into ac^gt 32 feet.
.e.
(c) 4. A train in 4 hours. b 14. if v = 50 meters per second 5000 feet per minute. c. 14.
2.
i.16 centimeters per second.
9
distance s passed over by a body moving with the uniform velocity v in the time t is represented by the formula
The
Find the distance passed over by A snail in 100 seconds. 12. Find the height of the tree. d. the three sides of a triangle are respectively 13. 13. then a 13.16
1
= 84. count the resistance of the atmosphere. An electric car in 40 seconds. How far does a body fall from a state of rest in T ^7 of a (c)
A
second ?
3.INTRODUCTION
E.
84 square
EXERCISE
1. b. and 15 feet.) Assuming g
.
S = | V(13-hl4-fl5)(13H-14-15)(T3-14-i-15)(14-13-f-15)
= V42-12-14. and 5 feet.
4. and 13 inches. if v
:
a. and c
13
and
15
=
=
=
.seconds.
5.
diameter of a sphere equals d
feet.
2 inches. This number cannot be expressed exactly.
is
H
2
units of length (inches.).14
square meters.
the area
etc.
on $ 500 for 2 years at 4 %.
If the
(b) 1 inch.
(c)
8000 miles.
of this formula
:
The The
interest on interest
$800
for 4 years at
ty%.
ELEMENTS OF ALGEBRA
If
the radius of a circle
etc.) Find the surface of a sphere whose diameter equals
(a)
7.
~
7n
cubic feet. Find the area of a circle whose radius is
It
(b)
(a) 10 meters.
then the
volume
V=
(a) 10 feet.
(c)
10
feet. the
3.
If the diameter of a sphere equals d units of length.).
to Centigrade readings:
(b)
Change the following readings
(a)
122 F. then
=p
n
*
r
%>
or
Find by means
(a)
(b)
6.14
4.
denotes the number of degrees of temperature indi8.
32 F.
(c)
5
F.
$ = 3.
meters.
. If cated on the Fahrenheit scale. (The number 3.
6
Find the volume of a sphere whose diameter equals:
(b)
3
feet. the equivalent reading C on the Centigrade scale may be found by the formula
F
C
y
= f(F-32).
square units (square inches.14 is frequently denoted by the Greek letter TT.
:
8000 miles.
fo
If
i
represents the simple interest of
i
p
dollars at r
in
n
years. and the value given above is only an
surface
$=
2
approximation.14d (square units).
(c)
5 miles.
CHAPTER
II
ADDITION. Or in the symbols of algebra
$4) =
Similarly. the fact that a loss of
loss of
+ $2.
.$6) + (-
$4) = (-
$10). In algebra. we define the sum of two numbers in such a way that these results become general. we call the aggregate value of a gain of 6 and a loss of 4 the sum of the two.
While
in arithmetic the
word sum
refers only to
the
result obtained
by adding positive numbers. Thus a gain of $ 2 is considered the sum of a gain of $ 6 and a loss of $ 4. or that
and
(+6) + (+4) = +
16
10. however.
of
$6 and a gain
$4
equals a
$2 may be
represented thus
In a corresponding manner we have for a loss of $6 and a
of
loss
$4
(. SUBTRACTION. or positive and negative numbers. AND PARENTHESES
ADDITION OF MONOMIALS
31. but we cannot add a gain of $0 and a loss of $4.
Since similar operations with different units always produce analogous results. In arithmetic we add a gain of $ 6 and a gain of $ 4. in algebra this word includes also the results obtained by adding negative.
+ (-9). the average of 4 and 8
The average The average
of 2. c = = 5. find the numerical values of a + b
-f c-j-c?. 10. = 5.
l-f(-2).
d = 5. subtract their absolute values and
. and the sum of the numbers divided by n.
23.
Thus. 12.
18.
(always) prefix the sign of the greater. 24.
-
0. d = 0.
5.16
32. add their absolute values if they have opposite signs.
The average
of two
numbers
is
average of three numbers average of n numbers is the
is one half their sum. '. 23-26.
EXERCISE
Find the sum
of:
10
Find the values
17.
33.
(-17)
15
+ (-14). if :
a
a
= 2.
. 5. 4.
6
6
= 3.3.
22.
4
is
3 J. c =
4.
of:
20.
of 2. + -12. the one third their sum.
(_
In Exs.
19.
is 0.
21. is 2.
ELEMENTS OF ALGEBRA
These considerations lead to the following principle
:
If two numbers have the same sign.
. 74. 38. if his yearly gain or loss during 6 years was $ 5000 gain.
:
34..
27. 6.
which are not
similar. 7 yards.
.
What number must be added to 9 to give 12? What number must be added to 12 to give 9 ? What number must be added to 3 to give 6 ? C* What number must be added to 3 to give 6? **j Add 2 yards. c = 0. -4.
AND PARENTHESES
d = l. 12. & = 15. 7 a.
&
28.
40. Find the average gain per year of a merchant. 72. .
^
'
37.
4
F.
-'
1?
a
26.
are similar terms.
35.
3 and 25. 1. 32. and 3 a.
= -23.
. and 4. 09.5.
2. and 3 F.
2.
.
:
Find the average temperature of Irkutsk by taking the average of the following monthly temperatures 12. 60.5. -11 (Centigrade)..
Similar or like terms are terms which have the same
literal factors. $7000 gain.
31. 55. 41. \\ Add 2 a.3. . or
and
. 66.
= 22. $500 loss.
:
48. 10. $3000 gain.
Dissimilar or unlike terms are terms
4 a2 6c and o
4 a2 6c2 are dissimilar terms. }/ Add 2 a.
33. and $4500 gain. 0. 32. = -13.
or 16
Va + b
and
2Vo"+~&.
42. 39.
13.
30. SUBTRACTION.7.
25.
:
and
1. Find the average temperature of New York by taking the average of the following monthly averages 30. 10. 5 and
12. 10.
Find the average of the following temperatures 27 F.
d=
3.ADDITION. 37. $1000 loss. .4. 34.
36.
6. c=14.
.
. 3. 43.
29.
5 a2 &
6 ax^y and
7 ax'2 y. 7 a. and 3 a.
and
-8
F.
sets of
numbers:
13.
'
Find the average of the following
34. affected
by the same exponents.7.13. and 3 yards. 6.
:
2 a2.
5Vm + w.
ELEMENTS OF ALGEBRA
The sum
of 3
of
two similar terms
x2
is
is
another similar term.
+ 6 af
.
2(a-f &). in algebra it may be considered b.
5 a2
.
13.
EXERCISE
Add:
1.
5l
3(a-f-6).
1
\
-f-
7 a 2 frc
Find the sum of
9.
-3a
.
2
.
12Vm-f-n.
Vm
-f.
ab
7
c
2
dn
6.
12
13
b sx
xY xY 7 #y
7.
sum of two such terms can only be them with the -f.
In algebra the word sum is used in a 36.
14
.
11
-2 a +3a -4o
2.
-f
4 a2. The indicated by connecting and a 2 and
a
is
is
-f-
a2
.
9(a-f-6). b wider sense than in arithmetic. While in arithmetic a denotes a difference only.18
35.ii.
11.sign. 7 rap2.
b
a
-f (
6).
The sum The sum
of a of a
Dissimilar terms cannot be united into a single term. or
a
6. and 4 ac2
is
a
2 a&
-|-
4 ac2. either the difference of a and b or the sum of a and
The sum
of
a.13 rap
25 rap 2.
10.
12(a-f b)
12.
The sum
x 2 and
f
x2 .
2 a&. Algebraic sum.
.
12
2 wp2 .
change the sign
of the subtrahend and
add.ADDITION.
.
State the other practical examples which show that the number is equal to the addition of a
40. SUBTRACTION.
6
-(-3) = 8.
To
subtract.
From
5 subtract
to
.
may
be stated in a
:
5 take form e. two numbers are given.
+b
3.3. The student should perform mentally the operation of chang8 2 6 from 6 a 2 fc.
41. the given number the subtrahend.
AND PARENTHESES
23
subtraction of a negative
positive number.
From
5 subtract
to
The number which added
Hence.
From
5 subtract
+ 3.2.
This gives by the same method. may be stated number added to 3 will give 5? To subtract from a the number b means to find the number which added to b gives a. the other number is required. Subtraction is the inverse of addition. Therefore any example in subtraction
different
.
3. In addition. called the minvend.
1.
2. ing the sign of the subtrahend thus to subtract 6 a 2 6 and 8 a 2 6 and find the sum of change mentally the sign of
.
(-
6)
-(-
= .g.
a.
3 gives
3)
The number which added
Hence. a-b =
x.
Ex. In subtraction. and the
required number the difference.
The
results of the preceding examples could be obtained
by the following
Principle. Or in symbols.
NOTE.
if
x
Ex.
Ex.
5
is
2.
3 gives 5
is
evidently 8. from What 3. and their algebraic sum is required.
7. the algebraic sum and one of the two numbers is
The algebraic sum is given.
Hence
the
it is
sign
may
obvious that parentheses preceded by the -f or be removed or inserted according to the fol:
lowing principles
44.6 b -f (.
45. may be written as follows:
a
-f ( 4.
a+(b-c) = a +b .
AND PARENTHESES
27
SIGNS OF AGGREGATION
43.c.a
-f-
= 4a
sss
7a
12
06
6. one occurring within the other.g. The beginner will find it most convenient
at every step to
remove only those parentheses which contain
(7 a
no others.
46.
66
2&-a + 6
4a
Answer.2 b . we may begin either at the innermost or outermost. If we wish to remove several signs of aggregation.b c = a a
&
-f-
-f.& c
additions
and sub-
+ d) = a + b
c
+ d. SUBTRACTION. Ex. I.
6
o+(
a
+ c) = a =a 6 c) ( 4-.
tractions
By using the signs of aggregation.
. Simplify 4 a f
+ 5&)-[-6& +(-25.
A
moved
w
may be resign of aggregation preceded by the sign inserted provided the sign of evei'y term inclosed is
E.c.a^6)]
-
}
.
If there is no sign before the first term within a paren*
-f-
thesis. A sign of aggregation preceded by the sign -f may be removed or inserted without changing the sign of any term.ADDITION.
& -f
c.
II.
changed.
(b
c)
a
=a
6 4-
c. the sign
is
understood.a~^~6)]} = 4 a -{7 a 6 b -[.
4a-{(7a + 6&)-[-6&-f(-2&.
2.
Nine times the square of the sum of a and by the product of a and b.
6.
The
difference of a
and
6.
of the cubes of
m and
n.
7.ADDITION.
Three times the product of the squares of The cube of the product of m and n.
4 xy
7 x* 4-9 x + 2.
terms
5.
13.
EXERCISE
AND PARENTHESES
16
29
In each of the following expressions inclose the last three terms in a parenthesis
:
1.
3.
first.
8.7-fa.
5.
The product The product
m
and
n.1. 9.
m and n.
12.
5 a2
2.2 tf .
II.
4.
The sum
of tKe squares of a
and
b.
6 diminished
.
'
NOTE.
y
-f-
8
.
EXERCISES
IN"
ALGEBRAIC EXPRESSION
17
:
EXERCISE
Write the following expressions
I.
and the subtrahend the second.
The sum
of the fourth powers of a of
and
6.
z
+ d.
3.4 y* .
The The
difference of the cubes of
m
and
n.
p + q + r-s.
In each of the following expressions inclose the last three in a parenthesis preceded by the minus sign
:
-27i2 -3^ 2 + 4r/. SUBTRACTION.
difference of the cubes of n and m. )X
6. The product of the sum and the difference
of
m and n.
m
x
2
4.
5^2
_ r . .
7.
a-\-l>
>
c
+
d.
10.
2m-n + 2q-3t. The minuend is always the of the two numbers mentioned.
The sum^)f
m
and
n.
The square of the difference of a and b.
and
c
divided by the
ference of a and
Write algebraically the following statements:
V 17.
difference of the cubes of a
and
b divided
by the
difference of
a and
6.
The sum
The
of a
and
b multiplied
b is equal to the difference of
by the difference of a and a 2 and b 2
.)
.
b.
x cube minus quantity 2 x2 minus 6 x plus
The sum
of the cubes of a.
a plus the prod-
uct of a and
s
plus the square of
-19. 6. (Let a and b represent the numbers. The difference of the squares of two numbers divided by the difference of the numbers is equal to the sum of the two numbers.
ELEMENTS OF ALGEBRA
The sum
x.
18.
6 is equal to the square of
b. 16.
d.30
14.
dif-
of the squares of
a and
b increased
by the
square root of
15.
weights. weights at A ? Express this as a multibalance. weight at A ? What is the sign of a 3 Ib. If the two loads balance.
If the
two loads balance.
A
A
A
1. what force
31
is
produced by tak(
ing away 5 weights from
B ? What therefore is
5)
x(
3) ?
. is
by
taking away 5 weights from
A?
5
X 3?
6.
If the two loads
what
What. what force is produced by the Ib. is 5 x ( 3) ?
7.CHAPTER
III
MULTIPLICATION
MULTIPLICATION OF ALGEBRAIC NUMBERS
EXERCISE
18
In the annexed diagram of a balance. 2. weight at B ?
If the
addition of five 3
plication example.
two loads balance.
3.
By what sign is an upward pull at A represented ? What is the sign of a 3 Ib. let us consider the and JB. applied at let us indicate a downward pull at by a positive sign.
5.
force is produced
therefore.
4. and forces produced at by 3 Ib. therefore. what force is produced by the addition of 5 weights at B ? What.
4 multi44-44-4 12.
Thus.4)-(.
48.9) x
11. a result that would not be obtained by other assumptions.
the multiplier is a negative number. and we may choose any definition that does not lead to contradictions. such as given in the preceding exercise. 4 multiplied by 3. Multiplication by a positive integer is a repeated addition.
examples were generally
method of the preceding what would be the values of
(
5x4. To take a number 7 times.4)-(-4) = +
12. This definition has the additional advantage of leading to algenumbers which are identical with those for positive numbers. becomes meaningless
if
definition.
(. (-5)X4.
ELEMENTS OF ALGEBRA
If
the signs obtained by the
true.32
8.
NOTE. or plied by 3.
. or
4x3 =
=
(_4) X
The preceding
3=(-4)+(-4)+(-4)=-12.
In multiplying integers we have therefore four cases
trated
illus-
by the following examples
:
4x3 = 4-12.
4
x(-8) = ~(4)-(4)-(4)=:-12.
4x(-3)=-12.
x
11.
however. 9 x
(-
11).
(
(.
(-
9)
x (-
11) ?
State a rule by which the sign of the product of
two
fac-
tors can be obtained. Practical examples^
it
however. thus. times is just as meaningless as to fire a gun
tion
7
Consequently we have to define the meaning of a multiplicaif the multiplier is negative.
9
9.
Multiplication
by a negative
integer is a repeated
sub-
traction.4) x
braic laws for negative
~ 3> = -(. make venient to accept the following definition
:
con-
49. 5x(-4).
Ex.3 a 2 + a8 . the student should
apply this test to every example.
If
Arranging according to ascending powers
2
a
. Since errors. To multiply two polynomials.
Multiply 2
+ a -a. Multiply 2 a .3 a 2 + a8
a
a = =-
I
1
=2
-f
2
a
4.
Check.
2. Since all powers of 1 are 1. the work becomes simpler and more symmetrical by arranging these expressions according to either ascending or descending powers.a6
=2
by numerical
Examples
in multiplication can be checked
substitution.
.3 ab
2
2 a2
10 ab
-
13 ab
+ 15 6 2 + 15 6 2
Product.M UL TIP LICA TION
37
58.
a2
+ a8 + 3 .4.3
b
by a
5
b. however.a6 4 a 8 + 5 a* . this method
tests only the values of the coefficients
and not the values of
the exponents.2 a2 6 a8
2 a*
*
-
2"
a2
-7
60.
59. If the polynomials to be multiplied contain several powers of the same letter.a
.
2a-3b a-66 2 a . multiply each term of one by each term of the other and add the partial products thus formed. as illustrated in the following example
:
Ex.
The most convenient way of adding the partial products is to place similar terms in columns. are far more likely to occur in the coefficients than anywhere else. 1 being the most convenient value to be substituted for all letters.1.3 a
3
2
by 2 a
:
a2 + l.
(5a-4)(4a-l). and are represented as
2 y and 4y 3 x.
.
or
The student should note
minus
signs.42
ELEMENTS OF ALGEBRA
of the result is obtained
product of 5 x
follows:
by adding the These products are frequently called the cross products.
11. ) (2
of a polynomial.
2
(2x y
(6
2
2
+ z )(ary + 2z ). 13.
7%e square of a polynomial is equal to the sum of the squares of each term increased by twice the product of each term with each that follows it.
3. 14.
(5a6-4)(5a&-3).
The middle term
or
Wxy-12xy
Hence in general.
7.
5.
((5a?
(10
12. (100 + 3)(100 + 4).
(2a-3)(a + 2).
2
10. the product of two binomials whose corresponding terms are similar is equal to the product of the first two
terms.
(x
i-
5
2
ft
x 2 -3 6 s).-f
2 a&
-f
2 ac
+ 2 &c.&
+ c) = a + tf + c
. 9.
2
(2m-3)(3m + 2).
2
2
+ 2) (10 4-3).
4. (4s + y)(3-2y).
65.
2 2 2 2 (2a 6 -7)(a & +
5).
:
25
2.
8.
that the square of each term is while the product of the terms may have plus always positive.
sum of the
cross products.
The square
2
(a 4. plus the product of the
EXERCISE
Multiply by inspection
1.
6. plus the
last terms. (3m + 2)(m-l).
is the process of finding one of two factors and the other factor are given. The dividend is the product of the two factors, the divisor the given factor, and the quotient is the required factor.
67.
Division
if
their product
is
Thus
by
-f
to divide
12.
12
by
+
3,
we must find
is
the
;
number which
3 gives
But
this
number
4
hence
_
multiplied
12 r +3
=4.
68.
Since
-f
a
-
-f b
-fa
_a
and
it
-f-
a
= -f ab = ab b = ab b = ab,
b
-f-
follows that
4-a
=+b
ab
a
ab
a
69.
Hence the law
:
of signs
is
the same in division as in
multiplication
70.
Like signs produce plus, unlike signs minus.
Law
of
,
a8 -5- a5
=a
3
for a 3
It follows from the definition that Exponents. X a5 a8
=
.
Or
in general, if
greater than
m n, a
-f-
and n are positive integers, and m ~ n an = a m a" = a'"-", for a
<
m
m
is
45
46
ELEMENTS OF ALGEBRA
71. TJie exponent of a quotient of two powers with equal bases equals the exponent of the dividend diminished by the exponent
of the divisor.
DIVISION OF MONOMIALS
7 3 72. To divide 10x y z by number which multiplied by number is evidently
2x y
6
2
,
we have
z
to
find
the
2x*y
gives 10 x^ifz.
This
Therefore,
the quotient
*
,
= - 5 a*yz.
is
Hence,
sign,
of two monomials of their
part
coefficients,
is the
a monomial whose
coefficient is the quotient
preceded by the proper
literal
and whose
literal
found
in accordance with the
quotient of their law of exponents.
parts
73. In dividing a product of several factors by a number, only one of these factors is divided by that number. Thus (8 12 20)-?-4 equals 2 12 20, or 8 3 20 or 8 12 5.
-
-
.
-
.
-
.
EXERCISE
Perform the divisions indicated
'
:
28
'
2
.
76-H-15.
-39-*- 3.
2
15
3"
7
7'
3.
-4*
'
4.
5.
-j-2
12
.
4
2
9
5 11
68
3 19 -j-3
5
10.
(3
38
-
-2 4 )^(3 4 .2 2).
56
'
11.
3
(2
.3*.5 7 )-f-(
2
'
12
'
2V
14
36 a
'
13
''
y-ffl-g
35
-5.25
-12 a
2abc
15
-42^
'
-56aW
'
UafiV
DIVISION
lg
47
-^1^. 16 w
7
20>
7i
9
_Z^L4L.
22.
10 iy.
132 a V* 14 1
*
01
-240m
120m-
40
6c
fl
/5i.
3J)
c
23.
2 (15- 25. a ) -=- 5.
25. 26.
(18
(
.
5
.
2a )-f-9a.
2
24.
(7- 26 a
2
)
-f-
13.
DIVISION OF POLYNOMIALS BY MONOMIALS
To divide ax-}- fr.e-f ex by x we must find an expression which multiplied by x gives the product ax + bx -J- ex.
74.
But
TT
x(a
aa?
Hence
+ b e) ax + bx + ex. + bx -f ex = a 4- b +
-\.
,
.
c.
a?
To divide a polynomial by a monomial, cfc'wde each term of the dividend by the monomial and add the partial quotients thus
formed.
3 xyz
EXERCISE
Perform the operations indicated
1.
:
29
2.
5.
fl
o.
(5*
_5* + 52)
-5.
52
.
3.
97
.
(2
(G^-G^-G^-i-G
(11- 2
4.
(8- 3
+
11 -3
+ 11
-5)-*- 11.
18 aft- 27 oc
Q y.
9a
4
-25 -2 )^-2
<?
2
.
+8- 5 + 8-
7) -*-8.
5a5 +4as -2a
2
-a
-14gV+21gy
Itf
15 a*b
-
12
aW + 9 a
2
2
3a
48
,
ELEMENTS OF ALGEBRA
22
4,
m n - 33 m n
4
s
2
-f
55
mV
- 39 afyV + 26 arVz 3
- 49 aW + 28 a -W - 14 g 6 c
4 4
15. 16.
2 (115 afy -f 161 afy
- 69
4
2
a;
4
?/
3
- 23 ofy
3
4
)
-5-
23 x2y.
(52
afyV - 39
4
?/
oryz
- 65 zyz - 26 tf#z)
-5-
13 xyz.
-f-
,
17.
(85 tf
- 68 x + 51 afy - 34 xy* -f 1 7
a;/)
- 17
as.
DIVISION OF A POLYNOMIAL BY A POLYNOMIAL
75.
Let
it
be required to divide 25 a
- 12 -f 6 a - 20 a
3
2
by
2 a 2 -f 3 a, divide
4
a, or, arranging according to
2
descending powers of
6a3 -20a
-f
25a-12
2 by 2a -
The term containing the highest power of a in the dividend (i.e. a 8 ) is evidently the product of the terms containing respectively the highest power of a in the divisor and in the quotient.
Hence the term containing the highest power
of a in the quotient is
If
the product of 3 a and 2
2
4 a
+
3, i.e.
6 a3
12 a 2
-f
9 a, be sub-
8 a 2 -f 16 a tracted from the dividend, the remainder is 12. This remainder obviously must be the product of the divisor and the rest of the quotient. To obtain the other terms of the quotient we have
therefore to divide the remainder,
8 a2
-f-
16 a
12,
2 by 2 a
4 a
+
3.
consequently repeat the process. By dividing the highest term in the new dividend 8 a 2 by the highest term in the divisor 2 a 2 we obtain
,
We
4,
the next highest term in the quotient. 4 by the divisor 2 a2 4 a Multiplying
-I-
+ 3, we
obtain the product
8 a2
16 a
12,
which subtracted from the preceding dividend leaves
the required quotient.
no remainder. Hence 3 a
4
is
DIVISION
The work
is
49
:
usually arranged as follows
- 20 * 2 + 3 0a-- 12 a 2 +
a3
25 a
{)
-
12
I
2 a2 8 a
-
4 a 4
a
_
12
+3
I
-
8 a? 4- 16
a-
76. The method which was applied in the preceding example may be stated as follows 1. Arrange dividend and divisor according to ascending or
:
descending powers of a common letter. 2. Divide the first term of the dividend by the first term of the divisor, and write the result for the first term of the quotient.
3.
Multiply this term of the quotient by the whole divisor, and
subtract the result
4.
from
it
the dividend.
the same order as the given new dividend, and proceed as before.
Arrange
the
remainder in
as a
expression, consider
5.
until the highest poiver
Continue the process until a remainder zero is obtained, or of the letter according to which the dividend
is less
was arranged
the divisor.
than the highest poiver of the same
letter in
77.
Checks.
Numerical substitution constitutes a very con-
venient, but not absolutely reliable check. An absolute check consists in multiplying quotient and divisor. The result must equal the dividend if the division
was
exact, or the dividend diminished by the remainder division was not exact.
The
first
member
or left side of an equation
is
that part
The secof the equation which precedes the sign of equality. is said to satisfy an equation.CHAPTER V
LINEAR EQUATIONS AND PROBLEMS
79.
A set of numbers which when substituted for the letters an equation produce equal values of the two members. in the equation 2 x 0. (a + ft) (a b) and b. ond member or right side is that part which follows the sign of
equality.
the
80.
82. The sign of identity sometimes used is = thus we may write
. y y or z) from its relation to
63
An
known numbers. second member is x
+
4
x
9.
which
is
true for all values
a2 6 2 no matter what values we assign to a Thus.r
-f9
= 20
is
true only
when
a.
x
20.
81.
. An identity is an equation of the letters involved.
hence
it
is
an equation
of
condition.
Thus.
ber
equation is employed to discover an unknown num(frequently denoted by x. An equation of condition is usually called an equation. An equation of condition is an equation which is true only for certain values of the letters involved.
y
=
7 satisfy
the equation x
y
=
13.
in
Thus x
12 satisfies the equation x
+
1
13.
83.
.
=11.
.
(rt+6)(a-ft)
=
2
-
b'
2
.
the
first
member
is
2 x
+
4.
. A
2
a.g. x
I.
90.e.
To
solve
an equation
to find its roots.
Like powers or
like roots
of equals are equal. 87. Consider the equation b Subtracting a from both members.
(Axiom
2)
the term a has been transposed from the left to thQ
right
member by changing
its
sign.54
84. the products are equal.
3.
A
linear equation or
which when reduced
first
to its simplest
an equation of the first degree is one form contains only the
as
9ie
power of the unknown quantity.
fol-
A
linear equation is also called a simple equation.
If equals be multiplied by equals.
.
but 4 does not equal
5. the remainders are
equal.
= bx
expressed by a letter or a combination of
c.
A
term may be transposed from
its sign.
one member to another by changing
x + a=. an^ unknown quantity which satisfies the equation is
a root of the equation.
5.
E.
85. the quotients are equal.
NOTE.
Transposition of terms. called axioms
1.
If equals be subtracted from equals.
86.
A numerical
equation is one in
which
all
.
ELEMENTS OF ALGEBRA
If
value of the
an equation contains only one unknown quantity. the
sums are
equal.b.
expressed in arithmetical numbers
literal
is
as (7
equation
is
one in which at least one of the
known
quantities as x -f a letters
88.
4.
Axiom
4
is
not true
if
0x4
= 0x5.
89.
2
= 6#-f7.
a.
If equals be divided by equals.
If equals be added
to equals.
the divisor equals zero.
The process
of solving equations depends upon the
:
lowing principles.
the
known quan
x) (x -f 4)
tities are
=
.2.
2.
9
is
a root of the equation 2 y
+2=
is
20.
Divide 100 into two
12.
ELEMENTS OF ALGEBRA
What must
be added to a to produce a sum b ?
:
Consider the arithmetical question duce the sum of 12 ?
What must
be added to 7 to pro-
The answer is 5.
9.
Divide a into two parts.58
Ex. Hence 6 a must be added
to a to give
5. is d.
one part equals
is 10.
14. and the smaller one
parts. or 12 7. greater one is g.
7.
is
a?
2
is
c?. so that one part Divide a into two parts. so that one part
The
difference between
is s. one yard will cost
100
-dollars. 11.
13.
33
2.
Find the greater one. one yard will cost -
Hence
if
x
-f
y yards cost $ 100.
5.
x
-f-
y yards cost $ 100
. 6.
If 7
2.
4.
two numbers
and the and the
2
Find the greater one.
$> 100 yards cost one hundred dollars. is b. so that
of c ?
is
p.
find the cost of one yard.
10. 15.
1.
What number divided by 3 will give the quotient a? ? What is the dividend if the divisor is 7 and the quotient
?
.
By how much does a exceed 10 ? By how much does 9 exceed x ? What number exceeds a by 4 ? What number exceeds m by n ? What is the 5th part of n ? What is the nth part of x ? By how much does 10 exceed the third part of a? By how much does the fourth part of x exceed b ? By how much does the double of b exceed one half Two numbers differ by 7. The difference between two numbers Find the smaller one.
6.
3.
Ex.
EXERCISE
1.
17.
a.
smaller one
16.
feet wider than the one
mentioned in Ex.
32.
and
B
is
y years old.
and spent
5 cents.
sum
If A's age is x years.
19. The greatest of three consecutive the other two.
What What What What
is
the cost of 10 apples at x cents each ?
is
is is
x apples cost 20 cents ? the price of 12 apples if x apples cost 20 cents ? the price of 3 apples if x apples cost n cents ?
the cost of 1 apple
if
. square feet are there in the area of the floor ?
How many
2 feet longer
29.
numbers
is x.
is
A A
is
# years
old.
smallest of three consecutive numbers
Find
the other two. find the of their ages 6 years hence. 33.
Find
21.
If
B
gave
A
6
25.
Find
35.
22.
A
feet wide.
How many
cents had he left ?
28. 28. b dimes.
How many
cents are in d dollars ? in x dimes ?
A has
a
dollars.
A
dollars.
and 4
floor of a room that is 3 feet shorter wider than the one mentioned in Ex.
and
c cents.LINEAR EQUATIONS AND PROBLEMS
18. rectangular field is x feet long and the length of a fence surrounding the field.
?/
31.
34.
59
What must
The
be subtracted from 2 b to give a?
is a. find the
has ra dollars.
24.
20. and B's age is y years.
A man
had a
dollars.
How many years
A
older than
is
B?
old. and B has n dollars. 28.
How many
cents
has he ?
27. amount each will then have.
y years
How
old was he 5 years ago ?
How
old will he be 10 years hence ?
23. Find the sum of their ages
5 years ago.
26.
Find the area of the Find the area of the
feet
floor of
a room that
is
and 3
30. A room is x feet long and y feet wide.
If a man walks n miles in 4 hours. What fraction of the cistern will be second by the two pipes together ?
44.
48.
If a
man walks
?
r miles per hour.
of 4. find the fraction.
of m.
A
cistern can be filled
in
alone
fills it
by two pipes.
how many
how many
miles will
he walk in n hours
38. The first pipe x minutes. and the second pipe alone fills it in
filled
y minutes.
Find x
% %
of 1000.
How many
x years ago
miles does a train
move
in
t
hours at the
rate of x miles per hour ?
41.
miles does
will
If a man walks r miles per hour. in how many hours he walk n miles ?
40.
b
To express in algebraic symbols the sentence: " a exceeds much as b exceeds 9.
m is the
denominator. What fraction of the cistern will be filled by one pipe in one minute ?
42. he walk each hour ?
39.
The numerator
If
of a fraction exceeds the denominator
by
3. and "by as much as" Hence we have means equals (=)
95.
Find the
number.
a. how many miles he walk in n hours ?
37.
-46. Find a
47.
The two
digits of a
number
are x and
y.
.50.60
ELEMENTS OF ALGEBRA
wil\
36.
A
cistern
is
filled
43. If a man walks 3 miles per hour.
per
Find 5 Find 6
45.
-.
A
was 20 years
old.
as
a exceeds
b
by as much
as c exceeds 9." we have to consider that in this by statement "exceeds" means minus ( ).
c
a
b
=
-
9.
% % %
of 100
of
x.
49.
Find
a.
How
old
is
he
now ?
by a pipe in x minutes.
EXERCISE
The The double The sum
One
34
:
Express the following sentences as equations
1.
of a and 10 equals 2
c.
The product
of the
is
diminished by 90 b divided by 7.LINEAR EQUATIONS AND PROBLEMS
Similarly.
equal to the
sum and the difference of a and b sum of the squares of a and
gives the
Twenty subtracted from 2 a
a.
80. Four times the difference of a and b exceeds c by as
d exceeds
9.
of a increased
much
8.
3.
In
many
word
There are usually several different ways of expressing a symbolical statement in words.
of x increased by 10 equals
x.
by one third of b equals 100.
6.
5.
cases it is possible to translate a sentence word by in algebraic symbols in other cases the sentence has to be changed to obtain the symbols. thus:
a
b
= c may
be expressed as follows
difference between a
:
The
and
b is
c.
same
result as 7
subtracted from
.
-80.
c. a is greater than b by b is smaller than a by
c.
8
-b ) + 80 = a
. etc.
=
2
2
a3
(a
-
80.
The double
as
7.
double of a
is
10.
9. the difference of the squares of a
61
and
b increased
-}-
a2
i<5
-
b'
2
'
by 80 equals the excess of a over
80
Or. 2.
a exceeds b by c.
c.
The
excess of a over b
is c.
third of x equals
difference of x
The
and y increased by 7 equals
a.
4.
18. and C have respectively 2 a. B's age
20. 14.
.
amounts. sum equals $20.
50
is
x % of
15. 3 1200 dollars.
5x
A sum of money consists of x dollars. pays to C $100.
as 17
is
is
above
a.
m is x %
of n.000.
A
gains
$20 and B
loses
$40.
is
If A's age is 2 x.
a. B's.62
10.
symbols
B.
(c)
If each
man
gains $500. 16. they have equal
of A's.
A
If
and
B
B together have $ 200 less than C.
12. In 3 years A will be twice as old as B.
first
00
x % of the
equals one tenth of the third sum. (e) In 3 years A will be as old as B is now. the first sum equals 6 % of the third sura. a second sum.
and C's age
4
a. a third sum of 2 x + 1 dollars. express in algebraic
3x
:
10.
x
is
100
x%
is
of 700.
17.
->. A is 4 years older than
Five years ago A was x years old.
#is5%of450. B.
(a)
(b)
(c)
A is twice as old as B. the
sum
and C's
money
(d)
(e)
will be $ 12.
x
4-
If A.
a. they have equal amounts. Express as
:
equations of the (a) 5
(b)
(c)
% a%
of the second
(d)
x c of / a % of
4
sum equals $ 90.
ELEMENTS OF ALGEBRA
Nine
is
as
much below a
13..
11. the first sum exceeds b % of the second sum by
first
(e)
%
of the first plus 5
%
of the second plus 6
%
of the
third
sum
equals $8000.*(/)
(g)
(Ji)
Three years ago the sum of A's and B's ages was 50. (d) In 10 years A will be n years old. express in algebraic symbols
:
-700.
6
%
of m. In 10 years the sum of A's. and C's ages will be 100. B's. and
(a)
(6)
A
If
has $ 5 more than B. of 30 dollars.
the
. etc.
Simplifying. denote the unknown
96.
number by x (or another letter) and express the yiven sentence as an equation. 2.
Uniting. the required
.
3
x
+
16
=
x
x
(x
-
p)
Or. Three times a certain no.
Uniting. In 15 years A will be three times as old as he was 5 years ago. Find A's present age.
Transposing.LINEAR EQUATIONS AND PROBLEMS
63
PROBLEMS LEADING TO SIMPLE EQUATIONS
The simplest kind of problems contain only one unknown number.
3 x or 60 exceeds 40
+ x = 40 + 40.
1.
equation is the sentence written in alyebraic shorthand. The equation can frequently be written by translating the sentence word by word into algebraic symbols in fact.
Let x = the number. = x x
3x
-40
3x
40-
Or. x + 15 = 3 x
3x 16
15.
be 30
. The solution of the equation (jives the value of the unknown number.
-23 =-30.
Check.
NOTE.
In 15 years
10.
number. Write the sentence in algebraic symbols.
6 years ago he was 10
.
but
30
=3
x
years. number of
yards. In order to solve them. be three times as old as he was 5 years ago.
Let x
The
(2)
= A's present age. x = 20.
A
will
Check. 15. verbal statement (1)
(1) In 15 years
A
will
may be expressed in symbols (2).
by 20
40 exceeds 20 by 20.
Ex. x= 15. The student should note that x stands for the number of and similarly in other examples for number of dollars.
. exceeds 40 by as much as 40 exceeds the no.
Transposing.
Dividing.
much
as 40 exceeds the number. 4 x = 80.
3z-40:r:40-z.
x+16 = 3(3-5).
Three times a certain number exceeds 40 by as Find the number.
Ex.
Four times the length of the Suez Canal exceeds 180 miles by twice the length of the canal.
Let x
3.
Find the width of the Brooklyn Bridge.
twice the number plus
7.
What number
7
%
of
350?
Ten times the width of the Brooklyn Bridge exceeds 800 ft.
Find the number whose double increased by 14 equals Find the number whose double exceeds 40 by 10.
3.
Find the number.
Dividing.
EXERCISE
1.
Find the number whose double exceeds 30 by as much
as 24 exceeds the number. A train moving at uniform rate runs in 5 hours 90 miles more than in 2 hours. exceeds the width of the bridge. Find the number.
. 14.
ELEMENTS OF ALGEBRA
56
is
what per cent
of 120 ?
= number
of per cent.
35
What number added
to twice itself gives a
sum
of
39?
44.
4. by as much as 135 ft. then the
problem expressed in symbols
W
or.
Six years hence a
12 years ago.2. How many miles per hour does it run ?
.
300
56.
A will
be three times as old as to-da3r
.64
Ex.
120.
How
old
is
man will be he now ?
twice as old as he was
9. % of
120.
13.
Hence
40
= 46f. How long is the Suez
Canal?
10.
A
number added
number.
Forty years hence
his present age.
Find
8.
to
42 gives a
sum
equal to 7 times the
original
6.
14 50
is
is
4
what per cent of 500 ? % of what number?
is
12.
5.
Uldbe
66
| x x
5(5 is
=
-*-.
47 diminished by three times a certain number equals 2. 11.
statements are given directly. In 1800 the population of Maine equaled that of Vermont. and
B
has $00.LINEAR EQUATIONS AND PROBLEMS
15.
five
If
A
gives
B
$200.
The problem consists of two statements I. while in the more complex probWe denote one of the unknown
x.
x.
numbers (usually the smaller one) by
and use one of the
given verbal statements to express the other unknown number in terms of x.000.
One number exceeds another by
:
and their sum
is
Find the numbers.
times as
much
as A.
1.
B
How
will loses $100.
How many
dol-
A has
A
to
$40.
97. During the following 90 years. then dollars has each ? many
have equal amounts of money.
Ex. Vermont's population increased by 180.
.
How many dollars
must
?
B
give to
18. Maine's population increased by 510.
make A's money
equal to 4 times B's
money
wishes to purchase a farm containing a certain He found one farm which contained 30 acres too many. and Maine had then twice as many inhabitants as Vermont. If A gains A have three times as much
16.000.
A
and
B
have equal amounts of money.
14. The other verbal statement. which gives the value of
8. two verbal statements must be given.
F
8.
is
the equation. One number exceeds the other one by II. Find
the population of Maine in 1800. and another which lacked 25 acres of the required number. and
as
15.
the second one. written in algebraic
symbols. Ill the simpler examples these two
lems they are only implied. B will have lars has A now?
17. If the first farm contained twice as many acres as
A man
number
of acres.
65
A
and
B
$200. The sum of the two numbers is 14. If a problem contains two unknown quantities.
how many
acres did he wish to
buy
?
19.
the smaller number.
Then.
x
3x
4-
and
B
will gain.
terms of the other.
x
x =14
8.
A will lose. the greater number.
o\
(o?-f 8)
Simplifying. B will have twice as
viz.
.
If
A gives
are
:
A
If
II.= The second statement written
the equation ^
smaller number. B will have twice as many as A.
Uniting. unknown quantity
in
Then. = 14.
expressed symbols is (14 x) course to the same answer as the first method. 26 = A's number of marbles after the exchange.
<
Transposing. Let
x
3x
express one
many as A.
= B's number of marbles. I. = 3. = A's number of marbles. A has three times as many marbles as B. 8 the greater number. the sum of the two numbers is 14. although in general the simpler one should be selected.
If
we
select the first one.
.
To
express statement II in algebraic symbols.
Another method for solving this problem is to express one unknown quantity in terms of the other by means of statement II viz.
.
in algebraic
-i
symbols produces
#4a. to
Use the simpler statement. 8 = 11.
x
= 8.
/
.
Dividing.
2x
a?
x
-j-
= 6.
26
= B's number of marbles after the exchange.
has three times as many marbles as B.
Let
x
14
I
the smaller number. consider that by the
exchange
Hence.
+
a-
-f
-f
8
= 14.66
ELEMENTS OF ALGEBRA
Either statement may be used to express one unknown number in terms of the other.
which leads
ot
Ex.
and
Let x
= the
Then x -+. 2. 25 marbles to B.
Statement
x
in
=
the larger number.
The two statements
I. A gives B 25 marbles.
Uniting.$3. 15 + 25 = 40.
by 44. 40 x .
Selecting the cent as the denomination (in order to avoid fractions).
dollars
and dimes
is
$3. 6 half dollars = 260 cents. Eleven coins.10.
The numbers which appear
in the equation should
always
be expressed in the same denomination. x = the number of half dollars.
6 dimes
= 60
= 310.
Find
the numbers.
Simplifying.
1.
50(11 660 50 x
-)+ 10 x = 310. have a value of $3. B's number of marbles. and the Find the numbers.
60.
*
98. their sum
+ +
10 x 10 x
is
EXERCISE
36
is five
v
v.
67
x
-f
25 25
Transposing. Check. then. The number of coins II.25 = 20. the number of dimes. cents. A's number of marbles.
Uniting. Dividing. consisting of half dollars and dimes.
Check.
50 x
Transposing.10. but 40 = 2 x 20.
(Statement II)
Qx
.
The sum of two numbers is 42. 11 x = 5.
x
from
I.
50. x = 15.
2. The value of the half
:
is 11.
Simplifying.
Never add the number number of yards to their
Ex.
x x
+
= 2(3 x = 6x
25
25). x = 6.75.
.LINEAR EQUATIONS AND PROBLEMS
Therefore.
*
'
.
Two numbers
the smaller.
is 70. Find the numbers. 3..
6 times the smaller. the
price. etc.
w'3.550 -f 310.5 x . of dollars to the number of cents. 3 x = 45.
Let
11
= the number of dimes. How many are there of each ?
The two statements are I.
differ
differ
and the greater and their sum
times
Two numbers
by
60.
greater
is
. the number of half dollars.240.. 45 . .
Dividing.
we
express the statement II in algebraic symbols.10..
?
Two
vessels contain together 9 pints. and four times the former equals five times the latter.
Find
Find two consecutive numbers whose sum equals 157.
7.
What
is
the altitude of each
mountain
12.000
feet. How many volcanoes are
in the
8. Mount Everest is 9000 feet higher than Mt. and twice the altitude of Mt.. the larger part exceeds five times the smaller part by 15 inches.
tnree times the smaller by 65.
How many
14 years older than B. How many inches are in each part ?
15. one of which increased by
9.
as the larger one.
ELEMENTS OF ALGEBRA
One number
is
six
times another number.
11. What are their ages ?
is
A A
much
line 60 inches long is divided into two parts. the number.
How many
hours does the day last ?
.
and in Mexico
?
A
cubic foot of aluminum. McKinley.
2 cubic feet of iron weigh 1600 foot of each substance.
and twice the greater exceeds Find the numbers. McKinley exceeds the altitude of
Mt.
6.
A's age is four times B's. the night in
Copenhagen
lasts 10 hours
longer than the day.68
4.
On December
21.
5. Find their ages.
3 shall be equal to the other increased by
10.
of volcanoes in
Mexico exceeds the number
of volcanoes in the United States by 2.
United
States. Twice 14. and B's age is as below 30 as A's age is above 40.
find the
weight of a cubic
Divide 20 into two parts.
cubic foot of iron weighs three times as much as a If 4 cubic feet of aluminum and
Ibs. and the
greater increased by five times the smaller equals 22.
would contain three times as
pints does each contain ?
much
13.
9. Everest by 11.
Two numbers
The number
differ
by
39.
it
If the smaller
one contained 11 pints more. and in 5 years A's age will be three times B's.
and the other
of x
problem contains three unknown quantities. = 48. then three times the sum of A's and B's money would exceed C's money by as much as A had originally.
has.
Let
x
II.
sum of A's and B's money would exceed much as A had originally.
first
According to
3 x
number
number
and according
to
80
4
x
=
the
express statement III by algebraical symbols. If A and B each gave $5 to C.
A
and B each gave $ 5
respectively.
. or 66 exceeds 58 by 8.
has. and B has three as A.
1. = number of dollars B had after giving $5. and C together have $80. let us consider the words ** if A and B each gave $ 5 to C. 19.
number
had. If A and B each gave $5 to C.
Ex.
If
4x
= 24.
69
If a
verbal statements must be given. 4 x = number of dollars C had after receiving $10. The solution gives
:
3x
80
Check. then
three times the
money by
I. and 68. III. x = 8.LINEAR EQUATIONS AND PROBLEMS
99. B.
the
the
number
of dollars of dollars of dollars
A
B
C
has. B. The third
verbal statement produces the equation.
5
5
Expressing in symbols Three times the sum of A's and B's money exceeds C's money by A's 3 x ( x _5 + 3z-5) (90-4z) = x."
To
x
8x
90
= number of dollars A had after giving $5.
8(8
+ 19)
to C. bers is denoted by x. number of dollars A had. II.
number
of dollars of dollars
B
C
had. they would have 3.
I.
times as
much
as
A. and C together have $80. three One of the unknown num-
two are expressed in terms by means of two of the verbal statements.
are
:
C's
The three statements
A. original amount. try to obtain
it
by a
series of successive steps. B has three times as much as A. Tf it should be difficult to express the selected verbal state-
ment
directly in algebraical symbols.
4 x -f 8 = 28.
III. and each sheep $ 15. according to III.
85 (x 15 (4 x
I
+ 4)
+
8)
= the number of sheep.
The
I. 90 x -f 35 x + GO x = 140 20 + 1185. 185 a = 925.
37
Find three numbers such that the second is twice the first. each cow $ 35. = the number of dollars spent for horses. the third five times the first. each horse costing $ 90. The number of sheep is equal to twice tho number of horses and
x 4
the
cows together. cows.
28 x 15 or 450
5 horses. 2 (2 x -f 4) or 4 x
Therefore. 28 2 (9 5).
+ 35 (x +-4) -f 15(4z-f 8) = 1185.
Dividing.
first
the third exceeds the second by and third is 20. sheep. = the number of dollars spent for cows. 9 -5 = 4 .70
ELEMENTS OF ALGEBRA
man spent $1185 in buying horses.
A
and the number of sheep was twice as large as the number How many animals of each kind did he buy ?
of horses and cows together.
first. = the number of dollars spent for sheep
Hence statement
90 x
Simplifying. and 28 sheep would cost 6 x 90 -f 9 + 316 420 = 1185. 90
may
be written. 2. 9 cows. number of sheep. x -f 4 = 9.
+
8
90 x
and.
The total cost equals $1185.
Find three numbers such that the second is twice the 2. + 35 x 4. number of cows.140 + (50 x x 120 = 185. x = 5.
x 35
-f
+
=
+
EXERCISE
1. and the difference between the third and the second is 15
2. number of horses.
and. and the sum of the
.
three statements are
:
IT.
Uniting. according to II. The number of cows exceeds the number of horses by 4. number of cows. and Ex. Let
then.
x
Transposing. The number of cows exceeded the number of horses by
4.
1 1
Check.
x
-j-
= the
number of horses.
and the third exceeds the
is
second by
5. increased by three times the second side.
is five
numbers such that the sum of the first two times the first. If twice
The sum
the third side.
twice the
6.
-
4.
the third
2. how many children
were present ?
x
11. men.
New York
delphia.000.000 more inhabitants than Philaand Berlin has 1.
first.
9. If the population of New York is twice that of Berlin. and children together was 37. the copper.
A
12. and 2 more men than women.
The
three angles of any triangle are together equal to
180.LINEAR EQUATIONS AND PROBLEMS
3.
and
of the three sides of a triangle is 28 inches.
If the second angle of a triangle is 20 larger than the and the third is 20 more than the sum of the second and
first.
first.
"Find three
is 4. and the sum of the first and third is 36. and is 5 years younger than sum of B's and C's ages was 25 years.
7.
A
is
Five years ago the What are their ages ?
C. equals 49 inches.
twice as old as B. the second one is one inch longer than the first.
what are the three angles ?
10. what is the length of each?
has 3.
13. In a room there were three times as many children as If the number of women.
Find three consecutive numbers whose sum equals
63.
v
.000.
the
first
Find three consecutive numbers such that the sum of and twice the last equals 22.
v
-
Divide 25 into three parts such that the second part first.000 more than Philadelphia (Census 1905). women. The gold. and the pig iron produced in one year (1906) in the United States represented together a value
.
71
the
Find three numbers such that the second is 4 less than the third is three times the second. what is the population of each city ?
8. and the third part exceeds the second by 10.
Since in uniform motion the distance is always the product of
rate
and time. statement "A and B walk from two towns 27 miles apart until they meet " means the sum of the distances walked by A and B equals 27 miles.
3x
+
4 (x
2)
=
27. it is frequently advantageous to arrange the quantities in a systematic manner.g.000. or time.000. number of miles A
x
x
walks.
start at the same hour from two towns 27 miles walks at the rate of 4 miles per hour.
= 35.
The copper had twice
the value of the gold. After how many hours will they meet and how
E. and 4 (x But the 2) for the last column.72
of
ELEMENTS OF ALGEBRA
$ 750. i. together.
California has twice as
many
electoral votes as Colorado. but stops 2 hours on the way.
B
many
miles does
A
walk
?
Explanation. such as length. = 5. Let x = number of hours A walks. First fill in all the numbers given directly.e. speed.
. width. and distance. we obtain 3 a.
Find the value of each.
how many
100. and quantities
area.
7
Uniting.
14. number of hours. and A walks at the rate of 3 miles per hour without stopping.
3z + 4a:-8 = 27. of 3 or 4 different kinds.000. 3 and 4.
A and B
apart.
has each state
?
If the example contains Arrangement of Problems. 8 x = 15.000 more than that
the copper.
Dividing. then x 2 = number of hours B walks.
and Massachusetts has one more than California and Colorado If the three states together have 31 electoral votes.
Hence
Simplifying.
of
arid the value of the iron
was $300.
3. A man bought 6 Ibs. and a second sum. but four men failed to pay their shares.
If the silk cost three times as
For a part he 7.55.
A
If its length
rectangular field is 2 yards longer than it is wide. How many pounds of each kind did he buy ?
8. paid 24 ^ per pound and for the rest he paid 35 ^ per pound. of coffee for $ 1. and its width decreased
by 2 yards. and in order to raise the required sum each of the remaining men had to pay one dollar more. and follows on horseback traveling at the rate of 5 miles per hour. How much did each man subscribe ?
sum
walking at the rate of 3 miles per hour.
1.74
ELEMENTS OF ALGEBRA
EXERCISE
38
rectangular field is 10 yards and another 12 yards wide. were increased by 3 yards.
Find the dimen-
A
certain
sum
invested at 5
%
%. twice as large.
A
sets out later
two hours
B
.
sions of the field.
2. how much did each cost per yard ?
6.
sum $ 50
larger invested at 4
brings the same interest Find the first sum.
Ten yards
$
42. and how far will each then have traveled ?
9. but as two of them were unable to pay their share. What are the
two sums
5. and the sum Find the length of their areas is equal to 390 square yards. and the cost
of silk
of the auto-
and 30 yards of cloth cost together much per yard as the cloth. After how many hours will B overtake A.
mobile.
A
of each.
as a
4. The second is 5 yards longer than the first. together bring $ 78 interest.
A sum
?
invested at 4 %.
Twenty men subscribed equal amounts
of
to raise a certain
money.
invested at 5 %. the area would remain the same.
Six persons bought an automobile. each of the others had to pay
$ 100 more.
Find the share of each.
how must B walk before he overtakes A ?
walking at the rate of 3 miles per hour.will they be 36 miles apart ?
11. Albany and travels toward New York at the rate of 30 miles per hour without stopping. After how many hours.
The
distance from
If a train starts at
. and another train starts at the same time from New York traveling at the rate of 41 miles an hour.
A
sets out
two hours
later
B
starts
New York to Albany is 142 miles. and B at the rate of 3 miles per hour.LINEAR EQUATIONS AND PROBLEMS
v
75
10.
A
and
B
set out
direction. traveling by coach in the opposite direction at the rate of 6 miles per hour. how many miles from New York will they meet?
X
12.
walking at the same time in the same If A walks at the rate
of 2
far
miles per hour. and from the same point. but
A has
a start of 2 miles.
we
shall not. a.
irrational. if this letter does not occur in any denominator.
The
factors of
an algebraic expression are the quantities
will give the expression. consider
105. An expression is integral with respect to a letter.
which multiplied together
are considered factors.
The prime
factors of 10 a*b are 2. as. but fractional with respect
103.
J Although Va'
In the present chapter only integral and rational expressions
b~
X
V
<2
Ir
a2
b'
2
2
?>
. if it is integral to all letters contained in it.
it is
composite.
An
expression
is integral
and rational with respect
and rational. if it does contain
some indicated root of
.
76
. at this
6
2
.
An
after simplifying.
a.
a factor of a 2
A
factor is said to be prime.CHAPTER
VI
FACTORING
101.
104. 6.
+ 62
is
integral with respect to a.
this letter. if it contains
no other
factors (except itself
and unity)
otherwise
. if.
a2
to 6. 5.
stage of the work.
-f-
db
6
to b.
vV
.
a-
+
2 ab
+ 4 c2
.
expression is rational with respect to a letter. it contains no indicated root of this letter
.
\-
V&
is
a
rational with respect to
and
irrational with respect
102.
109. 2.3 6a + 1). .
77
Factoring
is
into its factors. since (a + 6) (a 2 IP factored.
it
follows
that a 2
.g.
?/.
An
the process of separating an expression expression is factored if written in the
form of a product.
Hence
6 aty 2
= divisor x quotient.
1.62 + &)(a 2
. x. or that a
=
6)
(a
= a .9 x if + 12 xy\
2
The
greatest factor
common
2
to all terms
flcy*
is
8
2
xy'
.
Since factoring
the inverse of multiplication.
factors of 12
&V
is
are 3.
Factor G ofy 2
.
in the
form
4)
+3.
01.
It (a.
107.
2.
55.62
can be
&).
110.
y.
TYPE
I. for this result is a sum.3 sy + 4 y8).
x. 2.
Divide
6
a% .
POLYNOMIALS ALL OF WHOSE TERMS CONTAIN A COMMON FACTOR
(
mx + my+ mz~m(x+y + z).9 x2 y 8 + 12
3 xy
-f
by
3
xy\
and the quotient
But.)
Ex.
E.
Ex. dividend
is
2 x2
4
2
1/
.
.
or
Factoring examples may be checked by multiplication by numerical substitution.
The factors
of a
monomial can be obtained by inspection
2
The prime
108.
2 4 x + 3) is factored if written (x' would not be factored if written x(x and not a product.FACTORING
106.
it fol-
lows that every method of multiplication will produce a method
of factoring. 8) (s-1).9 x2^ + 12 sy* = 3 Z2/2 (2 #2 .
Factor
14 a*
W-
21 a 2 6 4 c2
+ 7 a2 6
2
c2
7
a2 6 2 c 2 (2 a 2
.
Hence z6 -? oty+12 if= (x -3 y)(x*-4 y ). Factor x? .a). can be factored. If q is negative.
77 as the product of 1 77.
determine whether
In solving any factoring example.4 .
.
or
77
l.G) = . of this type.1 1 a
tf
a 4.
but of these only
a:
Hence
2
. the two numbers have both the same sign as p.
Factor a2
. and (a .
tfa2 -
3.1 afy 8 The two numbers whose product is equal to 12 yp and whose sum equals 3 8 7 y are -4 y* and -3 y*.
4.
is
The two numbers whose product and -6.
Ex.FACTORING
Ex.
79
Factor a2
-4 x .11.
Ex.11 a
2
.6 = 20.. Hence fc -f 10 ax
is
10 a are 11 a
-
12 /. but only in a limited number
of ways as a product of two numbers.
. however.11 a + 30.11) (a
+
7).
11 a2 and whose sum The numbers whose product is and a.
11
7. or 11 and 7 have a sum equal to 4.30 = (a .
2
6.
3. the two numbers
have opposite
signs.
+
112.
and the greater one has the same sign
Not every trinomial
Ex. a 2 .
We may consider
1.4 x .
as p.
Factor
+ 10 ax . + 30 = 20.
Therefore
Check. it is advisable to consider the factors of q first.
If
30 and whose
sum
is
11 are
5
a2
11
a = 1.
EXERCISE
Besolve into prime factors :
40
4.5) (a .
m -5m + 6.
2. If q is positive.77 =
(a.
Since a number can be represented in an infinite number of ways as the sum of two numbers.
5.5) (a 6). 2 11 a?=(x + 11 a) (a. or 7 11.
. the student should first all terms contain a common monomial factor.
31 x
Evidently the
last
2
V A
6. 27 x 2. Hence only 1 x 54 and 2 x 27 need
be considered.
Ex.17 x
2o?-l
V A
5
-
13 a
combination
the correct one.
the
If p and r are positive.e-5
V A
x-1
3xl \/ /\
is
3
a.83 x
-f-
54.FACTORING
If
81
we consider that the
factors of -f 5
as
must have
is
:
like signs. 64 may be considered the
:
product of the following combinations of numbers 1 x 54.1).
all
it is not always necessary to write down combinations. 6 x 9.
The
and
factors of the first term consist of one pair only. the second terms of the factors have same sign as q.
The work may be shortened by the
:
follow-
ing considerations
1. X x 18. or
G
114.
and that they must be negative. exchange the
signs of the second terms of the factors. we have to reject every combination of factors of 54 whose first factor contains a 3.
Factor 3 x 2
. 18 x 3. 54 x 1. and r is negative. Since the first term of the first factor (3 x) contains a 3. 2 x 27.
sible
13 x
negative. and after a little practice the student possible should be able to find the proper factors of simple trinomials
In actual work
at the first trial. 3 x and x.
If
the factors
a combination should give a sum of cross products. which has the same absolute value as the term qx.
a. all pos-
combinations are contained in the following
6x-l
x-5 .
11 x
2x. 9 x 6.5) (2 x .5 . If p is poxiliw. none of the binomial factors can contain a monomial factor. but the opposite sign. then the second terms of
have opposite signs. viz.
. 2. If py? -\-qx-\-r does not contain any monomial factor.
3. the signs of the second terms are minus.
.13 x + 5 = (3 x .
of a 7 and a e b 7
. of the algebraic expressions. C.
EXERCISE
Find the H. F.
6.
5
7
34 2s
.
54
-
32
.
the algebraic factor of highest degree common expressions to these expressions thus a 6 is the II.
3. C. of a 4 and a 2 b is a2
The H. 12 tfifz.
+
8
ft)
and
cfiW is
2
a 2 /) 2
ft)
. of
two or more monomials whose factors
.
8
. F.
15
aW.
F.CHAPTER
VII
HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE
HIGHEST COMMON FACTOR
120.
The
highest
is
common
factor (IT. of
:
48
4. C. F. F. C. The H.
3
. The H.
121.
II
2
.
2.
2
2
. F.
The student should note
H. aW.
C. F.
5.
-
23 3
.
C.
33
2
7
3
22 3 2
.
.
122.) of
two or more
.
expressions which have no are prime to one another.
13 aty
39 afyV.
Two
common
factor except unity
The H. C.
5
2
3
.
25
W. F. F. F.
89
. is the lowest
that the power of each factor in the power in which that factor occurs in any
of the given expressions. C. and GO aty 8 is 6 aty.
5
s
7
2
5. find by arithmetic the greatest common factor of the coefficients. Thus the H.
are prime can be found by inspection. of
aW. If the expressions have numerical coefficients.
24
s
. and prefix it as a coefficient to H.
C. C. of 6 sfyz. of (a
and (a
+
fc)
(a
4
is
(a
+ 6)
2
.
C.
&)
2
M. of several expressions which are not completely factored.
Ex.
of 12(a
+
ft)
and (a
+ &)*( -
is
12(a
+ &)( . C. two lowest common multiples.C.
126. C.
etc. L.
M.
2 multiples of 3 x
and 6 y are 30 xz y. Obviously the power of each factor in the L.
Hence the L. which
also
signs. find by arithmetic their least common multiple and prefix it as a coefficient to the L.
300 z 2 y.
.
2
The The
L.M.
The
L. M. of as -&2 a2 + 2a&-f b\ and 6-a.
A
common
remainder. M. M.
C. ory is the L.
127. 60
x^y'
2
. C. of the
general.LOWEST COMMON MULTIPLE
91
LOWEST COMMON MULTIPLE
multiple of two or more expressions is an which can be divided by each of them without a expression
124.
M. M. is equal to the highest power in which it occurs in any of the
given expressions.
Ex. but opposite
. To find the L.
of 4 a 2 6 2 and 4 a 4
-4 a 68
2
. of tfy and xy*.
NOTE.
1.
Common
125. C. C. M.
M of the algebraic expressions.6 3 ).
= (a -f
last
2
&)'
is
(a
-
6) .M.6)2.(a + &) 2 (a
have the same absolute value.
=4 a2 62 (a2 . If the expressions have a numerical coefficient. C. thus.
4 a 2 &2
_
Hence.
The
lowest
common
multiple (L.
6
c6 is
C a*b*c*.
a^c8
3
. L.
of 3
aW. C.
128.
Find the L.
M.
2. C.
Find the L.
. each set of expressions has
In example ft). resolve each expression into prime factors and apply the method for monomials.) of
two or more
expressions is the common multiple of lowest degree. C.
If both terms of a fraction are multiplied or divided by the same number) the value of the fraction is not altered. the value of a fraction is not altered by multiplying or dividing both its numerator and its denominator by the same number. only positive integral numerators shall assume that the
all
arithmetic principles are generally true for
algebraic numbers. thus -
is identical
with a
divisor b the denominator.
131.
common
6
2
divisors of
numerator and denomina-
and z 8
(or divide the terms
.
and
i
x mx = my y
terms
A
1.CHAPTER
VIII
FRACTIONS
REDUCTION OF FRACTIONS
129. F.
and denominators are considered.
rni
Thus
132.
a b
= ma
mb
.
All operations with fractions in algebra are identical
with the corresponding operations in arithmetic.
an indicated quotient. Thus.
Remove
tor.ry ^
by
their H.
TT
Hence
24
2 z = --
3x
.
the product of two fractions is the product of their numerators divided by the product of their denominators.
A
-f-
fraction is
b. etc. as 8. however.
successively all
2
j/' .
a?.
Reduce
~-
to its lowest terms.
Ex.
130. C. but we
In arithmetic.
The dividend a is called the numerator and the The numerator and the denominator
are the terms of the fraction.
fraction
is
in its lowest
when
its
numerator
and
its
denominator have no
common
factors.
C.
2>
. we have
(a
+ 3) (a -8) (-!)'
NOTE. by the denominator of each fraction.C.
^
to their lowest
com-
The
L.
Reduce -^-.
Ex
-
Reduce
to their lowest
common
denominator.
M. we may use the same process as in arithmetic for reducing fractions to the lowest
common
denominator. C. and
135.
Since a
(z
-6 + 3)(s-3)O-l)'
6a. =(z
(x
+ 3)(z.
3 a\ and 4
aW
is
12 afo 2 x2 .M.M.~16
(a
+ 3) (x. C.
multiply each quotient by the corresponding numerator. and 6rar 3 a? kalr
.
we may extend this method
to integral expressions.
To reduce
to a fraction with the
denominator 12 a3 6 2 x2 numerator
^lA^L O r 2 a 3
'
and denominator must be multiplied by
Similarly.
we have
the quotients (x
1).
1).D.
.
-
of
//-*
2
.r
2
2
.96
134.
and the terms of
***.
by any quantity without altering the value of the fraction. of the denominators for the common denominator.
and
Tb reduce fractions to their lowest common denominator.
we have
-M^.3) (-!)'
=
.
mon
T denominator.
Ex.
-
by 4
6' .3)O -
Dividing this by each denominator.
ELEMENTS OF 'ALGEBRA
Reduction of fractions to equal fractions of lowest common Since the terms of a fraction may be multiplied
denominator.
TheL.
and
(a-
8).-1^22
' . Divide the L.
Multiplying these quotients by the corresponding numerators and writing the results over the common denominator.
.
1.
+
3). multiplying the terms of
22
.by 3 ^
A
2
' . take the L.
.
we may extend any
e.
2
a
Ex.102
ELEMENTS OF ALGEBRA
MULTIPLICATION OF FRACTIONS
140. and the product of the denominators for the
denominator.
integer.
2.
fractions to integral numbers.
Since -
= a. each
numerator and denomi-
nator has to be factored.)
Ex.
Simplify 1 J
The
expreeaion
=8
6
. Fractions are multiplied by taking the product of tht numerators for the numerator. multiply the
142. Common factors in the numerators and the denominators should be canceled before performing the multiplication.
-x
b
c
=
numerator by
To multiply a fraction by an
that integer.
expressed in symbols:
c
a
_ac b'd~bd'
principle proved for
b
141.
or.
!.
F J Simplify
. (In
order to cancel
common
factors.g.
and the principle of division follows
may
be expressed as
145.
1.104
ELEMENTS OF ALGEBRA
DIVISION OF FRACTIONS
143.
144.
.
8
multiply
the
Ex.
The The
reciprocal of a
is
a
1
-f-
reciprocal of J
is
|
|. :
a 4-1
a-b
* See page 272.y3
+
xy*
x*y~ -f y
8
y
-f
3
2/
x3
EXERCISE 56*
Simplify the following expressions
2
x*
'""*'-*'
:
om
2 a2 6 2
r -
3
i_L#_-i-17
ar
J
13 a& 2
5
ft2
'
u2
+a
.
The reciprocal of ?
Hence the
:
+*
x
is
1
+ + * = _*_. To divide an expression by a fraction. expression by the reciprocal of the fraction. To divide an expression by a fraction. Integral or mixed divisors should be expressed in fractional form before dividing.
Divide X-n?/
.
* x* -f xy 2
by
x*y
+y
x'
2
3
s^jf\ =
x'
2
x*
. invert the divisor and multiply it by the dividend.
The
reciprocal of a
number
is
the quotient obtained by
dividing 1
by that number. x a + b
obtained by inverting
reciprocal of a fraction
is
the fraction.
.
x
Or
Uniting.
and
12
= the number
over.
then
= 2 TT#. hence the question would be formulated After how many minutes has the minute hand moved 15 spaces more than the hour hand ?
Let then
x x
= the required number of minutes after 3 o'clock. When between 3 and 4 o'clock are the hands of
a clock together
?
is
At
3 o'clock the hour hand
15 minute spaces ahead of the minute
:
hand.114
35.
12.
A would do
each day ^ and
B
j.
Find
R in terms of C and
TT. = 16^. 2 3
.
. 2.
~^ = 15
11 x
'
!i^=15.
Multiplying by
Dividing.
100
C. 1. Ex.
is
36.
days by x and the piece of work while in x days they would do
respectively
ff
~ and and hence the sentence written in algebraic symbols ^. A can do a piece of work in 3 days and B in 2 days.
ELEMENTS OF ALGEBRA
(a) Find a formula expressing degrees of Fahrenheit terms of degrees of centigrade (<7) by solving the equation
(F)
in
(ft)
Express in degrees Fahrenheit 40
If
C. = the number of minute spaces the minute hand moves
over.
C
is
the circumference of a circle whose radius
R.
PROBLEMS LEADING TO FRACTIONAL AND LITERAL EQUATIONS 152.
of minute spaces the
hour hand moves
Therefore x
~ = the number of minute spaces the minute hand moves
more than the hour hand.20
C.
.180. In how many days can both do it working together ?
If
we denote
then
/-
the required
number
by
1.minutes after x=
^
of
3 o'clock.
Ex..
= 100 + 4 x. what is
the rate of the express train
?
180
Therefore.
or 1J. then
Ox
j
5
a
Rate Hence the rates can be expressed. hours more than the express train to travel 180 miles. the required
number
of days. The speed of an express train is $ of the speed of an If the accommodation train needs 4 accommodation train.
180
Transposing.
Solving.
32
x
= |."
:
Let
x -
= the
required
number
of days.
fx
xx*
=
152
+4
(1)
Hence
=
36
= rate
of express train.
in
Then
Therefore.
Explanation
:
If
x
is
the rate of the accommodation train.FRACTIONAL AND LITERAL EQUATIONS
A
in symbols the following sentence
115
more symmetrical but very similar equation is obtained by writing ** The work done by A in one day plus the work done by B in one day equals the work done by both in one day.
the rate of the express train.
But
in
uniform motion Time
=
Distance
.
Clearing. 4x = 80.
Ex.
= the
x
part of the
work both do
one day. 3." gives the equation /I). and the statement. u The accommodation train needs 4 hours more than the express train.
The sum
10 years hence the son's age will be
of the ages of a father and his son is 50.
ceeds the smaller by
4.
its
Find the number whose fourth part exceeds part by 3.
of his present age.
length in the ground. How much money had he at first?
12
left
After spending ^ of his
^ of his money and $15.
and 9
feet above water. and one half the greater Find the numbers.
J-
of the greater
increased by ^ of the smaller equals
6.
by 6. -|
Find their present ages.
Twenty years ago A's age was |
age. to his daughand the remainder.
is oO.
Find a number whose third and fourth parts added
together
2.
money and $10.
Two numbers
differ
l to s of the smaller.
make
21. to his son. which was $4000. and J of the greater Find the numbers. A man lost f of his fortune and $500.
by 3.
Find two consecutive numbers such that
9.
is
equal
7.116
ELEMENTS OF ALGEBRA
EXERCISE
60
1.
fifth
Two numbers
differ
2.
are the
The sum of two numbers numbers ?
and one
is
^ of the other. How
did the
much money
man
leave ?
11. a man had How much money had he
at first?
.
Find A's
8.
ex-
What
5. and found that he had \ of his original fortune left.
3.
9
its
A
post
is
a fifth of
its
length in water.
A man left ^ of his property to his wife. and of the father's age. one half of What is the length
of the post ?
10
ter.
At what time between 4 and
(
5 o'clock are the hands of
a clock together?
16.
At what time between 7 and
8 o'clock are the hands of
?
a clock in a straight line and opposite
18.
?
In
how many days can both do
working together
23.
152. ounces of gold and silver are there in a mixed mass weighing
20 ounces in
21.
air.FRACTIONAL AND LITERAL EQUATIONS
13. ^ at 5%.
investments. what is the
14. An ounce of gold when weighed in water loses -fa of an How many ounce. and B in 4 days.
at 4J % and P> has invested $ 5000 They both derive the same income from their How much money has each invested ?
20. 3. and B In how many days can both do it working together
in
?
12 days. Ex. If the rate of the express train is -f of the rate of the accommodation train. 1.) (
An express train starts from a certain station two hours an accommodation train.
. Ex. In how many days can both do it working together ? ( 152. and has he invested if
his animal interest therefrom is
19.
117
The speed
of an accommodation train
is
f of the speed
of an express train.
A can
A
can do a piece of work in 2 days. and after traveling 150 miles overtakes the accommodation train. A can do a piece of work in 4 clays.
and losing
1-*-
ounces when weighed in water?
do a piece of work in 3 days.)
At what time between 7 and 8
o'clock are the
hands of
a clock together ?
17. what is the rate of the express train? 152.
A man
has invested
J-
of his
money
at
the remainder at
6%. Ex. and
it
B in 6 days.
A has invested capital
at
more
4%.
after
rate of the latter ?
15.)
22. 2. If the accommodation train needs 1 hour more than the express train to travel 120 miles. and an ounce of silver -fa of an ounce.
How much money
$500?
4%.
.
ELEMENTS OF ALGEBRA
The
last three questions
and their solutions differ only two given numbers.g. and n = 3. Find the numbers if m = 24 30. and apply the
method of
170. n x
Solving. 3. A in 6. The problem to be solved.= -.
6
I
3
Solve the following problems
24. .
. 2.414. Find three consecutive numbers whose sum equals m.
they can both do
in 2 days. B in 16.
26. Hence. B in 30.
25.e. by taking for these numerical values two general algebraic numbers. In how
in the numerical values of the
:
many days
If
can both do
we
let
x
= the
it working together ? required number of days. Ex.
To
and
find the numerical answer. Then
ft
i. if
B
in 3 days. A in 6. it is possible to solve all examples of this type by one example. Answers to numerical questions of this kind may then be found by numerical substitution. e. therefore.009 918. is 57.=
m
-f-
n
it
Therefore both working together can do
in
mn
-f-
n
days.
:
In
how many days
if
can
A
and
it
B
working together do a
piece of
work
each alone can do
(a)
(6)
(c)
in the following
number
ofdavs:
(d)
A in 5.
is 42. A in 4. is A can do a piece of work in m days and B in n days.118
153.
make
it
m
6
A can do this work in 6 days Q = 2. B in 12.
B in 5.
we
obtain the equation
m m
-. m and n.
Find three consecutive numbers whose sum
Find three consecutive numbers whose sum
last
:
The
two examples are
special cases of the following
problem 27.
If each side of a square were increased by 1 foot.
119
Find two consecutive numbers the difference of whose
is 11.
squares
30.
The
one:
31.
. 88 one traveling 3 miles per hour. d miles the first traveling at the rate of m. respectively (a) 60 miles.
is ?n
. 2 miles per hour. (b) 8 and 56 minutes. (b) 149. respectively.000.
33.
squares
29. After how many hours do they rate of n miles per hour. Find the side of the square.001.
34. two pipes together ? Find the numerical answer.721.
is (a)
51. and how many miles does each travel ?
32. solve the following ones Find two consecutive numbers the difference of whose squares
:
find the smaller number.
same hour from two towns.
Find two consecutive numbers -the difference of whose
is 21. (b) 35 miles. (c) 16. the rate of the
first. (a) 20 and 5 minutes. the second at the apart. 3 miles per hour.
last three
examples are special cases of the following
The
difference of the squares of
two consecutive numbers
By using the result of this problem.
and
the rate of the second are.
meet. 5 miles per hour.
Two men
start at the
first
miles
apart. After how many hours do they meet. and the second 5 miles per hour. 3J miles per hour. and how many miles does each travel ? Solve the problem if the distance. 2 miles per hour.FRACTIONAL AND LITERAL EQUATIONS
28. the area would be increased by 19 square feet.
by two pipes in m and n minutes In how many minutes can it be filled by the respectively.
:
(c)
64 miles. the
Two men start at the same time from two towns. (d) 1. if m and n are.
4J-
miles per hour.
A cistern can
be
filled
(c)
6 and 3 hours.
Thus the
written a
:
ratio of a
b
is
.
A
ratio
is
used to compare the magnitude of two
is
numbers.
1. a ratio
is
not changed
etc.
term of a ratio
a
the
is
is
the antecedent.
The
first
156.
is
numerator of any fraction
consequent. all principles
relating
to
fractions
if its
may
be af)plied to ratios.) The ratio of 12 3 equals 4.or a *
b
The
ratio is also frequently
(In most European countries this symbol is employed as the usual sign of division. 158.
Ex.
b
is
a
Since a ratio
a
fraction. b is the consequent. etc.
The
ratio -
is
the inverse of the ratio -.
" a Thus.
antecedent.
In the ratio a
:
ft.
:
:
155.
b.
b.
.
terms are multiplied or divided by the same number.
:
A somewhat shorter way
would be to multiply each term by
120
6.
the symbol
being a sign of division. 6 12 = .
Simplify the ratio 21 3|. the antecedent.
The
ratio of
first
dividing the
two numbers number by the
and
:
is
the quotient obtained by
second.
E. instead of writing
6 times as large as
?>.5.CHAPTER X
RATIO AND PROPORTION
11ATTO
154. the denominator
The
the
157."
we may
write
a
:
b
= 6. the second
term the consequent.g.
16a2 :24a&.
Transform the following
unity
15.
b is the
mean
b.
9.
proportional between a
and
c.
term
is
the fourth proportional to the
:
In the proportion a b = c c?.
:
ratios so that the antecedents equal
16:64.
:
a-y
. and the last term the third proportional to the first and second
161.
1.
10.
:
is
If the means of a proportion are equal.
terms.
17.
62:16.
and
c
is
the third proportional to a and
.
8^-
hours.
61
:
ratios
72:18.
6. b and c the means.
In the proportion a b
:
=
b
:
c.
two
|
ratios.
7f:6J. the second
and fourth terms of a proportion are the and third terms are the means.
3
8.
= |or:6=c:(Z are
The
first
160.
A
proportion
is
a statement expressing the equality of
proportions.
3:4.
7|:4 T T
4
.
4|-:5f
:
5.
12. The last term d is the fourth proportional to a.
4. b.
159. and c.
5 f hours
:
2.
equal
2.
J:l.
$24: $8. a and d are the extremes.
11.
27 06: 18 a6. either mean the mean proportional between the first and the last terms.RATIO
Ex.
extremes.
18.
AND PROPORTION
ratio 5
5
:
121
first
Transform the
3J so that the
term will
33
:
*~5
~
3
'4*
5
EXERCISE
Find the value of the following
1.
16.
:
1.
16 x*y
64 x*y
:
24 48
xif. The last
first three.
3:1}.
Simplify the following ratios
7.
3.
3 4. Hence the number of men required to do some work. then 8 men can do it in 3 days.
164.
ad =
be. if the ratio of any two of the first kind is equal \o the inverse ratio of the corresponding two of
the other kind.
If
(Converse of
nq. In any proportion product of the extremes. is equal to the ratio of the corresponding two
of the other kind.
t/ie
product of the means
b
is
equal
to the
Let
a
:
=c
:
d.
q~~ n
. " we " NOTE.
pro-
portional. 6 ccm. of a proportion.
!-. of iron weigh . are
: : :
inversely proportional. if the ratio of any two of the first kind.
Clearing of fractions.e.
ccm. i.122
162.
If 6 men can do a piece of work in 4 days.
briefly.
ELEMENTS OF ALGEBRA
Quantities of one kind are said to be directly proper
tional to quantities of another kind. of iron weigh 45 grams.) b = Vac.
163. then G ccm.
The mean proportional
of their product. and we
divide both
members by
we have
?^~ E. = 30 grams 45 grams.__(163. or 8 equals the inverse ratio of 4 3.'*
Quantities of one kind are said to be inversely proportional to quantities of another kind.
163. If the product of two numbers is equal to the product of two other numbers^ either pair may be made the means. a b
:
bettveen two
numbers
is
equal to
the square root
Let the proportion be
Then Hence
6
=b = ac.
:
:
directly proportional
may say. Hence the weight of a mass of iron is proportional to its volume. and the
other pair the extremes.30 grams.
:
c.)
mn = pq. and the time necessary to do it. Instead of u
If 4
or 4 ccm.
2
165.
and
the time necessary for it.
ELEMENTS OF ALGEBEA
State the following propositions as proportions : T (7 and T) of equal altitudes are to each.126
54.
56.
(b)
The time a
The length
train needs to travel 10 miles.
A
line 7^.
what
58. and the area
of the rectangle.
and the speed
of the train.
the squares of their radii
(e)
55. under a pressure of 15 pounds per square inch has a volume of
gas
is
A
16 cubic
feet. othei
(a) Triangles
as their basis (b
and
b'). (e) The distance traveled by a train moving at a uniform rate.
(d)
The sum
of
money producing $60
interest at
5%. State whether the quantities mentioned below are directly or inversely proportional (a) The number of yards of a certain kind of silk. the volume of a
The temperature remaining
body of gas inversely proportional to the pressure.
What
will be the
volume
if
the pressure
is
12 pounds per square inch ?
. and the
:
total cost.
1
(6) The circumferences (C and C ) of two other as their radii (R and A").
57.inches long represents
map corresponds to how many miles ?
The
their radii. The number of men (m) is inversely proportional to the number of days (d) required to do a certain piece of work. (c) The volume of a body of gas (V) is
circles are to
each
inversely propor-
tional to the pressure (P).
areas of circles are proportional to the squares of If the radii of two circles are to each other as
circle is
4
:
7.
and the area of the smaller
is
8 square inches.
(c)
of a rectangle of constant width.
(d)
The
areas
(A and
A') of two circles are to each other as
(R and R').
A
line 11 inches long
on a certain
22 miles. and the time.
the area of the larger? the same.
Divide 108 into two parts which are to each other
7. it is advisable to represent these unknown numbers by mx and nx. 11 x -f 7 x = 108. 2. AB = 2 x.
Let
A
B
AC=1x. x = 6. 7 x = 42 is the second number.
11
x
x
7
Ex.000
168. = the second number.
Hence
or
Therefore
Hence
and
= the first number. 4 inches long. When a problem requires the finding of two numbers which are to each other as m n.
. 11 x = 66 is the first number.
x=2.
4
'
r
i
1
(AC): (BO) =7: 5.
as 11
Let
then
:
1. so that
Find^K7and BO.
:
Ex.
Therefore
7
=
14
= AC.
2 x
Or
=
4.RATIO AND PROPORTION
69.
127
The number
is
of miles one can see from an elevation of
very nearly the mean proportional between h and the diameter of the earth (8000 miles). What is the greatest distance a person can see from an elevation of 5 miles ? From h miles
the
Metropolitan
Tower (700
feet high) ?
feet
high) ?
From Mount
McKinley (20.
is
A line AB. produced to a point C.
Then
Hence
BG = 5 x. 18 x = 108.
14.
The
total area of land is to the total area of
is
water as
7 18. How
The
long are the parts ? 15. 12.)
.
What
are the parts ?
5. and 15 inches. The three sides of a triangle are respectively a. If c is divided in the ratio of the other two. Brass is an alloy consisting of two parts of copper and one part of zinc. cubic feet of oxygen are there in a room whose volume is 4500
:
cubic feet?
8.
Divide 20 in the ratio 1 m.
12. and the longest is divided in the ratio of the other two. and c inches.
m
in the ratio x:
y
%
three sides of a triangle are 11. 6.128
ELEMENTS OF ALGEBRA
EXERCISE
63
1.
:
Divide 39 in the ratio 1
:
5.
11.
A line 24 inches
long
is
divided in the ratio 3
5.
13. How many gen.
Divide 44 in the ratio 2
Divide 45 in the ratio 3
:
9.
7. How many ounces of copper and zinc are in 10 ounces of brass ?
6.
of water?
Divide 10 in the ratio a
b.
:
197.
How many
7.
consists of 9 parts of copper and one part of ounces of each are there in 22 ounces of gun-
metal ?
Air is a mixture composed mainly of oxygen and nitrowhose volumes are to each other as 21 79.000 square miles.
Gunmetal
tin.
How many
grams of hydrogen are contained in 100
:
grams
10. Water consists of one part of hydrogen and 8 parts of
If the total surface of the earth
oxygen.
:
4. 9. what are
its
parts ?
(For additional examples see page 279. find the number of square miles of land and of water.
3.
:
Divide a in the ratio 3
Divide
:
7. 2.000.
Hence. y = 1.e. =. there is only one solution. From (3) it follows y 10 x and since
by the same values of x and
to be satisfied
y.
y
(3)
these
unknown numbers can be found.
a?
(1)
then
I.
2 y = .y=--|.CHAPTER XI
SIMULTANEOUS LINEAR EQUATIONS
169.
An
equation of the
first
unknown numbers can be the unknown quantities. etc. the equation is satisfied by an infinite number of sets Such an equation is called indeterminate.
If
satisfied
degree containing two or more by any number of values of
2oj-3y =
6. y =
5
/0 \ (2)
of values. which substituted in (2) gives y both equations are to be satisfied by the same Therefore. such as
+ = 10. is x = 7. values of x and y.
Hence
2s -5
o
= 10 _ ^
(4)
= 3.
The
root of (4)
if
K
129
.
However. x = 1. expressing a y.
if
there
is
different relation
between x and
*
given another equation.-. if
.-L
x
If
If
= 0.
the equations have the two values of
y must be equal.
6x
.
21 y
.
A
system of two simultaneous equations containing two
quantities is solved by combining them so as to obtain
unknown
one equation containing only one
173.24. The first set of equations is also called consistent.
are simultaneous equations. for they express the x -f y 10. for they cannot be satisfied by any value of x and y.
of elimination
most frequently used
II. 26 y = 60.
= . and 3 x + 3 y =. 3.
The process of combining several equations so as make one unknown quantity disappear is called elimination. for they are 2 y = 6 are But 2 x 2.
By By
Addition or Subtraction.
unknown
quantity.26.
Solve
-y=6x
6x
-f
Multiply (1) by
2. same relation.
ELIMINATION BY ADDITION OR SUBTRACTION
175. 30 can be reduced to the same form -f 5 y Hence they are not independent.
cannot be reduced to the same form.
~ 50.
Substitution. Independent equations are equations representing different relations between the unknown quantities such equations
.130
170. y
I
171. 6 and 4 x y not simultaneous.
172. y = 2.
E. the last set inconsistent. Any set of values satisfying 5 x + 6 y = 60 will also satisfy the equation 3 x -f.
4y
.
(3)
(4)
Multiply (2) by
-
Subtract (4) from (3).
Therefore.
174.
to
The two methods
I.
x
-H
2y
satisfied
6 and 7 x 3y = by the values x = I.
ELEMENTS OF ALGEBRA
A
system
of simultaneous equations is
tions that can be satisfied
a group of equa by the same values of the unknown
numbers.
viz.X.3 y = 80.
+2+
6
= 8. The digit in the tens' place is | of the sum of the other two digits.
2
= 6.
to express
it is difficult
two of the required
digits in
terms
hence we employ 3
letters for the three
unknown
quantities.
x
:
z
=1
:
2.
1
=
2.
(1)
100s
+ lOy + z + 396 = 100* + 10y + x.
M=i.
symbols:
x
+
y
+z-
8.
Let
x
y z
= the
the digit in the hundreds' place.
# 4.
Check. however. z + x = 2 n.
.
= 2 m.y
125
(3)
The solution of these equations gives x Hence the required number is 125. and if 396 be added to the number. 1. the first and the last digits
will be interchanged.
+
396
= 521.
=
l.
Ex.)
it is advisable to represent a different letter.
The
three statements of the problem can
now be
readily expressed in
.
.
(
99.
and Then
100
+
10 y
+z-
the digit in the units' place. The sum of three digits of a number is 8. + z = 2p.
unknown quantity by
every verbal statement as an equation.
2
= 1(1+6). as many verbal statements as there are unknown quantities.
Obviously
of the other
. Problems involving several unknown quantities must contain. 1 digit in the tens place. either directly or implied. the number. Simple examples of this
kind can usually be solved by equations involving only one
unknown
every
quantity.
y
*
z
30. and to express
In complex examples. y
31.SIMULTANEOUS LINEAR EQUATIONS
143
x
29.
Find the number.2/
2/
PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS
183.
4
x
= 24.
Ex.
x 3
= 24. 5_
_4_
A.
8
= xy + x xy = xy -f 3 x 2 y = 2. starts 2 hours after B and overtakes A at the same How many miles has A then traveled? instant as B.
3+1 5+1
4_2.144
Ex.
6
x 4
= 24.
ELEMENTS OF ALGE13KA
If both numerator and denominator of a fraction be
.
x
y
= the = the
x
denominator
. y = 3. who travels 2 miles an hour faster than B.
(1) (2)
12.
xy
a:
2y 4y
2. C.
=
Hence the
fraction
is
f. and C travel from the same place in the same B starts 2 hours after A and travels one mile per hour faster than A. the fraction
Let and
then y
is
reduced to
nurn orator. From (3)
Hence xy
Check.
Find the
fraction.
increased by one.
Or
(4)-2x(3).
+
I
2
(1)
and
These equations give x
Check. x 3x-4y = 12.
direction.
2.
By
expressing the two statements in symbols.
(3) C4)
=
24 miles. = 8.
Since the three
men
traveled the
same
distance.
3. B. the distance traveled by A.
.
we
obtain. the fraction is reduced to | and if both numerator and denominator of the reciprocal of the fraction be dimin-
ished by one.
3
xand y
I
1
(2)
5.
2.
= the
fraction.
its value added to the denominator.
Five times a certain number exceeds three times another 11. and four times the first digit exceeds the second digit by 3.
?
What
9. Four times a certain number increased by three times another number equals 33. If the denominator be doubled. Find the number. and the second increased by 2 equals three times the first. Find the numbers. and
its
denomi-
nator diminished by one. If
9 be added to the number. if its numerator and its denominator are increased by 1. A fraction is reduced to J. Find the number. the fraction is reduced
fraction. fraction is reduced to \-. it is reduced to J.
7.
If 4 be
Tf 3 be
is J. both terms.
If 27 is
10.
number by
the
first
3. Find the numbers.
2.
The sum
18
is is
and
if
added
of the digits of a number of two figures is 6. the value of the fraction is fa. and the two digits exceeds the third digit by 3.
the
number
(See Ex. and the fourth 3. Find the fraction.
to
L
<>
Find the
If the
numerator and the denominator of a fraction be If 1 be subtracted from increased by 3.)
added to a number of two digits.
1.
5.
added to the numerator of a fraction. the Find the fraction.
tion ?
8. and the second one increased by 5 equals twice
number. the fraction equals .
The sum
of the first
sum
of the three digits of a number is 9.
Find the numbers.
183. to the number the digits will be interchanged. and the numerator increased by 4.
Half the sum of two numbers equals 4. the last two digits are interchanged.
.
If the
numerator of a fraction be trebled. and twice the numerator What is the fracincreased by the denominator equals 15. the digits will be interchanged.
6.
part of their difference equals
4.SIMULTANEOUS LINEAR EQUATIONS
EXERCISE
70
145
1.}.
Find
the rates of interest. What was the amount of each investment ?
A man
%
5%. and B's age is \ the sum of A's and C's ages. respectively ?
16.146
ELEMENTS OF ALGEBRA
11.
and 5 years ago
their ages is 55. Three cubic centimeters of gold and two cubic centimeters of silver weigh together 78 grains. A man invested $750. and in 5 years to $1125.
in 8 years to $8500.000
is
partly invested at
6%. What was the amount of each investment ?
15. and 4 %.
Ten years ago A was B was as
as old as
B
is
old as
will be 5 years hence .
A
sum
of $10. and the 5% investment brings $15 more interest than the 4 % investment.
and
money and
17. Ten years ago the sum of their ages was 90.
.
partly at
5 %. How 6 %. bringing a total yearly interest of $530.
13. the rate of interest ?
What was
the
sum
of
A sum
of
money
at simple interest
amounted
in 2 years
to $090. 12. Find their present ages. and The 6 investment brings $ 70 more interest than the 5
%
%
4%
investments together. Twice A's age exceeds the sum of B's and C's ages by 30. Find the weight of one cubic centimeter of gold and one cubic centimeter of silver.
much money
is
invested at
A sum
of
money
at simple interest
amounted
in 6 years
to $8000. If the sum of
how
old
is
each
now ?
at
invested $ 5000.
14. partly at 5% and partly at 4%. a part at 6 and the remainder bringing a total yearly interest of $260. the annual interest would be $ 195. If the rates of interwere exchanged. 5 %.
What was
the
sum and
rates
est
The sums of $1500 and $2000 are invested at different and their annual interest is $ 190. the rate of interest?
18. now.grams.
19. Two cubic centimeters of gold and three cubic centimeters of silver weigh
together 69 J. and partly at 4 %.
SIMULTANEOUS LINEAR EQUATIONS
147
20.
1
NOTE. receiving $ 100 for each horse.
BC=7. and angle BCA = 70.
24. but if A would double his pace. and F '(see diagram). and F. BC = 7 inches. angle c = angle d. If angle ABC = GO angle BAG = 50.
A
r
^
A
circle is inscribed in triangle
sides in D. then AD = AF. BD = HE.
.
23. and angle e angle/.
.
25. are taken so
ABC.
the three sides of a triangle E. andCL4 = 8. and F.
An C touch ing the sides in D. what are the angles of the triangle ?
22. and AC = 5 inches.
is
the center of the circum-
scribed circle. he would walk it in two hours less than
than
to travel
B
B. and sheep.
respectively. and GE = CF.
the length of
NOTE. If one angle exceeds the sum of the other two by 20. and e.
It takes
A two hours
longer
24 miles.
Find the parts of the
ABC touching the three sides if AB = 9.
E. what is
that
=
OF.
On
/). three
AD = AF. for $ 740. A farmer sold a number of horses.
In the annexed diagram angle a = angle b. B find angles a. How many did he sell
of each if the total
number
of animals
was 24?
21. BE. and CF?
is
a circle
inscribed in the
7<7.
Find their
rates of walking. cows. ED = BE. The number of sheep was twice the number of horses and cows together. c. and CE If AB = G inches. and their difference by GO .
triangle
Tf
AD. and $15 for each sheep. $ 50 for each cow.
points. The sum of the 3 angles of a triangle is 180.
and
respectively represented
Dare
and
by
(3 7 4). and point the origin.
the ordinate of point P.
.
?/.
It'
Location of a point. and PN _L YY'.
PN are given. and ordinates abore the x-axis are considered positive .
The
of
Coordinates. and r or its equal OA is
.
PN.
or its equal
OM.CHAPTER
XII*
GRAPHIC REPRESENTATION OF FUNCTIONS AND
EQUATIONS
184.
186.
first
3). B. jr. YY' they-axis.
The
abscissa
is
usually denoted by
line XX' is called the jr-axis. and whose ordinate is usually denoted by (X ?/).
two fixed straight lines XX' and YY' meet in at right angles. PM..
(2. hence
The
coordinates lying in opposite directions are negative.
(2.
2).
(3. then
the position of point is determined if the lengths
of
P
P3f and
185. Abscissas measured to the riyht of the origin. and PJ/_L XX'.
lines
PM
the
and P^V are
coordinates
called
point P. the ordinate by ?/.
-3). Thus the points A.
is
The point whose abscissa is a.
is the abscissa. (7.
* This chapter
may
be omitted on a
148
reading.
4). the mutual dependence of the two quantities may be represented
either by a table or
by a diagram.GRAPHIC REPRESENTATION OF FUNCTIONS
The
is
149
process of locating a point called plotting the point.
Plot the points:
(4.
2. 1). -4). (4.
=3?
is
If a point lies in the avaxis.
.
12. 0). (-4.
6.
(-1.
(-3. 0). 3).
the quadrilateral whose vertices are respectively
(4. 0).1).
1).
two variable quantities are so related that
changes of the one bring about definite changes of the other.
8. 3).and(l.
What
is
the locus of
(a?.
4)
from the
origin ?
7.
all all
points
points
lie
lie
whose abscissas equal zero ?
whose ordinates equal zero?
y) if y
10. (See diagram on page 151.
4)
and
(4.
(-5. which of its coordinates
known ?
13.)
EXERCISE
1.
2J-). (0.
3.
-2). -!).2). Graphic constructions are greatly facilitated by the use of cross-section paper. (-2.
(4.
.
What
are the coordinates of the origin ? If
187.
4.
Draw
the triangle whose vertices are respectively
(-l.
6. (0. (-4.
and measure
their
distance.
Plot the points
:
(0. (4. -2).
Plot the points
(6.
4).
Where do Where do
Where do
all
points
lie
whose ordinates
tfqual
4?
9. paper ruled with two sets of equidistant and parallel linos intersecting at right angles. 11.
0).
71
2).3).e.
What
Draw
is
the distance of the point
(3.
Plot the points: (-4. (4.
whose coordinates are given
NOTE. i.(!. -3).
Graphs.
may be represented graphby making each number in one column the abscissa.
representation does not allow the same accuracy of results as a numerical table. 10
.
Thus the average temperature on May
on April 20.
1. but it indicates in a given space a great many more
facts than a table.
188. Thus the first table produces 12 points. or the curved line the temperature. B.
may be found
on Jan.
ure the ordinate of F. in like manner the average temperatures for every value of the time.
By representing
of points.
ically
each representing a temperature at a certain date. A.150
ELEMENTS OF ALGEBRA
tables represent the average temperature
Thus the following
of
New
volumes
1
Y'ork City of a certain
to 8 pounds. and the amount of gas subjected to pressures from
pound
The same data.. and the corresponding number in the adjacent column the ordinate of a point.
we meas1
.
A graphic
and
it
impresses upon the eye
all
the peculiarities of
the changes better and quicker than any numerical compilations.
ABCN
y
the so-called graph of
To
15
find
from the diagram the temperature on June
to be 15
. C.
.
from January 1 to December 1. D. we obtain an uninterrupted sequence
etc.
15. however.
The engineer. Whenever a clear. as the prices and production of commodities.
physician. the rise and fall of wages. uses them. concise representation of a
number
of numerical data
is
required. the graph
is
applied. Daily papers represent ecpnoniical facts graphically. (b) July 15. (d) November 20. (c) January 15.GRAPHIC REPRESENTATION OF FUNCTIONS
151
i55$5St5SS 3{utt|s33<0za3
Graphs are possibly the most widely used devices of applied matheThe scientist uses them to compile the data found from experiments.
. etc. the merchant.
EXERCISE
From the diagram
questions
1. the
matics.
:
72
find approximate answers to the following
Determine the average temperature of New York City on (a) May 1. and to deduce general laws therefrom.
1 ?
does
the
temperature
increase from
11.?
is
is
the average temperature of
New York
6.
How
much.
How much warmer
1 ?
on the average
is it
on July 1 than
on
May
17. At what date is the average temperature highest the highest average temperature?
?
What What
is
4.
ELEMENTS OF ALGEKRA
At what date
(a) G
or dates
is
New York
is
C.
is
ture
we would denote the time during which the temperaabove the yearly average of 11 as the warm season.
From what
date to what date does the temperature
increase (on the average)?
8. At what date is the average temperature lowest? the lowest average temperature ?
5. (c)
the average temperature oi 1 C. 1?
11
0. from what date to what date would it extend ?
If
. (freezing
point) ?
7. 1 to Oct.
Which month
is
is
the coldest of the year?
Which month
the hottest of the year?
16.
When
What
is
the temperature equal to the yearly average of
the average temperature from Sept.
15..
is
10.
When
the average temperature below
C. (d) 9 0.
During what months
above 18 C. ?
9.
June
July
During what month does the temperature increase most
?
rapidly
12.
During what month does the temperature change least?
14. ?
-
3.. (1)
10
C.
on
1 to
the
average.152
2..
During what month does the temperature decrease most
rapidly ?
13.
in a similar manner as the temperature graph was applied in examples 1-18.
Represent graphically the populations
:
(in
hundred thou-
sands) of the following states
22. One meter equals 1.
Draw
. From the table on page 150 draw a graph representing the volumes of a certain body of gas under varying pressures.
Hour
Temperature
.GRAPHIC REPRESENTATION OF FUNCTIONS
18.
a temperature chart of a patient.
153
1?
When is the average temperature the same as on April
Use the graphs of the following examples for the solution of concrete numerical examples. Construct a diagram containing the graphs of the mean temperatures of the following three cities (in degrees Fahren-
heit)
:
21.
Draw
a
graph for the
23.09 yards.
19.
NOTE. transformation of meters into yards.
20.
29.
books from
for printing. if each copy sells for $1.. x
7 to 9. and $. Represent graphically the cost of butter from 5 pounds if 1 pound cost $. e.
x*
x
19.50 per copy
(Let 100 copies = about \.
(Assume ir~
all circles
>2
2
. 2 .
28. represent his daily gain (or loss).
Show
graphically the cost of the
REPRESENTATION OF FUNCTIONS OF ONE VARIABLE
189. if x assumes
successively the
tively
values
1.5
grams. 1 to 1200 copies.
to
27.
2
is
called
x
2 xy
+ 7 is a function of x.
+7
If
will
respec-
assume the values 7. 3.
to 20 Represent graphically the weight of iron from cubic centimeters. 2 x -f 7 gradually from 1 to 2. Represent graphically the distances traveled by a train in 3 hours at a rate of 20 miles per hour.
A
10 wheels a day. etc.
If
dealer in bicycles gains $2 on every wheel he sells.) On the same diagram represent the selling price of the books.
190. 2 8 y' + 3 y is a function of x and
y.)
T
circumferences of
25. etc.. the value of a of this quantity will change. the daily average expenses for rent.154
24.inch.
then
C
irJl..
26.
2.g.
from
R
Represent graphically the = to R = 8 inches.
4. An expression involving one or several letters a function of these letters. binding. amount to $8. if he sells 0.
ELEMENTS OF ALGEBRA
If
C
2
is
the circumference of a circle whose radius
is J2.
The
initial cost of
cost of manufacturing a certain book consists of the $800 for making the plates.
function
If the value of a quantity changes. gas.50.
3. if 1 cubic centimeter of iron weighs 7.
x increases will change gradually from
13.50. 9.
.
The values of func192. (2. hence
various values of x
The values of a function for the be given in the form of a numerical table. to con struct the graph x of x 2 construct a series of -3 points whose abscissas rep2 resent X) and whose ordi1
tions
. plot points which
lie
between those constructed above.
1
the points (-3. 4). while 7 is a constant.
and join
the
points in order.
a*. be also represented by a graph.
may
. construct
'.
may. (1.
etc. 4).
Ex.1).
155
-A
variable is a quantity
whose value changes in the
same
discussion. and (3.
To
obtain the values of the functions for the various values of
the
following arrangement
be found convenient
:
. however. If a more exact diagram
is
required.
(1^.e.2 x
may
4 from x
=
4. E.
9).
values of x2
nates are the corresponding i.
Graph
of a function.
Draw the graph of x2 -f.
3
(0.0). Thus the table on page 1G4 gives the values of the functions x 2 x3 and Vsr.
it is
In the example of the preceding article.
-J). as
1.
9).
2
(-1.
is
A
constant
a quantity whose value does not change in the
same discussion.g.
Q-.
2).
is
supposed to change. x a variable. to
x = 4. for x=l.
. 2.1).GRAPHIC REPRESENTATION OF FUNCTIONS
191.
(-
2. 3 50.
Draw
y
z x
the graph of
= 2x-3.
rf
71
.
(To avoid
very large ordinatcs. Thus 4x + 7. the scale unit of the ordinatcs is taken smaller than that of the x. straight line produces the required graph..-.
It can be
proved that the
graph is a straight
of a function of the first degree
line.)
For brevity. j/=-3.
A
Y'
function of the
first
degree is an integral
rational function
involving only
the
power of the variable. y = 6.4).. -1).
. and joining in order produces
the graph
ABC.
4J. (4.20). = 4.. 4).
= 0.. Thus in the above example.156
ELEMENTS OF ALGEBRA
Locating the points(
4. hence two points are sufficient for the construction
of these graphs. 2 4 and if y = x -f. 5). as y. (-2.
Ex.
If
If
Locating
ing
by a
3) and (4.2 x
.
r
*/
+*
01
. etc. or ax + b -f c are funclirst
tions of the first degree. the function
is
frequently represented
by a single letter.
if
/*
4
>
1i >
>
?/
=
193.
2.
7
.
194.
(-3. and join(0.
if c
Draw
the locus of this equation
= 12.. it is evidently possible Thus to find to find graphically the real roots of an equation.
to Fahrenheit readings
:
Change
10
C.
If
two variables x and y are inversely proportional.
GRAPHIC SOLUTION OF EQUATIONS INVOLVING ONE
UNKNOWN QUANTITY
Since we can graphically determine the values of x make a function of x equal to zero.
Show
any convenient number).
ELEMENTS OF ALGEBRA
Degrees of the Fahrenheit
(F.24.. Represent 26. we have to measure the abscissas of the intersection of the
195.158
24. the abscissas of 3. then
cXj
where
c is a constant.
. 32 F.
A body
moving with a uniform
t
velocity of 3 yards per
second moves in
this
seconds a distance d
=3
1.e.. Therefore x = 1.
25.where x
c is
a constant. what values of x make the function x2 + 2x 4 = (see 192).) scale by the formula
(a)
Draw
the graph of
C = f (F-32)
from
to
(b)
4 F F=l.
y=
formula graphically. 14 F.
1
C. 9 F.
that
graph with the o>axis.
that the graph of two variables that are directly proportional is a straight line passing through the origin (assume
for c
27.
C.. then
y = .. If two variables x and y are directly proportional.
From
grade equal to
(c)
the diagram find the number of degrees of centi-1 F. i.) scale are
expressed in
degrees of the Centigrade (C.24 or x =
P and
Q.
y y
2. 0). y = -l. locate points
(0.
?/
=4
AB.2 y ~ 2.
i.
first
degree. = 0. and join the required graph.
represent graphically equations of the form y function of x ( 1D2).
1)
and
0).
Hence we may
join (0.
y=
A
and construct
x
(
-
graphically.
4)
and
(2.
?/. solve for
?/. we can construct the graph or locus of any
Since
we can
=
equation involving two
to the above form.
Ex. Represent graphically
Solving for
y ='-"JJ y. Draw the locus of 4 x + 3 y = 12.1. because their graphs are straight lines.
If the given equation is of the we can usually locate two
y.
3x
_
4
.
Ex.
unknown
quantities.
If
x
=
0.e.
T
. that can be reduced
Thus
to represent
x
-
-
-L^-
\
x
=2
-
graphically.
199.
Hence.
Graph
of
equations involving two
unknown
quantities.
NOTE. 2).
fc
= 3.160
ELEMENTS OF ALGEBRA
GRAPHIC SOLUTION OF EQUATIONS INVOLVING TWO UNKNOWN QUANTITIES
198.
X'-2
Locating the points
(2.
and joining by a straight
line.
4) and
them by
straight line
AB
(3.
== 2.
if
y
=
is
0.2.
Hence
if
if
x
x
-
2.
(f
. Equations of the first degree are called linear equations.
Thus
If
in
points without solving the equation for the preceding example:
3x
s
.
produces the
7*
required locus.
.
(2)
.
Solve graphically the equations
:
(1)
\x-y-\.15.
3.
parallel have only one point of intersection. we obtain the roots. viz. The coordinates of every point of the graph satisfy the given equation.
To
find the roots of
the system.
202.
The
every
coordinates
of
point in satisfy the equation
(1).
and CD.
equation
x=
By measuring
3.
AB
y
= .
Graphical solution of a linear system.GRAPHIC REPRESENTATION OF FUNCTIONS
161
200.
By
the
method
of
the preceding article construct the graphs
AB
and
and
CD
of
(1)
(2) respectively.1=0.
Since two straight lines which are not coincident nor simultaneous
Ex. P.
201. the point of intersection of the coordinate of P.
203. and every set of real values of x and y satisfying the given equation is represented by a point in
the locus.
AB
but only one point
in
AB
also satisfies
(2). linear equations have only one pair of roots. The roots of two simultaneous equations are represented by the coordinates of the point (or points) at which their
graphs intersect.57.
V25
5.
Solve graphically the
:
fol-
lowing system
= =
25. e. etc.0).
Inconsistent equations.
4.
5. parallel graphs indicate inconsistent equations.
y equals
3.y~ Therefore. 5.
Using the method of the preceding para. (-2. 0) and (0.
Solving (1) for y. 3). and joining by a
straight line.
There can be no point of
and hence no
roots.
3.
obtain the graph (a circle)
AB C
joining.
In general.
-
4. construct CD the locus of (2)
of intersection.g.
4.
and
+ 3).5. 4. = 0. 0.
4.
P
graphs meet in two and $. (-4. which consist of a
pair of parallel lines.
Locating the
points
(5.5.
Locating two points of equation (2). 2. they are inconsistent.
1. AB the locus of (1). 3.
(1)
(2)
cannot be satisfied by the same values of x and y.9.
3x
2 y = -6. 1. i.
2 equation x
3).
4.. the point
we
obtain
Ex. there are two pairs of By measuring the coordinates of
:
P and Q we find
204.0.
Measuring the coordinates
of P.
The equations
2
4
= 0.
3. Since the two
-
we obtain DE. we of the
+
y*
= 25.162
ELEMENTS OF ALGEBRA
graph. This is clearly shown by the graphs of (1) arid (2).
4. 0.e. and
.
(1)
(2)
-C.
intersection.
x2
.
4.
4.
2.
. (4. if x equals
respectively
0. the graph
of
points
roots.
numbers.
\/"^27=-3.
Evolution
it is
is the operation of finding a root of a quan the inverse of involution.
Every odd root of a quantity has
same sign as
and
2
the
quantity. and
all
other numbers are. Since even powers can never be negative.
tity
.
a)
4
= a4
.CHAPTER XIV
EVOLUTION
213.
= x means
= 6-.
or x
&4 .
4
4
. and ( v/o* = a.
for (-f 3) 2
(
3)
equal
0. called real
numbers.
V9 = +
3.
Thus
V^I is an imaginary number.
or
-3
for
(usually written
3)
. etc.
\/a
=
x means x n
=
y
?>
a. (_3) = -27.
V
\/P
214.
1.
2.
It follows
from the law of signs
in evolution that
:
Any
even root of a positive. which can be simplified no further.
27
=y
means
r'
=
27. quantity
may
the
be either 2wsitive
or negative.
215. for (+ a) = a \/32 = 2. or y
~
3.
109
. for distinction. it is evidently impossible to express an even root of a negative quantity by Such roots are called imaginary the usual system of numbers.
>
13.
10. the that 2 ab -f b 2
=
we have then to consider sum of trial divisor 2 a.
2
49a 8 16 a 4
9.e.
term a of the root
is
the square root of the
first
The second term
of the root can be obtained
a. let us consider the relation of a -f.
The work may be arranged
2
:
a 2 + 2 ab
+ W \a + b
. it is not known whether the given
expression is a perfect square.
15. In order to find a general method for extracting the square root of a polynomial. i.
ELEMENTS OF ALGEBEA
4a2 -44a?> + 121V2 4a
s
.
#2
a2
-
16.
The
term
a'
first
2
.
the given expression is a perfect square. 11.2 ab + b
.
+ 6 + 4a&. a -f.2 ac .172
7.
14.
a-\-b
is
the root
if
In most cases.
a2
+ & + c + 2 a& .
second term 2ab by the double of
by dividing the the so-called trial divisor.
8
.b 2 2 to its square.
2
.
12.
and
b. and b (2 a -f b).72 aW + 81 &
4
. however.
2ab
.2 &c.
multiplied by b must give the last two terms of the
as follows
square.
.
mV-14m??2)-f 49.
2 2
218.
.
Arrange the expression according to descending powers root of 10 x 4 is 4 # 2 the lirst term of the root. 8 /-.
double of this term
find the next
is
the
new
trial divisor. As there is no remainder.EVOLUTION
Ex. 8 a 2 .
.
*/''
.
The square
. 8 a 2 Second complete divisor.24 a + 4 -12 a + 25 a8
s
. Second trial divisor.
First complete divisor. and consider Hence the their sum one term.
\
24 a 3
4-f
a2
10 a 2
Second remainder.
. the first term of the answer. the required root
(4
a'2
8a
+
2}. By doubling 4x'2 we obtain 8x2 the trial divisor. 8 a 2
-
12 a
+4
a
-f 2.
-
24 a
3
+
25 a 2
-
12 a
+4
Square of 4 a First remainder.
1.
219.
2. 1.
10 a 4
8
a. we obtain the next term of the root 3 y 3 which has to be added to 2 the trial divisor. 6 a.
First trial divisor.
Arranging according to descending powers of
10 a
4
a.
. We find the first two terms of the root by the method used in Ex.
Ex.
. 8 a 2
2.
by division we
term of the
root.24 afy* -f 9 tf.
4 x2
3
?/
8 is
the required square foot.
and so
forth.
is
As
there
is
no remainder.
173
x*
Extract the square root of 1G
16x4
10 x*
__
.
Explanation. Multiply the complete divisor Sx' 3y 3 by Sy 8 and subtract the product from the remainder. The process of the preceding article can be extended to polynomials of more than three terms. 24# 2 y 3 by the trial divisor Dividing the first term of the remainder.
Extract the square root of
16 a 4
.
of x. 2 Subtracting the square of 4x' from the trinomial gives the remainder '24 x'2 + y.
Ex. Thus the square root of 96'04' two digits.
square root of arithmetical numbers can be found to the one used for algebraic
Since the square root of 100 is 10.
= 80. the square root of 7744 equals 88.
The
is
trial divisor
=
160. Hence the root is 80 plus an unknown number.EVOLUTION
220. two figures.176..
Find the square root of 524. and we may apply the method used in algebraic process.
the
consists of
group is the first digit in the root.
Find the square root of 7744. the first of which is 4. etc. and the complete divisor
168. the first of which is 8. of a number between 100 and 10. Hence if we divide the digits of the number into groups. the first of which is 9 the square root of 21'06'81 has three digits.000. etc.
Ex.
a 2 = 6400. and the first remainder is. Therefore 6 = 8. which may contain one or two). the integral part of the square root of a number less than 100 has one figure. then the number of groups is equal to the number of digits in the square root.
2.
first
.000 is 100.
a
f>2'41 '70
6
c
[700
+ 20 + 4 = 724
2 a
a2 = +6=
41)
00 00
1400
+ 20 = 1420
4
341 76
28400
=
1444
57 76
6776
.
From
A
will
show the
comparison of the algebraical and arithmetical method given below identity of the methods.
7744 80 6400
1
+8
160
+ 8 = 168
1344
1344
Since a
2 a
Explanation.
175
The
by a method very similar
expressions. of 1.
the preceding explanation it follows that the root has two digits. beginning at the
and each group contains two digits (except the last.000 is 1000.1344.
1.000. of 10. and the square root of the greatest square in
units.
As
8
x 168
=
1344.
Find the square root of
6/.10.7 to three decimal places.0961
are
'.
EXERCISE
Extract the square roots of
:
82
.70
6.
places.
we must
Thus the groups
1'67'24. annex a cipher.
The groups
of 16724.1 are
Ex.
Roots of common fractions are extracted either by divid-
ing the root of the numerator by the root of the denominator.688
4
45 2 70
2 25
508
4064
6168 41)600
41344
2256
222. or by transforming the common fraction into a decimal. and if the righthand group contains only one digit.
12.
ELEMENTS OF ALGEKRA
In marking
off groups in a number which has decimal begin at the decimal point.
3.1T6
221.GO'61.
in .
If a 2 4. Find the side of each field.
solve for v.
'
4.
EXERCISE
1.
3. opposite the right angle is called the hypotenuse (c in the diagram).
r. and the first exceeds the second by 405 square yards.b 2 If s
If
=c
.
27.
26.
4.180
on
__!_:L
ELEMENTS OF ALGEBRA
a.
9
&
-{-
c#
a
x
+a
and
c. If the hypotenuse
whose angles
a
units of length.
22
a.
.
2
.
find a in terms of 6
.
29. and the two other sides respectively
c
2
contains
c
a and b units.
2. 25.
is
one of
_____
b
The side right angle.
and the sum
The
sides of
two square
fields are as
3
:
5.
108.
Three numbers are to each other as 1 Find the numbers.
:
6.
Find the side
of each field. then
Since such a triangle
tangle.)
of their squares
5. 24.
and their product
:
150.
2
.
84
is
Find a positive number which
equal to
its
reciprocal
(
144).
228. its area contains
=a
2
-f-
b2
.
Find
is
the number.
A
right triangle is a triangle.
If s
= 4 Trr
'
2
. is 5(5.
.
A
number multiplied by
ratio of
its fifth
part equals 45.
solve for
r.
solve for d.
=
a
2
2
(' 2
solve for solve for
= Trr
.
may
be considered one half of a
rec-
square units.
If 2
-f 2 b*
= 4w
2
-f c
sol ve for
m. The sides of two square fields are as 7 2.
If 22
= ~^-.
2
:
3.
The
two numbers
(See
is
2
:
3.
If
G=m m
g
.
2a
-f-
1
23.
28.
and they con-
tain together 30G square feet.
Find the numbers.
and the two smaller
11.
-J-
=
12.
of a right triangle Find these sides. (b) 100 feet?
=
.
add
(|)
Hence
2
.
Find the
radii. Find the unknown sides and the area.
181
The hypotenuse
of a right triangle
:
is
35 inches.
is
and the other
two
sides are equal.
The area $
/S
of a circle
2
. the radius of a sphere whose surface equals
If the radius of a sphere is r. A body falling from a state of rest.
The area
:
sides are as 3
4. passes in t seconds 2 over a space s yt Assuming g 32 feet.
Find these
10.
the formula
= Trr
whose radius equals r is found by Find the radius of circle whose area S
equals (a) 154 square inches.7 x -f 10 = 0. we have
of
or
m = |.2
7
.
8 = 4 wr2 Find 440 square yards.
The hypotenuse
of a right triangle is 2. let us compare x 2
The
left
the perfect square x2
2
mx -f m
to
2
. x* 7 x=
10.
.
7r
(Assume
and their
=
2 7
2
.)
13.
Solve
Transposing.
4. (b) 44 square feet.
member can be made a complete square by adding 7 x with another term. and the third side is 15 inches.
radii are as 3
14.QUADRATIC EQUATIONS
7.
.
The following
ex-
ample
illustrates the
method
or
of solving a complete quadratic
equation by completing the square.
24.)
COMPLETE QUADRATIC EQUATIONS
229.
Two
circles together contain
:
3850 square
feet. in how many seconds will a body fall (a) G4 feet.
8.
.
9. its surface
(Assume
ir
=
2 . and the
other two sides are as 3
4.
Method
of completing the
square. The hypotenuse of a right triangle is to one side as 13:12.
2m. To find this term.
make x2
Evidently 7 takes the place 7x a complete square
to
to
which corresponds
m
2
.
Find the
sides.
sides.
.c
= 0.
231.
= 12.
any quadratic equation may be obtained by 6.184
ELEMENTS OF ALGEBRA
45
46.
Solving this equation we obtain
by the method of the preceding
2a
The
roots of
substituting the values of a.
ao.
x
la
48.
Solution
by formula. and c in the general answer.
article.
2
Every quadratic equation can be
reduced to the general form. -\-bx-\.
=8
r/io?.
o^
or
-}-
3 ax == 4 a9
7 wr
.
=0.
49.
2x
3
4.
Problems involving quadratics have
lems of this type have only one solution.
54.
56.
and whose sum
is
is 36.
-2.
of their reciprocals is
4. and consequently many prob-
235. Find the number.
7.
PROBLEMS INVOLVING QUADRATICS
in general two answers.
and whose
product
9.
6. and the difference Find the numbers.
5.
52.
189
the equations whose roots are
53.
:
3.
area
A
a perimeter of 380
rectangular field has an area of 8400 square feet and Find the dimensions of the field.
Find two numbers whose difference
is 40.
Twenty-nine times a number exceeds the square of the 190. -5.0.
-4. The
11.
The sum
of the squares of
two consecutive numbers
85.
2.
What
are the
numbers
of
?
is
The product
two consecutive numbers
210.0.
Find
the numbers.
-2. and equals 190 square inches. -2.
57.
two numbers is 4.
8.
its
sides of a rectangle differ by 9 inches.9.3.
1.1.
3.
number by 10.
88
its reciprocal
A
number increased by three times
equals
6J.3.
G.QUADRATIC EQUATIONS
Form
51. but frequently the conditions of the problem exclude negative or fractional answers.
The
difference of
|.
Find the number.
3.
0.
58.
55.
is
Find two numbers whose product
288.
EXERCISE
1.
. Find the sides.2.
2.
1.
Divide CO into two parts whose product
is 875.
-2.3. feet.
Find a number which exceeds
its
square by
is
-|.
If a train had traveled 10 miles an hour faster. start together on voyages of 1152 and 720 miles respectively.
17. he would have received 12 apples less for the same money.
of a rectangle is to the length of the recthe area of the figure is 96 square inches.
19.
The diagonal
:
tangle as 5 4.
watch for $ 24.
14. and the line BD joining
two opposite
vertices (called "diagonal")
feet. and the slower reaches its destination one day
before the other. a distance One steamer travels half a mile faster than the two hours less on the journey. Find the rate
of the train. and lost as many per cent Find the cost of the watch. Two vessels. one of which sails two miles per hour faster than the other.
Two steamers
and
is
of 420 miles.
If he
each horse ?
.
c equals 221
Find
AB and AD.10.
13.
ply between the same two ports. it would have needed two hours less to travel 120 miles.190
12. ABCD.
ELEMENTS OF ALGEBRA
The length
1
B
AB of a rectangle. dollars. What did he pay for each
apple ?
A man bought a certain number of horses for $1200. sold a horse for $144.
A man
cent as the horse cost dollars. A man bought a certain number of apples for $ 2.
A man
A man
sold a
as the watch cost dollars.
and gained as many per Find the cost of the horse.
other.
15.
.
as the
16. At what rates do
the steamers travel ?
18.
watch cost
sold a watch for $ 21. had paid $ 20 less for each horse. What did he pay for
21.
vessel sail ?
How many
miles per hour did the faster
If 20. exceeds its widtK AD by 119 feet. and Find the sides of the rectangle. he would have received two horses more for the same money. and lost as many per cent Find the cost of the watch. he had paid 2 ^ more for each apple.
Find and CB.
or x
= \/l = 1. and the unknown factor of one of these terms is the square of the unknown factor of the
other.
is
On the prolongation of a line AC.
237. so that the rectangle.
Find
TT r (Area of a circle . and working together.
Find the side of an equilateral triangle whose altitude
equals 3 inches. Equations in the quadratic form can be solved by the methods used for quadratics.
and the area of the path
the radius of the basin.
=9
Therefore
x
=
\/8
= 2.
A rectangular
A
circular basin is surrounded
is
-
by a path 5
feet wide.
27. 23 inches long. In how many days can B do the work ?
=
26. is surrounded by a walk of uniform width.
(tf. contains B 78 square inches.
^-3^ = 7.QUADRATIC EQUATIONS
22.
EQUATIONS IN THE QUADRATIC FORM An equation is said to be in the quadratic form
if it
contains only two unknown terms. A needs 8 days more than B to do a certain piece of work.
B
AB
AB
-2
191
grass plot.) 25.
Ex.
of the area of the basin. How many eggs can be bought for $ 1 ?
236. how wide is the walk ?
23.
1.
By formula. constructed with and CB as sides. a point taken. as
0. If the area of the walk is equal to the area of the plot. the two men can do it in 3 days.
24. The number of eggs which can be bought for $ 1 is equal to the number of cents which 4 eggs cost.
Solve
^-9^ + 8 =
**
0. 30 feet long and 20 feet wide.I) -4(aj*-l)
2
= 9.
.
no
Fractional and negative exponents.
a m a" = a m+t1 .
must be
*The symbol
smaller than. (a ) s=a m = aw bm
a
."
means "is greater than"
195
similarly
means "is
. ~ a m -f.
II.
provided
w > n.
we may choose
for such
symbols any definition that
is
con-
venient for other work. we let these quantities be what they must be if the exponent law of multiplication is generally true. 4~ 3 have meaning according to the original definition of power.
It is. The following four fundamental laws for positive integral exponents have been developed in preceding chapters
:
I.a" = a m n
mn . and
.
m
IV.*
III. very important that all exponents should be governed by the same laws.
Then the law
of involution.
the direct consequence of the defiand third are consequences
FRACTIONAL AND NEGATIVE EXPONENTS
243.
We assume.
>
m therefore. (a m ) w
.
for all values
1
of
m and n.
244.CHAPTER XVI
THE THEORY OF EXPONENTS
242.
= a""
<
. while the second of the first. that a
an
= a m+n
. instead of giving a formal definition of fractional and negative exponents.
The
first
of these laws
is
nition of power.
(ab)
. however. such as 2*. hence.
3*.
^=(a^)
3*
3
.
= a.
To
find the
meaning
of
a fractional exponent.
23. since the raising to a positive integral power is only a repeated multiplication. or zero exponent
equal
x. at. 4~ .
245.
a\
26.
laws.196
ELEMENTS OF ALGEBRA
true for positive integral values of n. ml.
28. as.
Hence
Or
Therefore
Similarly.
24.
-
we
find
a?
Hence we
define a* to be the qth root of of.
e.
Write the following expressions as radicals :
22.g.
31.
'&M
A
27.
29. fractional. (xy$.
30.
disappear.
Assuming these two
8*.
we
try to discover the
let the
meaning of
In every case we
unknown quantity
and apply to both members of the equation that operation which makes the negative.
a*.
0?=-^.
etc.
Let
x
is
The operation which makes the fractional exponent disappear evidently the raising of both members to the third power.
n 2 a. 25.
. a .
(bed)*.
m$.
a?*.
Let
x=
or".g.2
=
a2
. Factors
may
be transferred
from
the
numerator
to
the
denominator of a fraction.
by changing the sign of
NOTE.
Or
a"#
= l. in which
obtained from the preceding one by dividing both
members by
a.
248. e.
each
is
The
fact that a
if
=
we
It loses its singularity
1 sometimes appears peculiar to beginners.
Multiplying both members by
a". or
the exponent.
a
a
a
= =
a a a
a1
1
a.
an x = a.
a8 a
2
=
1
1
. consider the following equations.198
247.
ELEMENTS OF ALGEBRA
To
find the
meaning
of a negative exponent.
cr n.
etc.
.
vice versa.
34.
6
35.
V ra
4/
3
-\/m
33.
If
powers of
a?.2 d
.
2.
we wish to arrange terms according to descending we have to remember that.
1 Multiply 3 or
+x
5 by 2 x
x.
40.
Arrange in descending powers of
Check.
powers of x arranged are
:
Ex.
1. The
252. the term which does not contain x may be considered as a term containing #.
lix
=
2x-l
=+1
Ex.
Divide
by
^
2a
3 qfo
4.
1.202
ELEMENTS OF ALGEBRA
32.
48.
47. Monomial surdn of the same order may be divided by multiplying the quotient of the coefficients by the quotient of the
surd factors.
is
1
2.
Ex.
(3V5-2V3)(2V3-V3).
53.
43.
44.
it
more convenient to multiply dividend and divisor by a factor which makes the divisor rational.V5) ( V3 + 2 VS).
46.y.
Ex.
60.
(3V3-2Vo)(2V3+V5).
Va
-v/a.
ELEMENTS OF ALGEHRA
(3V5-5V3)
S
.
E.
V3 .
(5V2+V10)(2V5-1).
.
51.
all
monomial surds may be divided by
method.
a fraction.
(5V7-2V2)(2VT-7V2).
(V50-f 3Vl2)-4-V2==
however. a
VS
-f-
a?Vy
= -\/ -
x*y
this
Since surds of different orders can be reduced to surds of
the same order.
(2
45.
52.214
42.
49.
268.
-v/a
-
DIVISION OF RADICALS
267. the quotient of the surds
is
If.
73205. arithTo find.
VTL_Vll '
~~"
\/7_V77
.by the usual arithmetical method. To show that expressions with rational denominators are simpler than those with irrational denominators. the rationalizing factor
x
' g
\/2.
The
2.
we have
to multiply
In order to make the divisor (V?) rational.
/~
}
Ex. metical problems afford the best illustrations.g. called rationalizing the
the following examples
:
215
divisor.
.73205
we
simplify
JL-V^l
V3
*>
^>
division
Either quotient equals . is illustrated
by
Ex. the by 3 is much easier to perform than the division by
1.
Divide 12 V5
+ 4V5 by V..RADICALS
This method.
Divide 4 v^a by
is
rationalizing factor
evidently \/Tb
hence.
is
Since \/8
12 Vil
=
2 V*2.57735.
1. e.
.
by V7.
4\/3~a'
36
Ex. we have
V3
But
if
1.
.
+ 4\/5 _ 12v 3 + 4\/5 V8 V8
V2 V2
269. Evidently. however.
Divide
VII by v7.
3.
Hence
in arithmetical
work
it
is
always best to
rationalize the denominators before dividing.
Ex. if n is even.
2 8 (3 a )
+8=
+
288.
Factor
consider
m
m
6
n9
.
actual division
n.g.
2.
By
we obtain the other
factors.
1. If n is a Theorem that
1.y n is divisible by x -f ?/. xn y n y n y n = 0.
x* -f-/
= (x +/)O . if n For ( y) n -f y n = 0.
xn -f.230
285.
ELEMENTS OF ALGEBRA
positive integer."
.
The
difference of
two even powers should always be
considered as a difference of two squares. For substituting y for x. if w is odd.
ar
+p=
z6
e.
Two
special cases of the preceding propositions are of
viz.
and have
for
any positive integral value of
If n
is
odd.
is
odd.
It
y is
not divisible by
287.
it
follows from the Factoi
xn y n is always divisible by x y.
2.
We may
6
n 6 either a difference of two squares or a
dif-
* The symbol
means " and so forth to.
286.
Factor 27 a* -f
27 a 6
8.
-
y
5
=
(x
-
can readily be seen that #n -f either x + y or x y.
:
importance.
2
Ex.xy +/).
while the
remaining terms do not
cancelj the root is infinity.
(1)
is
an
identity. the answer is indeterminate.
the
If in an equation
terms containing
unknown quantity
cancel.000
a.
be the numbers.
306.g. without exception.
TO^UU"
sufficiently small. however
x approaches the value
be-
comes
infinitely large.decreases
X
if
called infinity. cancel.
ELEMENTS OF ALGEBRA
Interpretation of ?
e.
The
solution
x
=-
indicates that the problem
is indeter-
If all terms of an minate.
Interpretation of
QO
The
fraction
if
x
x
inis
infinitely large.
(1).
of the second exceeds the product of the first
Find three consecutive numbers such that the square and third by 1.x'2 2 x =
1.
as
+ l.
and becomes infinitely small. equation.
is satisfied
by any number.e.
By making x
any * assigned
zero.
= 10.
great.increases
if
x
de-
x
creases.
I.
i.
Hence such an equation
identity.
x
-f 2. or that x may equal any finite number.
Or.
Hence any number will satisfy equation the given problem is indeterminate.
1.
and
.
.
customary to represent this result
by the equation ~
The symbol
304.
ToU"
^-100 a.e.
1.
oo is
= QQ.
Let
2.
(a:
Then
Simplifying.can be
If
It is
made
larger than
number.242
303.
+
I)
2
x2
'
-f
2x
+
1
-x(x + 2)= . or infinitesimal) This result is usually written
:
305.
(1)
= 0.
creases. it
is
an
Ex.i
solving
a problem
the result
or oo indicates that the
all
problem has no solution.
.
The
~~f
fraction .
i.
and the side of one increased by the side of the other e.
the
The mean proportional between two numbers sum of their squares is 328. Find two numbers whose product whose squares is 514.
14. increased by the edge of the other. equals 4 inches. Find these sides.244
3.
.
13.
255 and the sum of
5. Find the numbers. and
its
The diagonal
is
is
perimeter
11.
9.
and the sum of
(
228. Find the dimensions of the
field. The sum of the areas of two squares is 208 square feet.
is
the breadth
diminished by 20 inches.
The area of a
nal 41 feet. 148 feet of fence are required.
rectangle is 360 square Find the lengths of the sides. The volumes of two cubes differ by 98 cubic centimeters. To inclose a rectangular field 1225 square feet in area.
two numbers Find the numbers. Two cubes together contain 30| cubic inches. 12.
and the diago(Ex.)
53 yards.
of a right triangle is 73. Find the side of each square.
10. the area becomes
-f%
of
the original area.
8. But if the length is increased by 10 inches and
12.
Find the
sides of the rectangle.quals 20 feet.
Find the other two
sides.
The hypotenuse
is
the other two sides
7. Find the edge of each cube.)
The area
of a right triangle is 210 square feet.
6.
146 yards.
of a rectangular field
feet.
Find the
sides. and the edge of one. Find the edges. p. 190.
103.
is
is
17 and the
sum
4.
and
is
The area of a rectangle remains unaltered if its length increased by 20 inches while its breadth is diminished by 10 inches.
ELEMENTS OF ALGEBRA
The
difference between
is
of their squares
325.
is 6.
and the
hypotenuse
is 37. and the edge of one exceeds the edge of the other by 2 centimeters.
and the equal to the surface of a sphere Find the radii.
differ by 8 inches. Find the number.
and
if
the digits will be interchanged.
by the product of 27 be added to the number.)
17. irR *. their areas are together equal to the area of a circle whose radius is 37 inches.SIMULTANEOUS QUADRATIC EQUATIONS
15. Find the radii.
.) (Area of circle
and
=
1
16.
The
radii of
two spheres
is
difference of their surfaces
whose radius = 47T#2. (Surface of sphere
If a
number
of
two
digits be divided
its digits.
245
The sum of the radii of two circles is equal to 47 inches.
the quotient
is 2.
is
20 inches.
a
+
d. to produce the 3d term.
added to each term to obtain the next one.
.
The terms
ARITHMETIC PROGRESSION
308.7.
to each
term produces the next term.
. The first is an ascending...
-f
. the second a descending.
Hence
/
= a + (n .
to
A series
is
a succession of numbers formed according
some
fixed law. progression. to produce the 4th term. 15 is 9 -f. a
11. is derived from the preceding by the addition of a constant number. 16..
The common
Thus each
difference is the
number which added
an A. except the first.. . 3.
The common differences are respectively 4. (n 1) d must be added to a.) is a series.
17.
+
2 d. each term of which.. a + 2 d.11 246
(I)
Thus the 12th term of the
3
or 42.. P. the first
term a and
the
common difference d being given.
Since d
is
a
-f
3
d..1) d..
309. to produce the nth term.. P. a
3d.. An arithmetic progression (A. 3 d must be added to a.
.
of a series are its successive numbers.
To
find the
nth term
/
of an A. 12. and d. 2 d must be added to a.
11.
of the following series is
3. 19. P. The progression is a. a -f d.
10.
:
7.
a.
series 9.CHAPTER XX
PROGRESSIONS
307.
-4.
P.
1. P.
2
EXERCISE
1.
5.-..
9..
Or
Hence
Thus
from
(I)
= (+/). 3.
2.3 a = -l.
= -2..
Find the 101th term of the
series 1.
6.PROGRESSIONS
310.. -10...
2*=(a + Z) + (a + l) + (a + l)
2s = n
*
. 21. 5. -7.. a = 2..
5. 5.
-|.. 7. 6.
= 99.
.
3. d = 3. -4^.
.
= I + 49 = *({ +
.' cZ == .
if
a = 5. .
first
2
Write down the
(a)
(6)
(c)
6 terms of an A.
.
115..
Which
(6)
(c)
of the following series are in A. 5.. 19.
series
.
of the series 10.
.. 99) = 2600. 2J. ?
(a) 1.
(d) 1J. 8..
1.4.. 2
sum
of the first 60
I
(II)
to find the
' '
odd numbers..
7.-
(a
+ + (a +
l)
l).
Find the 10th term of the
series 17.
8. d
. 2.. P.
247
first
To
find the
sum s
19
of the first
n terms of an A.
1-J.
the last term
and the common difference d being given..
-3.
Find the 12th term of the
-4.
Find the nth term of the
series 2.16.
= a + (a
Reversing the order. -24. the
term
a.-.
series 2. 3. 4.
Find the 7th term of the Find the 21st term
series
.
Adding.
.
9.
Find the 5th term of the
4.
6
we have
Hence
. 8.8. 6. 3.
7. 29.
19.
>
2-f
2.
. striking hours only.
3. 15.
1+2+3+4H
Find the sum of the
first
n odd numbers.
1.
13.
17. 16.
-.
4. 31. and for each than for the preceding one. 1|.
18.
15.
ELEMENTS OF ALGEBRA
last
term and the sum of the following series :
.
(x +"l) 4.
1J.
to 7 terms.
.
+ 2-f-3 + 4 H
hlOO. 12.
.
6.
.
(i)
(ii)
.
23.
+ 3.
to 10 terms.
:
3.
8.
Q^) How many times
in 12 hours ?
(&fi)
does a clock.7 -f
to 12 terms. strike
for the first yard.
. 11. 2J.248
Find the
10. and a yearly increase of $ 120.
7.
.
Sum
the following series
14.
16.
to 15 terms.
11.
20.
21.
rf.
1.
\-n..5
H + i-f
-f-
to 10 terms.1 -f 3.
'.
to 20 terms. 7.
to 8 terms. hence if any three of them are given. P. Jive quantities are involved.(#
1
2) -f (x -f 3) H
to
a terms. In most problems relating to A. 11.
22.
to 20 terms. 11.
. the other two may be found by the solution of the simultaneous equations
.
33.
to 16 terms.
$1
For boring a well 60 yards deep a contractor receives yard thereafter 10^ more How much does he receive all
together ?
^S5 A bookkeeper accepts a position at a yearly salary of $ 1000. How much does he receive (a) in the 21st year (6) during the first 21 years ?
j
311.
to 20 terms.
2.
12.
15.
I. P. Find a Given a = 7. Find w. n
has the series
^
j
.
I
Find
I
in terms of a.250
ELEMENTS OF ALGEBRA
EXERCISE
116
:
Find the arithmetic means between
1.
= 16.
7.
6?
9.
14.
4. = 17.
. 74.
Between 4 and 8
insert 3 terms (arithmetic
is
means)
so
that an A.
10. Find d. n = 20.
15.
12.
y and #-f-5y. Find d and Given a = 1700. = ^ 3 = 1. 78.
and
s. d = 5. s = 70. ceding one. and all his savings in 5 years amounted to $ 6540.
13. n = 17. of 5 terms
6. = 52. n = 13. n = 4. n = 16.
16.
has the series 82. Given a = 4.
11.
How many terms How many terms
Given d = 3.
m
and
n
2.
3.
A
$300
is
divided
among 6 persons
in such a
way
that each
person receives $ 10 did each receive ?
more than the preceding
one.
17. = 83. Given a = . Find a and Given s = 44.
How much
.
a+
and
b
a
b
5.
Between 10 and 6
insert 7 arithmetic
means
.
f?
. Given a = |. = 1870. Find n.
8.3. How much did he save the first month?
19.
a x
-f-
b
and a
b. s == 440. Given a = 1. = 45.
produced. Find d. Find?.
T?
^.
f
J 1 1
/
.
man saved each month $2 more than in the pre 18. n.
The progression is a.
E. 36. +1.
2 a.
The
314. To find the sum s of the first n terms term a and the ratio r being given.)
is
a series each term of
which.
or 81
315.
|. -I.
.. or.
If
n
is less
:
than unity.
.
Therefore
Thus the sum
= ^ZlD.
..
4-
(1)
. rs =
s
2
-..
-2.
2
arn
(2)
Subtracting (1) from
(2).
of a G.
r
n~ l
..
fl
lg[(i)
-l]
==
32(W -
1)
= 332 J. P.
ratios are respectively 3.
s(r
1)
8
= ar"
7*
JL
a. <zr . called the ratio. P.
the following form 8
nf +
q(l-r")
1
r
.
4.
(II)
of the
8 =s
first
6 terms of the series 16.
.. 36.PROGRESSIONS
251
GEOMETRIC PROGRESSION
313.
is
16(f)
4
.
(I)
of the series 16.
g==
it is
convenient to write formula' (II) in
*. P. the first
= a + ar -for ar -f ar Multiplying by r.
and
To
find the
nth term
/ of
a G. 36.
.
Hence
Thus the 6th term
l
= ar
n~l
. ar.. except the
multiplying
derived from the preceding one by by a constant number..
4. is
it
(G.g..arn ~ l . 24.
A geometric progression
first.
NOTE.
ar8
r. a?*2 To obtain the nth term a must evidently be multiplied by
. 12.. 108. 24.. the first term a and
the ratios r being given.
80.
7.*.
.
series
Find the llth term of the Find the 7th term of the
ratio is
^.
. if any three of them are given.
Find the 7th term of the Find the 6th term of the
Find the 9th term of the
^._!=!>. 9.
(it.
Write down the first 5 terms of a G. 144.
is 3.5.288.252
ELEMENTS OF ALGEBRA
316.
i 288.
.
l.
72. 36. whose and whose second term is 8.
Find the 6th term of the
series J.
|. 3.18.
Evidently the total
number
of terms is 5
+ 2. 20.
0.
-fa. 4.
288.
. 676.
.l..
9.72.
or
7.
\
t
series
.
r^2.4. 9. Jive quantities are in.
..
Hence n
=
7.
2
term
3..
576.
117
Which
(a)
of the following series are in G. the other two be found by the solution of the simultaneous equations :
may
(I)
/=<!/-'.
.
To
insert 5 geometric
means between 9 and 576. f...
+-f%9 %
. 72..
f. whose and whose common ratio is 4.-.
676
t
Substituting in
= r6 = 64.
Write down the first 6 terms of a G.
10.18. 144. ?
(c)
2.
first
term
is
125 and
whose common
.
(b) 1.
EXERCISE
1..
. P..
144.
And the
required
means are
18.
.
is 16.
Hence the
or
series is
0. I
= 670.
6. In most problems relating to G.
.
a
=
I. hence.
-fa.
(d) 5.
series 5. 18.
8.
first
5.
4.
first
term
4. 1.. P.... 25. whose
. P.
volved .
Find the 5th term of a G. P.54.
series
.
series 6. P.6.5..
36.
.
.
Ex. + 5.
. 36.
6 in each row the lowest row has 2 panes of glass in each window more than the middle row. was three times that of the younger. and 5 h. the
sum
of the ages of all three is 51. side were one foot longer.
+
a.
. and the father's present age is twice what the son will be 8 years
hence.
dimension
182. and the middle row has 4 panes in each window more than the upper row there are in all 168 panes of glass.56. Four years ago a father was three times as old as his son is now. Find the age
5 years older than his sister
183. z 2
-92.
187.
186. How many are there in each window ?
.
3 gives the
174.
.
188.
train.
two boys is twice that of the younger.
7/
191.
A
each
177.
What
is
the distance?
if
square grass plot would contain 73 square feet more Find the side of the plot.
ELEMENTS OF ALGEBRA
A A
number increased by
3. z 2 + x .
176.
-ll?/-102.
180.
178.
younger than his Find the age of
the father.
number divided by
3.
The age
of the elder of
it
three years ago of each. aW + llab-2&. + 11 ~ 6. Find the dimensions of the floor.
An
The two
express train runs 7 miles an hour faster than an ordinary trains run a certain distance in 4 h.
and | as old as his Find the age of the
Resolve into prime factors
:
184.
.
A
the
boy
is
as old as his father
and
3 years
sum
of the ages of the three is 57 years.
respectively.
190. 189.
3 gives the
same
result as the
numbet
multiplied by
Find the number.
179.
power one of the two Find the power of each.
181.
sister
.
if
each
increased 2 feet.
13 a + 3.-36.
A
boy
is
father. 12 m.
is
What are their ages ? Two engines are together
more than the
of 80 horse
16 horse power
other.
2
2
+
a
_ no. A house has 3 rows of windows.
Find the number.
The length
is
of a floor exceeds its width
by 2
feet.
+x-
2.266
173.
father. 4 a 2
y-y
-42. 15 m. 10x 2 192. x*
185.
same
result as the
number
diminished by
175. the ana of the floor will be increased 48 square feet.
he takes 7 minutes longer than in going. x
1
a
x
x1
ab
1
1
a
x
a
c
+
b
c
x
a
b
b
~
c
x
b
416
417.(c rt
a)(x
-
b)
=
0. and was out 5 hours. Find the number.
418 ~j-o.
mx ~
nx
(a
~
mx
nx
c
d
d
c)(:r
lfi:r
a
b)(x
. In a
if
and
422. 18 be subtracted from the number.
-f
a
x
-f
x
-f c
1
1
a-b
b
x
415. the order of the digits will be inverted.a)(x b
b)
(x
b
~
)
412. Find the number of miles an hour that A and B each walk.
a
x
)
~
a
2 b
2
ar
a
IJ a. How long is each road ?
423.
(x
-f
ELEMENTS OF ALGEBRA
a)(z
-
b)
=
a
2 alb
=
a
(x
-f
b)(x
2
.
hour. and at the rate of 3^ miles an hour.
a
x
a
x
b
b
x
c
b
_a
b
-f
x
414. far did he walk all together ?
A
.
2 a
x
c
x
6
-f c
a
+
a
+
a
+
6
-f
walks 2 miles more than B walks in 7 hours more than A walks in 5 hours. Tn 6 hours
.c) .278
410.
421.
A man
drives to a certain place at the rate of 8 miles an
Returning by a road 3 miles longer at the rate of 9 miles an hour.
420.
down again
How
person walks up a hill at the rate of 2 miles an hour.
-
a)
-2
6 2a.
(x
.
4x
a
a
2 c
6
Qx
3 x
c
419.
A
in 9 hours
B walks
11 miles
number of two digits the first digit is twice the second.(5 I2x
~r
l
a)
. 411.
. Find the sum and the rate of
interest. also a third of the greater exceeds half the less by 2. What is that fraction which becomes f when its numerator is doubled and its denominator is increased by 1. and the other number least. and 5 times the less exceeds the greater by 3. A sum of money at simple interest amounted in 10 months to $2100. Find the
fraction. Find the numbers.
least
The sum
of three
numbers
is
is
21.282
ELEMENTS OF ALGEBRA
476.
A
number
consists of
two
digits
4. and becomes when its denominator is doubled and its numerator increased by 4 ?
j|
478.
if
the
sum of
the digits be multiplied by
the digits will be inverted. There are two numbers the half of the greater of which exceeds the less by 2. age.
latter
would then be twice the
son's
A
and B together have $6000. Find their ages.
485.
Find
the number. In a certain proper fraction the difference between the nu merator and the denominator is 12. A sum of money at simple interest amounts in 8 months to $260. Find the numbers.
half the
The greatest exceeds the sum of the greatest and
480. Of the ages of two brothers one exceeds half the other by 4 is equal to an eighth of 482.
A
spends \ of his. and in 18 months to $2180.
486. and if each be increased by 5 the Find the fraction. had each at first?
B
B
then has
J
as
much
spends } of his money and as A.
487. If 31 years were added to the age of a father it would be also if one year were taken from the son's age
.
to
. Find the principal and the rate of
interest.
by 4.
thrice that of his son
and added to the father's.
483. Find two numbers such that twice the greater exceeds the by 30. fraction becomes equal to |. How much money
less
484. and a fifth part of one brother's age that of the other. the Find their ages.
If 1
be added to the numerator of a fraction
it
if 1
be added to the denominator
it becomes equal becomes equal to ^. and in 20 months to $275.
477.
481.
479.
whose difference
is
4.
years.
L. A vessel can be filled by three pipes.
530. A can do a piece of work in 12 days B and C together can do the same piece of work in 4 days A and C can do it in half the time in which B alone can do it. A boy is a years old his mother was I years old when he was born. in 28 minutes.
. Find the numbers. AB=6. Tf and run together. if and L. and 23 pounds of lead lose 2 pounds. (a) How many pounds of tin and lead are in a mixture weighing 120 pounds in air. and BE. the first and second digits will change places.
sum of the reciprocals of of the reciprocals of the first of the reciprocals of the second and
the
sum
528. Tu what time will it be filled if all run
M
N
N
t
together?
529. In
circle
A ABC. BC = 5. B and C and C and A in 4 days. How long will B and C take to do
.
and B together can do a piece of work in 2 days.
touches
and
F respectively.
AC
in /).
and CA=7. CD. if L and Af in 20 minutes.
532. his father is half as old again as his mother was c years ago.
. M.
E
533.
and third equals \\ the sum third equals \.
it
separately
?
531. they would have met in 2 hours. Find the present ages of his father and mother. When weighed in water. and one overtakes the other in 6 hours. In how many days can each alone do the same work?
526. An (escribed) and the prolongations of BA and BC in Find AD. 90.
527. Two persons start to travel from two stations 24 miles apart. What are their rates of
travel?
. N. if the number be increased by Find the number.REVIEW EXERCISE
285
525. If they had walked toward each other. and losing 14 pounds when weighed in water? (b) How many pounds of tin and lead are in an alloy weighing 220 pounds in air and 201 pounds in water ?
in 3 days. A number of three digits whose first and last digits are the same has 7 for the sum of its digits. it is filled in 35 minutes. Throe numbers are such that the
A
the
first
and second equals
. 37 pounds of tin lose 5 pounds.
Draw the graph of y 2 and from the diagram determine
:
+
2 x
x*.
-
7.
2. z 2
-
x x
-
5. 2
-
x
-
x2
.
The values of y.
-
3 x. The value of x that produces the greatest value of y.
550.
from x
=
2 to x
= 4. the function. 3 x
539. formation of dollars into marks. then / = 3 and write
=
3.10 marks. GERMANY. One dollar equals 4. x 8
549.
Draw
the graphs of the following functions
:
538. 2 x
+
5. 2|.
of
Draw
a graph for the trans-
The number
in
of
workmen Draw
required to finish a certain piece
the graph
work
D
days
it
is
from
D
1 to
D=
12.3
Draw
down
the time of swing for a
pendulum
of length
8 feet. i.
542. x*
-
2
x.
b. If
to
feet is the length of a
seconds.
-
3 x.
546.
c.
.
e. Represent the following table graphically
TABLE OF POPULATION (IN MILLIONS) OF UNITED STATES.
+
3. 2 541.
d.
a. The values of x if y = 2. x
2
+
x.
548.e. FRANCE.
How
is
t /
long will
I
take 11
men
2
t' .
x*. the time of whose swing a graph for the formula from / =0
537.
to
do the work? pendulum. x *-x
+
x
+
1.
540.
. The greatest value of the function.
545. AND BRITISH ISLES
535.
547. The roots of the equation 2 + 2 x x z = 1.
543. if x = f 1. x 2 544. 536.286
ELEMENTS OF ALGEBRA
:
534.
25 might have bought five more for the same money.l
+
8
-8
+
ft)'
(J)-*
(3|)*
+
(a
+
64-
+ i.
722. Find two consecutive numbers whose product equals 600. he
many
312?
he had waited a few days until each share had fallen $6.40 a 2* 2 + 9 a 4 = 0.
The area
the price of 100 apples by $1.
What two numbers
are those whose
sum
is
47 and product
A man
bought a certain number of pounds of tea and
10 pounds more of coffee. 725. If a pound of tea cost 30 J* more than a pound of coffee. 721.
729.292
709. paying $ 12 for the tea and $9 for the coffee.
217
. What number exceeds its reciprocal by {$. if 1 more for 30/ would diminish
720. A man bought a certain number of shares in a company for
$375.
sum is a and whose product equals J. and working together they can build it in 18 days. 12
-4*+
-
8. In how many days can A build the wall?
718. what is the
price of the coffee per
pound ?
:
Find the numerical value of
728.
716.
___ _ 2* -5 3*2-7
715. Find two numbers whose 719. Find four consecutive integers whose product is 7920. 727. 2n n 2 2 -f-2aar + a -5 = 0.
714
2
*2
'
+
25
4
16
|
25 a2
711. How shares did he buy ?
if
726. **-13a: 2
710.44#2 + 121 = 0.
A
equals CO feet. 724.
723. Find the price of an apple.
a:
713. Find the altitude of an equilateral triangle whose side equals a.
in value.
The
difference of the cubes of
two consecutive numbers
is
find them. 16 x* .
of a rectangle is 221 square feet and its perimeter Find the dimensions of the rectangle.
.
ELEMENTS OF ALGEBRA
+36 = 0. 3or
i
-16 . 717. needs 15 days longer to build a wall than B.
rate each
man
ran in the
first
heat.
two squares equals 140 feet. In the first heat B reaches the winning post 2 minutes before A.
944. Find the length and breadth of the first rectangle. z(* + y + 2) = 76.
943. The difference of two numbers cubes is 513. (y + *) = .102. and also contains 300 square feet.
942. feet. The perimeter of a rectangle is 92 Find the area of the rectangle.
y(
934. and 10 feet broader. The
sum
of the circumferences of
44 inches.300
930. s(y
932. (3 + *)(ar + y + z) = 96.
.
feet.
ELEMENTS OF ALGEBRA
(*+s)(* + y)=10. = ar(a? -f y + 2) + a)(* + y
933. Find the sides of the rectangle. Find the numbers.
the difference of their
The
is
difference of
their cubes
270.
two numbers Find the numbers. the area of the new rectangle would equal 170 square feet.
two squares is 23 feet.
931.
much and A then
Find at what
increases his speed 2 miles per hour.
937.
the
The sum
of the perimeters of
sum
of their areas equals 617 square feet.square inches.
34
939.
and the
difference of
936. and B diminishes his as arrives at the winning post 2 minutes before B. If each side was increased by 2 feet. In the second
heat
A
.
The sum
of the perimeters of
sum
of the areas of the squares is 16^f feet.
diagonal
940.
935. a second rec8 feet shorter. and the Find the sides of the and
its
is
squares.
is
3
.
is 3.
2240. Tf there had been 20 less rows. A and B run a race round a two-mile course. y(x + y + 2) = 133.
Assuming
= -y. there would have been 25 more trees in a row. 152. *(* + #) =24.
and the sum of
their areas 78$.
is 20. Find the side of each two
circles is
IT
square. (y
(* + y)(y +*)= 50. How many rows are there?
941.
and the sum of their cubes
is
tangle
certain rectangle contains 300 square feet.
+ z)=18.
The sum of two numbers Find the numbers. A plantation in rows consists of 10.000 trees.
A
is
938.
find
the radii of the two circles. + z) =108. The diagonal of a rectangle equals 17 feet.
Find the number.
. If the breadth of the rectangle be decreased by 1 inch and its
is
length increased by 2 inches. A number consists of three digits whose sum is 14.
951. set out from two places. each block. The diagonal of a rectangular is 476 yards.
Find the
eter
947. the digits are reversed. at Find the his rate of traveling.
Find the number. and if 594 be added to the number. and
its
perim-
948. The area of a certain rectangle is 2400 square feet.
. The sum of the contents of two cubic blocks
the
of the heights of the blocks is 11 feet.
953.
950. was 9 hours' journey distant from P. and that B. A rectangular lawn whose length is 30 yards and breadth 20 yards is surrounded by a path of uniform width.
overtook
miles.
sum
Find an edge of
954.
whose
946. and travels in the same direction as A. and the other 9 days longer to perform the work than if both worked together. at
the same time
A
it
starts
and
B
from
Q
with the design to pass through Q. Find two numbers each of which
is
the square of the other. The area of a certain rectangle is equal to the area of a square side is 3 inches longer than one of the sides of the rectangle. What is its area?
field is
182 yards.
P and
Q. Find in what time both will do it.
952.
949.
. Find its length and breadth. the difference in the lengths of the legs of the Find the legs of the triangle.
unaltered. that
B
A
955. When
from
P
A
was found that they had together traveled 80 had passed through Q 4 hours before. Two men can perform a piece of work in a certain time one takes 4 days longer.
is
407 cubic feet.REVIEW EXERCISE
301
945. the square of the middle digit is equal to the product of the extreme digits.
Two
starts
travelers. the area lengths of the sides of the rectangle. triangle is 6. A certain number exceeds the product of its two digits by 52 and exceeds twice the sum of its digits by 53.
A and
B. Find the width of the path if its area is 216 square yards. its area will be increased 100 square feet. The square described on the hypotenuse of a right triangle is 180 square inches. if its length is decreased 10 feet and its breadth increased 10 feet. distance between P and Q.
001
4.
Find the number of grains which Sessa should have received.-.
all
A
perfect number
is
a number which equals the sum
divisible.
1. The
term.
990.
How many
sum
terms of 18
+
17
+
10
+
amount
. Insert 8 arithmetic means between
1
and
-.
992. 5
11. who rewarded the inventor by promising to place 1 grain of wheat on
Sessa for the
the 1st square of a chess-board. and the sum of the first nine terms is equal to the square of the sum of the first two.
989.)
the last term
the series
a perfect number.REVIEW EXERCISE
978. Find the value of the infinite product 4
v'i
v7-!
v^5
.2
. named Sheran. is 225.
987.
. and of the second and third 03. Find the sum of the series
988. then this
sum multiplied by
(Euclid. Insert 22 arithmetic means between 8 and 54..04
+
.
What
2 a
value must a have so that the
sum
of
+
av/2
+
a
+
V2
+
.
first
984. and the
common
difference.
If
of
2
of
integers + 2 1 + 2'2
by which
is
it
is
the
sum
of
the series
2 n is prime.
986.
985.-.---
:
+
9
-
-
V2
+
.
v/2
1
+ +
+
1
4
+
+
3>/2
to oo
+
+
.
of n terms of an A. The 21st term of an A. The Arabian Araphad reports that chess was invented by amusement of an Indian rajah.
to infinity
may
be 8?
.
Find
n. The sum
982..
Find four perfect
numbers. 2 grains on the 2d.1
+
2. P.
to oo.
0. such that the product of the and fourth may be 55. doubling the number for each successive square on the board. and so on...
980.+ lY L V. P. Find the
first
term.001
+
.3 '
Find the 8th
983..
to
n terms. P. Find four numbers in A..
to 105?
981. 4 grains on the 3d.01
3.
"(.
303
979.
of n terms of 7
+
9
+ 11+
is
is
40.
prove that they cannot be in A. ft.
999.
is 4. The other travels 8 miles the first day and After how increases this pace by \ mile a day each succeeding day. The side of an equilateral triangle equals 2. 995.
The sum and sum
. (a) after 5 strokes.
and the
fifth
term
is
8 times
the second . after how strokes would the density of the air be xJn ^ ^ ne original
density ?
a circle is inscribed.304
ELEMENTS OF ALGEBRA
993.
third circle touches the second circle and the
to infinity. The sum and product of three numbers in G. The
fifth
term of a G. are
45 and 765
find the
numbers.
994. in this circle a square. In an equilateral triangle second circle touches the first circle and the sides AB and AC. Find (a) the sum of all circumferences.
ABC
A A
n
same
sides.
1000. are 28 and find the numbers. The sides of a second equilateral triangle equal the altitudes of the first.
pump removes
J
of the
of air
is
fractions of the original amount contained in the receiver. Two travelers start on the same road.
. Under the conditions of the preceding example. Insert 3 geometric means between 2 and 162. 1003. many days will the latter overtake the former?
. and G. One of them travels uniformly 10 miles a day. P. P. P. inches. find the series.
998. P. the sides
of a third triangle equal the altitudes of the second. areas of all triangles. Insert 4 geometric means between 243 and 32.
of squares of four
numbers
in
G. at the same time.
and
if
so forth
What
is
the
sum
of the areas of all circles. Each stroke of the piston of an air
air contained in the receiver. are unequal. in this square a circle.
1001. (I) the sum of the perimeters of
all
squares. In a circle whose radius is 1 a square is inscribed. and so forth to infinity. (6) after n
What
strokes?
many
1002. and so forth to Find (a) the sum of all perimeters. (6) the sum of the infinity.
997.
512
996. P. c. If a.
AB =
1004.
Half
leather. without the sacrifice of scientific accuracy and thoroughness. given. The author
has emphasized Graphical Methods more than is usual in text-books of this grade. physics. To meet the requirements of the College Entrance Examination Board. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume.
A
examples are taken from geometry.
xiv+563
pages. great many
work. The introsimpler and more natural than the
methods given
In Factoring.25
lamo. and the Summation of Series is here presented in a novel form.
i2mo. which has been retained to serve as a basis for higher work. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
The Exercises are superficial study of a great many cases.ELEMENTARY ALGEBRA
By ARTHUR SCHULTZE.10
The treatment of elementary algebra here is simple and practical. especially
duction into Problem
Work
is
very
much
Problems and Factoring. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE.
Half leather. but none of the introduced illustrations is so complex as to require the expenditure of
time for the teaching of physics or geometry. than by the
.
$1. book is a thoroughly practical and comprehensive text-book. The more important subjects
tions. etc. save Inequalities. and commercial life.
HEW TOSS
.
not
The Advanced Algebra is an amplification of the Elementary. Particular care has been bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner. Ph.D.
xi 4-
373 pages. 64-66 FIFTH AVBNTC. proportions and graphical methods are introduced into the first year's course. very numerous and well graded there is a sufficient number of easy examples of each kind to enable the weakest students to do some work.
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix.
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THE MACMILLAN COMPANY
PUBLISHERS. so that the Logarithms.
HatF leather. comparatively few methods are
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Half leather. than by the superficial study of a great many cases. great many
A
examples are taken from geometry.D. especially
duction into Problem
Work
is
very
much
Problems and Factoring. etc. book is a thoroughly practical and comprehensive text-book. 64-66
7HTH
AVENUE. The author
grade. proportions and graphical methods are introduced into the first year's course. save Inequalities. there is a sufficient number of easy examples of each kind to enable the weakest students to do some work. but the work in the latter subject
has been so arranged that teachers
who
wish a shorter course
may omit
it
ADVANCED ALGEBRA
By ARTHUR SCHULTZE.
not
The Advanced Algebra is an amplification of the Elementary. but none of the introduced illustrations is so complex as to require the expenditure of
time for the teaching of physics or geometry. without
Particular care has been the sacrifice of scientific accuracy and thoroughness. All subjects now required for admission by the College Entrance Examination Board
have been omitted from the present volume. The Exercises are very numerous and well graded.
xi
-f-
373 pages.
HEW YOKE
.ELEMENTARY ALGEBRA
By ARTHUR Sen ULTZE.
xiv+56a
pages.25
i2mo. Logarithms. which has been retained to serve as a basis for higher work.
$1.
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PUBLISHBSS. and commercial life.
In Factoring. The more important subjects
which have been omitted from the body of the work Indeterminate Equahave been relegated to the Appendix. so that the tions. bestowed upon those chapters which in the customary courses offer the greatest difficulties to the beginner.
$1. but these few are treated so thoroughly and are illustrated by so many varied examples that the student will be much better prepared for further
work. Ph.
has emphasized Graphical Methods more than is usual in text-books of this and the Summation of Series is here presented in a novel form. physics. To meet the requirements of the College Entrance Examination Board.
12010.10
The treatment
of elementary algebra here
is
simple and practical. The introsimpler and more natural than the
methods given heretofore.
of Propositions has a
Propositions easily understood are given first and more difficult ones follow .
PLANE AND SOLID GEOMETRY
F.
Cloth. State: . Many proofs are presented in a simpler and manner than in most text-books in Geometry 8.
lines.
NEW YORK
. guides him in putting forth his efforts to the best
advantage. 7 he
. 6.
xii
+ 233 pages. Cloth. Attention is invited to the following important features I.
Algebraic Solution of Geometrical Exercises is treated in the Appendix to the Plane Geometry . The Schultze and Sevenoak Geometry is in use in a large number of the leading schools of the country. 10. SCHULTZE. Preliminary Propositions are presented in a simple manner . more than 1200 in number in 2. Proofs that are special cases of general principles obtained from the Exercises are not given in detail.
. Pains have been taken to give Excellent Figures
throughout the book. iamo. The Analysis of Problems and of Theorems is more concrete and practical than in any other
distinct pedagogical value.r and.
KEY TO THE EXERCISES
in
Schultze and Sevenoak's Plane and Solid Geometry.
text-book in Geometry
more
direct
ositions
7.
xtt-t
PLANE GEOMETRY
Separate.
ments from which General Principles may be obtained are inserted in the " Exercises. Difficult Propare made somewhat? easier by applying simple Notation .10
L.
i2mo.
izmo. 9.
SEVENOAK.
and no attempt has been made
to present these solutions in such form
that they can be used as models for class-room work. at the
It
same
provides a course which stimulates him to do original time. 64-66 FIFTH AVENUE.D.
Half
leather..
THE MACMILLAN COMPANY
PUBLISHERS.
By ARTHUR SCHULTZE and
370 pages.
wor. The numerous and well-graded Exercises the complete book. aoo pages. 4. under the heading Remarks".
$1.
80 cents
This Geometry introduces the student systematically to the solution of geometrical exercises.10
By ARTHUR
This key will be helpful to teachers who cannot give sufficient time to the Most solutions are merely outsolution of the exercises in the text-book. Hints as to the manner of completing the work are inserted The Order 5. Ph. $1. These are introduced from the beginning 3.
370 pages.
Students
to
still
learn
demon-
strations instead of learning
how
demonstrate. $1. making mathematical teaching less informational and more disciplinary.
causes of the inefficiency of mathematical teaching. and Assistant Professor of Mathematics in New York University
of
Cloth. New York City.
.
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64-66 Fifth Avenue. enable him to
" The chief object of the speak with unusual authority.
.
12mo.
methods of teaching mathematics the first propositions in geometry the original exercise parallel lines methods of the circle attacking problems impossible constructions applied problems typical parts of algebra.
.
New York
DALLAS
CHICAGO
BOSTON
SAN FRANCISCO
ATLANTA
.
. " is to contribute towards book/ he says in the preface.The Teaching
of
Mathematics
in
Secondary Schools
ARTHUR SCHULTZE
Formerly Head of the Department of Mathematics in the High School Commerce. and not from the information that it imparts. a great deal of mathematical spite
teaching
is
still
informational.
. of these theoretical views.
. Typical topics the value and the aims of mathematical teach-
ing
.25
The
author's long
and successful experience as a teacher
of mathematics in secondary schools and his careful study of the subject from the pedagogical point of view."
The treatment
treated are
:
is concrete and practical.
. Most teachers admit that mathematical instruction derives its importance from the mental training that it But in affords.
.
The book deserves the attention
of history teachers/'
Journal of Pedagogy.
New York
SAN FRANCISCO
BOSTON
CHICAGO
ATLANTA
.
but in being fully illustrated with
many excellent maps.
$1.AMERICAN HISTORY
For Use
fa
Secondary Schools
By ROSCOE LEWIS ASHLEY
Illustrated. Studies and Questions at the end of each chapter take the place of the individual teacher's lesson plans. and a full index are provided. which put the main stress upon national development rather than upon military campaigns. " This volume
etc.
Cloth. which have been selected with great care and can be found in the average high school library. Maps.
is
an excellent example of the newer type of
school histories.
diagrams.
An exhaustive system of marginal references. The author's aim is to keep constantly before the
This book
pupil's mind the general movements in American history and their relative value in the development of our nation.
Topics. diagrams.40
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i2mo. All
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Synopses & Reviews
Publisher Comments
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Now, there is a complete introduction to numerical methods and visualization with the latest, most powerful version of MATLAB, Version 6.0. Dr. Shoichiro Nakamura introduces the skills and knowledge needed to solve numerical equations with MATLAB, understand the computational results, and present them graphically.
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About the Author
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Table of Contents
Preface.
1. MATLAB Primer.
Before Starting Calculations. How to Do Calculations. Branch Statements. Loops with for/end or while/end. Reading and Writing. Array Variables. Unique Aspect of Numbers in MATLAB. Mathematical Functions of MATLAB. Functions That Do Chores. Developing a Program as an M-File. How to Write Your Own Functions. Saving and Loading Data. How to Make Hard Copies. | 677.169 | 1 |
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