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2) {4, 5} 2. The product (2a 3)(2a 3) can be written as (3) {4, 5, 6} (4) {5, 6} (1) 2a2 9 (2) 4a2 9 (3) 4a2 9 (4) 4a2 12a 9 3. When 0.00034 is written in the form 3.4 10n, the value of n is (4) 4 (1) 3 (2) 4 (3. When (1) 5, x equals (2) 1 (3) 1 2 5. The mean of the set of even integers from 2 to 100 is (1) 49 (2) 50 (3) 51 (4) 1 (4) 52 6. The probability that 9 is the sum of the numbers that appear when two dice are rolled is (1) 4 6 (2) 2 36 (3) 2 6 (4) 4 36 7. If the circumference of a circle is 12 centimeters, then the area of the circle is (1) 36 square centimeters (2) 144 square centimeters (3) (4) 36 p 144 p square centimeters square centimeters 8. Which of the following is not an equation of a function? (1) y 3x 2 (2) y x2 3x 1 9. The value of 10P8 is (1) 80 (2) 90 (3) y2 x (4) y x (3) 1,814,400 (4) 3,628,800 10. Which of the following is an equation of a line parallel to the line whose equation is (1) 2x y 7 y 5 22x 1 4? (2) y 2x 7 (3) 2x y 7 (4) y 2x 7 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitu- 730 Statistics tions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. In a bridge club, there are three more women than men. How many persons are members of the club if the probability that a member chosen at random, is a woman is? 3 5 12. Find to the nearest degree the measure of the smallest angle in a right tri- angle whose sides measure 12, 35, and 37 inches. Part
III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. The lengths of the sides of a triangle are in the ratio 3 : 5 : 6. The perimeter of the triangle is 49.0 meters. What is the length of each side of the triangle? 14. Huy worked on an assignment for four days. Each day he worked half as long as he worked the day before and spent a total of 3.75 hours on the assignment. a. How long did Huy work on the assignment each day? b. Find the mean number of hours that Huy worked each day. Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The perimeter of a garden is 16 feet. Let x represent the width of the garden. a. Write an equation for the area of the land, y, in terms of x. b. Sketch the graph of the equation that you wrote in a. c. What is the maximum area of the land? 16. Each morning, Malcolm leaves for school at 8:00 o’clock. His brother Marvin leaves for the same school at 8:15. Malcolm walks at 2 miles an hour and Marvin rides his bicycle at 8 miles an hour. They follow the same route to school and arrive at the same time. a. At what time do Malcolm and Marvin arrive at school? b. How far is the school from their home? INDEX A Abscissa, 76 Absolute value, 6–7, 55 Absolute value function, 382–386 transformations of, reflection, 386 scaling, 386 translation, 385–386 Accuracy, 31 Acute angle, 249 Acute triangle, 263 Addition, 38 of algebraic expressions, 168–171 of algebraic fractions, 550–554 associative property of, 47–48 commutative property of, 47 distributive property of multiplication over, 48–49 of radicals, 487–489 of like radicals, 487 of unlike radicals, 488–489 of signed numbers, 54–58 with opposite signs, 56–58 with same signs, 54
–55 verbal phrases involving, 89 Addition method in solving systems of linear equations, 416–420 Addition property of equality, 118 of inequality, 147–148 of zero, 49–50 Additive identity, 49–50 Additive inverse (opposite), 50 Adjacent angles, 250 Adjacent side of an angle, 307 Algebraic equation, 565 Algebraic expression(s), 89 addition of, 168–171 evaluating, 100–102 subtraction of, 168–171 terms in, 95–96, 168 like, 123, 168 unlike, 123, 168 translating verbal phrases into, 89–90, 92–93 writing in words, 98–99, Algebraic fraction(s), 540 addition of, 550–554 division of, 548–549 multiplication of, 545–547 subtraction of, 550–554 Algebraic solution of quadratic-linear Area system, 529–531 Alternate exterior angle(s), 259 parallel lines and, 260 Alternate interior angle(s), 258 parallel lines and, 259 Angle(s), 248–249 acute, 249 adjacent, 250 alternate exterior, 259 alternate interior, 258 base, 265 classifying triangles according to, 263 complementary, 250–251 congruent, 252 consecutive, 272 corresponding, 259 cosine of, 317 definition of, 248 of depression, 313 of elevation, 313 exterior, 258 interior, 258 on the same side of transversal, 259 linear pair of, 252 measuring, 248 obtuse, 249 opposite, 272 pairs of, 250–254 perpendicular, 249 parallel lines and, 258–261 parallelograms and, informal proofs, 274–275 polygons and, 275 of quadrilaterals, 273 right, 248 sides of an, 248 sine of, 317 straight, 249 sum of the measures of, for polygons, 275 for triangles, 263–265 supplementary, 251 tangent of, 308 types of, 248–249 vertical, 252–253 Angle bisector, 271 Approximation, 20–21 of numbers, 469 rational, 20, 477–478 of irregular polygons, 279–280 surface, 282–284 Arithmetic averages in, 680 order of operations in, 38–43 Arithmetic mean, 681 Ascending order, polynomials in, 170 Associative property of addition, 47–48 of multiplication, 48 Average(s). See also Central tendency in arithmetic, 680 numerical, 681 in statistics, 680–681 Axiom
(postulate), 246 Axis, graphing line parallel to, 352–353 Axis of symmetry of the parabola, 510 B Bar (fractional line), 41 Base(s), in geometry, angles of an isosceles triangle, 265 of a prism, 282 of a trapezoid, 275 of an isosceles triangle, 265 in number theory, division of powers with the same, 186–187 of exponents, 39–40 multiplication of powers with the same, 173–174 in percent, 227 powers of, 95–96 Bias, 662 Biased object, 579 Bimodal data, 683 Binary operation, 38 Binomial(s), 169 division of polynomial by, 200–201 multiplication of, 454–455 Bisector angle, 271 perpendicular, 271 Bivariate statistics, 710–720 causation, 711–715 correlation, 711–715 line of best fit, 715–720 extrapolation, 717 interpolation, 717 732 Index Bivariate statistics, line of best fit cont. regression line, 715 scatter plot, 710–715 time series, 715 Blinding, 663 single-blind experiments, 663 double-blind experiments, 664 Box-and-whisker plot, 699–701 Brackets, as grouping symbol, 41 C Calculated probability, 585. See also Theoretical probability Calculator, algebraic expressions, entering variables, 101–102, 130 equations, intersect feature, 526–527 solving systems of equations, 414, 526–527 testing for equality, 129–130, 218 graphing functions on, 349–350, 352 absolute value, 384 CALC, 513–514 exponential, 390 quadratic, 512–515 TBLSET, 515 TRACE, 349–350, 414 vertex, finding, 512–514 ZOOM, 349–350 number theory, additive inverses (opposites), 50 decimals, 13–14 converting to a fraction, 52, 209–210 FRAC, 52, 209–210 exponents, 40, 67 fractions, comparing, 12 converting to a decimal, 13–14 mixed, 57–58 grouping symbols, 42–43 multiplicative inverses (reciprocals), 51 negative symbol, 61 numerical expressions, 3–4 reciprocals, 51 roots, general, 472–475 square roots, 18–19 scientific notation, 193–
195 squaring numbers, 18, 470–472 probability, combinations, 642 factorials, 628–629 permutations, 631–634 statistics, 1-Var Stats, 684, 695 box-and-whisker plot, 700 five statistical summary, 700 frequency histogram, 676–678 line of best fit, 715–716 mean, 684 median, 684 scatter plot, 711–712 trigonometry, cosine ratios, 319–320 degree mode, 309 sine ratios, 319–320 tangent ratios, 308–311 Cancellation, 542 Cancellation method, 545 Cards, standard deck of, 580 Causation, 711 Check, in solving equations, 119 Census, 662 Center of a sphere, 286 Central tendency (average), 690–695 and linear transformations, 686 mean, 681 for grouped data, 691–692 for intervals, 694–695 median, 681–682 for grouped data, 691 for intervals, 694 mode, 682–683 for grouped data, 690 for intervals (modal interval), 693 Certainty, 591 Closure, property of, 45–47 Coefficient, 95 Combination(s), 639–644, 646 comparing permutations and, 639–641 relationships involving, 641–644 Commas in verbal phrases, 90 Common denominator, 551 Common fraction, 13 Common monomial factor, 447 Commutative property of addition, 47, 170 of multiplication, 47 Compass, 270 Complement of a set, 73 Complementary angles, 250–251 Completeness property of real numbers, 25 Composite number, 39, 443 Compound event, 609 Compound interest, 389 Computations with more than one operation, 41 Conditional equation, 117 Conditional probability, 619–623 Cone, volume of, 288 Congruent line segments, 265 Congruent angles, 252 Consecutive angles, 272 Consistent system of equations, 410–411 Constant of variation, 222 Constant ratio, 222 Construction, geometric, 270 angle bisector, 271 congruent angles, 270 congruent line segments, 270 perpendicular bisector, 271 Continued ratio, 209–210 Control group, 663 Convergence, 579 Coordinate axes, 75 Coordinate plane, 75 locating a point on, 76–77 Coordinates of a point, 76 finding on a plane, 77–78 Correlation, 7
11–715 Corresponding angles, 259 parallel lines and, 259–260 Cosine of an angle, 317 Cosine ratio, 318 applications of, 323–324 finding, on calculator, 319–320 Counting numbers, 2, 11 Counting principle, 610–611, 612 Cross-multiplication, 217 Cross-product, 217 Cube root, 472 Cumulative frequency, 702–703 Cumulative frequency histogram, 703–707 Cumulative relative frequency, 578–579 Cylinder(s) right circular, 283 surface area of, 283 D Data, 661 collection of, 661–665 bias, 662 sampling, 662 techniques of, 662–663 grouped, 668–670, 690–695 organization of, 667–672 frequency distribution table, 668 grouped data, 668–670 cumulative frequency, 702–703 types of, univariate, 710 bivariate, 710 qualitative, 661 quantitative, 661 visualization of data, interpreting graphs of data, 664–665 types of, box-and-whisker plot, 699–701 cumulative frequency histogram, 703–707 scatter plot, 711–714 stem-and-leaf diagram, 670–672 histogram, 675–678 Decimal(s) equivalent, 586–587 expressing as rational numbers, 15 expressing rational numbers as, 13–14 in ordering real numbers, 26 periodic, 14 repeating, 14, 17, 587 terminating, 14, 586 Decimal fraction, 13 Decimal notation, changing scientific notation to, 193–195 Decrease, percent of, 230 Degree, 248 Denominator common, 551 least common (LCD), 552 Dependent system of equations, 412–414 Dependent events, 617 Dependent variable, 341 Depression, angle of, 313 Descending order, polynomials in, 170 Descriptive statistics, 661 Diagram stem-and-leaf, 670–672 tree, 609–611, 627 Diameter of a sphere, 286 Die, 580, 584 rolling of fair, 596 Difference of two perfect squares, factoring, 452–453 Digits, set of, 3 significant, 29–32, 144–145, 195, 280, 284, 289, 304 Dimensional analysis, 234 Direct measurement, 301 Direct variation, 222–224 constant of, 222 equation of, 375 graphing, 374–376 Directly proportional, 222
Disjoint sets, 72 Distance formula, 136 Distributive property, 48–49 of multiplication over addition or subtraction, 168, 178, 183 Division, 4, 38 of fractions, 548–549 of a monomial by a monomial, 197–198 of a polynomial by a binomial, 200–201 by a monomial, 198–199 of powers that have the same base, 186–187 of signed numbers, 68–70 of square-root radicals, 494–495 verbal phrases involving, 90 Division property of equality, 118 of a fraction, 541 Domain, 89, 151 of a function, 341 of an inequality, 151 of an open sentence, 104–105 of a relation, 339 Double-blind experiment, 664 Double root, 506 E Elevation, angle of, 313 Empirical probability, 576–581 Empirical study, 577, 579 Empty set, 3, 72 Endpoint(s), 247–248 of a ray, 247 of a segment, 247 Equality addition property of, 118 division property of, 118 multiplication property of, 118 properties of, 117–119 subtraction property of, 118 Equally likely outcomes, 586 Equation(s), 117 conditional, 117 equivalent, 117 left side of, 117 right side of, 117 root of, 117 simplifying each side of, 122–126 solving linear, with more than one operation, 117–121 with the variable in both sides, 128–132 solution set of, 117 systems of, linear, 416–420, 422–424 quadratic-linear, 529–531 writing linear given slope and one point, 402 given the intercepts, 407–409 given two points, 404–405 Equiangular triangle, 263 Equilateral triangle, 265, 266 Equivalent decimals, 586–587 Equivalent equations, 117 Equivalent fractions, 541 Equivalent inequalities, 151 Equivalent ratios, 208–209 Error, 227 in geometric calculations, 289–290 percent of, 227–229 relative, 228 Estimation, radicals and, 477–478 Evaluating an algebraic expression, 100–102 Event(s), 584 compound, 609 dependent, 617 favorable, 584 independent, 611–612, 618 mutually exclusive, 600–603 probability of an, 592 singleton, 585, 590 unfavorable, 584 Everywhere dense, 12 Experimental design, 663–664 Experiment(s), 579, 663
double-blind, 664 in probability, 579–581 single-blind, 663 Exponent(s), 39, 40, 95–96 negative integral, 189–190 zero, 188–189 Exponential decay, 389 Exponential function, 387–391 exponential decay, 389–391 exponential growth, 388–391 Exponential growth, 388 Expression. See Algebraic expression(s); numerical expressions Exterior angle(s), 258 alternate, 259 Extraneous solutions, 565 Extrapolation, 717 Extremes, of a proportion, 216 F Face, of a solid, 282 Factorial n, 628 Factorial symbol (!), 628 Factoring, 443–445 a number, 443 over the set of integers, 443–445 a polynomial, 447, completely, 461–463 the difference of two perfect squares, 452–453 trinomials, 457–460 solving quadratic equations by, 503–507 Factor(s), 39, 443 common monomial, 447 Index 733 greatest common, 444–445 greatest common monomial, 447 of a term, 95 Fair and unbiased objects, 579 Favorable event, 584 Finite set, 3, 339–340 First-degree equations in one variable, 122 graphing, 346–350, 370–373 parallel to axes, 352–353 solving with more than one operation, 117–121 with the variable in both sides, 128–132 systems of, consistent, 410–411 dependent, 412–414 inconsistent, 411 independent, 411 solving, graphically, 410–414 using substitution, 422–424 using the addition method, 416–420 writing given slope and one point, 402 given the intercepts, 407–409 given two points, 404–405 First-degree inequalities in two variables, graphing, 378–381 First quartile, 698 Five statistical summary, 699 FOIL method, 183 in multiplication of binomials, 454 in multiplication of polynomials, 183 Formula(s), 107, 143 distance, 136 in problem solving, 134–137 for surface area, 293 transforming, 143–145 for volume, 293 writing, 107–108 Fractional coefficients solving equations with, 556–559 solving inequalities with, 562–563 Fractional equation(s), 565 solving, 565–567 Fractional expression, 540 Fraction line, as grouping symbol, 41 Fraction(s) addition of, 550–554 algebra
ic, 540 common, 13 decimal, 13 division of, 548–549 division property of, 541 equivalent, 13–14, 541, 216 lowest terms, 541 multiplication of, 545–547 multiplication property of a, 542 reducing to lowest terms, 541–543 subtraction of, 550–554 writing probability of events as, 586 Frequency distribution table, 668 Frequency histogram, 676 cumulative, 703–707 734 Index Function, 341. See also Absolute value function; Exponential function; Linear function; Quadratic function domain of a, 341 range of a, 341 G Geometric calculations, error in, 289–290 Geometry angles in, 248–249, 258–261 areas of irregular polygons in, 279–280 half-lines in, 247 line segments in, 247 lines in, 246–247 parallel lines in, 258–261 perpendicularity in, 249 planes in, 246 points in, 246 quadrilaterals in, 272–276 rays, 247–248 surface areas of solids in, 282–284 triangles in, 262–267 undefined terms in, 246 volumes of solids in, 286–290 Graph(s), interpreting, 664–665 of ordered pairs, 76, 610 of a point, 6 of a polygon, 78–79 and root finding, 522–524 of sets, intersection, 153–154 union, 154–155 and solving systems of equations, linear, 410–414 quadratic-linear, 525–527 types of, absolute value, 382–386 direct variation, 374–376 exponential, 387–391 inequalities, 378–381 linear, 346–350, 352–353, 370–373 quadratic, 508–516 Graphic solution of a quadratic-linear system, 525–527 Histogram, 675–678 cumulative frequency, 703–707 frequency, 676 Horizontal number line, 6 Hypotenuse, 263, 301, 307 I Identity, 117 additive, 49–50 multiplicative, 50, 179 Identity element of addition, 49–50 of multiplication, 50 Impossibility, 591 Inconsistent system of equations, 411 Increase, percent of, 230 Independent events, 611–612, 618 Independent system of equations, 411 Independent variable, 341 Index, of a radical, 473 Indirect measurement, 301 Inequalities domain (replacement set) of, 151 equivalent, 151 finding and graphing the solution
of, 151–155 with fractional coefficients, 562–563 graphing first-degree, in one variable, 151–155 in two variables, 378–381 systems of, 431–434 in problem solving, 157–159 properties of, 146–149 addition, 147–148 multiplication, 148–149 order, 146–147 transitive property, 147 solution set of, 151 symbols of, 7–8 verbal phrases involving, 157–159 Infinite set, 3, 340–344 Integer(s), 2–8 ordering, 6 set of, 4–5 subsets of, 5 Graphing calculator. See also Calculator Greatest common factor (GCF), Intercept form of a linear equation, 367 Intercept(s) 444–445 Greatest common monomial factor, 447 Greatest possible error (GPE), 29 Grouped data, 668–670 calculator solution for, 692–693 mean of a set of, 691–692 measures of central tendency and, 690–695 median of a set of, 691 mode of a set of, 690 Grouping symbols expressions with, 41–43 multiplication and, 179–180 Group mode, 693 Group, 669 H Half-line, 247 Half-plane, 378 of a line, 366–367 writing equation given, 407–409 Interest, compound, 389 Interior angle(s), 258 alternate, 258 on the same side of the transversal, 258 Interpolation, 717 Intersection of sets, 71–72, 597 graphing, 153–154 Interval, 669 containing the mean, 694–695 containing the median, 694 containing the mode (modal interval), 693 Interval notation, 151–152 Inverse operation, using, in dividing signed numbers, 68–69 Inverse additive, 50 multiplicative, 51–52, 548 Irrational numbers, 17–23 basic rules for radicals that are, 478–481 radicals and, 476–481 set of, 17–18 Irregular polygons, area of, 279–280 Isosceles trapezoid, 276 Isosceles triangle, 265–266 base angles of, 265 base of, 265 legs of, 265 vertex angle of, 265 L Latitude, 75 Leaf, in a stem-and-leaf diagram, 670 Least common denominator (LCD), 552 Left member of an equation, 117 Left side of an equation, 117 Legs of isosceles triangles, 265 of right
triangles, 263, 307 Letters, using, to represent numbers, 89–90 Length of a line segment, 247 Like radicals, 487 addition of, 487 subtraction of, 487 Like terms (similar terms), 123, 168 Line, in geometry, 246–247. See also Linear function Line of best fit, 715–720 Line segment, 247 Linear equation, 347. See also Firstdegree equation Linear function, graphing, using slope, 370–373 using solutions, 346–350 intercepts of, 366–367 x-intercept, 366–369 y-intercept, 366–369 slope of, 355–360 finding, 355–357 parallel, 258–261, 352–353, 363 perpendicular, 249, 364–365 transformations of, reflection, 372 scaling, 372 translation, 372 Linear growth, 387 Linear pair of angles, 252 Linear regression. See Bivariate statistics Linear system of equations, 410 solving, addition method, 416–420 graphically, 410–415 substitution method, 422–424 types of, consistent, 411 dependent, 412 inconsistent, 411 independent, 411 using to solve verbal problems, 426–428 Linear transformation of a set of data, 686 Line segment(s), 247 congruent, 265 length of, 247 List of ordered pairs, 609 Longitude, 75 Lower quartile, 698 Lowest terms, 212 reducing fractions to, 541–543 Lowest terms fraction, 541 M Mathematics, defined, 2 Maximum, 511 Mean, 681 arithmetic, 681 interval containing, 694–695 of a set of grouped data, 691–692 Measure of a line segment, 247 Measurement(s) accuracy of, 31 changing units of, 234–236 direct, 301 indirect, 301 numbers as, 28–32 precision of, 30–31 Measures of central tendency. See Central tendency Median, 681–682 interval containing, 694 of set of grouped data, 691 Member of a set (∈), symbol for, 338 Minimum, 509 Minus symbol, 61–62 Modal interval, 693 Mode, 682–685 of set of grouped data, 690 Monomial, 169 division of a, by a monomial, 197–198 division of a polynomial by a, 198–199 finding the principal square root of a, 482–483 multiplication of a monomial by a, 177–178 multiplication of a polynomial by a, 178–179 square of a, 449
Monomial square roots, division of, 494–495 Multiplication, 3–4, 38 associative property of, 48 of binomials, 454–455 commutative property of, 47 distributive property of, over addition, 49, 168, 178, 183 of fractions, 545–547 grouping symbols and, 179–180 of monomial by a monomial, 177–178 of polynomials, 183–184 by a monomial, 178–179 of powers that have the same base, 173–174 of signed numbers, 64–67 of square-root radicals, 491–492 of sum and difference of two terms, 450–451 verbal phrases involving, 89–90 Multiplication property of equality, 118 of a fraction, 542 of inequality, 148–149 of one, 50 of zero, 52 Multiplicative identity, 50, 179 Multiplicative inverse (reciprocal), 51–52, 548 Mutually exclusive events, 600–603 N n factorial, 628 Natural numbers, 2 Negative integral exponent, 189–190 Negative numbers, 4 absolute value of, 7 Negative slope, 358 Nonlinear function, 388 No slope, 358 Not a member of a set (βˆ‰), symbol for, 338 Notation decimal, 193–195 interval, 151–152 scientific, 191–195 set-builder, 338–344 Null set, 3 Numerical coefficient, 95 Number line, 2, 6 ordering real numbers on, 25 rational, 11–12 real, 25 standard, 6 Number pairs, graphing, 75–79 Number(s), 2 approximations of, 469 composite, 39, 443 counting, 2, 4, 5, 11 factoring, 443 irrational, 17–20, 469, 470 as measurements, 28–32 natural, 2 negative, 4 positive, 4 prime, 39, 443 rational, 11–15, 470–475 real, 2, 25–26 representing two, with the same variable, 125–126 symbols for, 2 using letters to represent, 89–90 whole, 3 writing, in scientific notation, 192–193 Number systems, 1–36 integers in, 2–8 irrational numbers in, 17–20 measurements and, 28–32 rational numbers in, 11–15 real numbers in, 2, 25–26 Numeral, 2 Numerical average, 681 Numerical expression(s), 3–4 with grouping symbols, 41–43 simplifying, 3,
42–43 O Obtuse angle, 249 Obtuse triangle, 263 One, multiplication property of, 50 Open sentence, 104–105 Index 735 Operation(s) binary, 38 computations with more than one, 41 order of, 38–43 properties of, 45–52 with sets, 71–73 solving equations using more than one, 117–121 Operational symbols, 3 Opposite angles, 272 Opposite rays, 248 Opposite (additive inverse), 4–5, 50 Opposite side of an angle, 307 Ordered pair(s), 38, 75 graph of, 610 list of, 609 Order of operations, 38–43 Order property of real numbers, 146–147 Ordinate, 75 Origin, 75 Outcome(s), 584 equally likely, 586 P Palindrome, 37 Parabola, 509 axis of symmetry of the, 510 Parallel lines, 258–261 alternate exterior angles and, 260 alternate interior angles and, 259 corresponding angles and, 259–260 slope of, 363 to the x-axis, 352 to the y-axis, 353 Parallelograms, 272 family of, 273–274 informal proofs for statements about angles in, 274–275 Parentheses as grouping symbol, 41 Percent, 227 of decrease, 230 of error, 227–229 of increase, 230 Percentage, 227 Percentiles, 701–702 Perfect squares, 449, 471 factoring the difference of two, 452–453 square root of, 471 Perfect square trinomial, 455 Perimeter, 134 Periodic decimals, 14 Permutations, 627–634, 646 calculator and, 631–634 comparing with combinations, 639–641 with repetition, 636–637 representing, 629–630 symbols for, 631 that use some of the elements, 630–631 Perpendicular bisector, 271 Perpendicularity, 249 Perpendicular lines, 249 slope of, 364–365 Pi (p), 20 Placebo, 663 Placebo effect, 663 736 Index Placeholder, 89 Plane, 246 points on, 75–76 finding the coordinates of, 77–78 Plane divider, 378 Plot box-and-whisker, 699–701 scatter, 710, 711 Point(s), 246 coordinates of, 75 finding the coordinates of, on a plane, 77–78 locating, on the coordinate plane, 76–77 on a plane, 75–76 writing
equation given slope and one, 402 writing equations given two, 404–405 Polygon(s), 78, 262 angles and, 275 area of irregular, 279–280 graphing, 78–79 sides of, 262 vertices of, 262 permutations and, 627–634 as sums, 605–606 theoretical, 584–587 with two or more activities with replacement, 617–618 without replacement, 617 uniform, 586 writing as fractions, 586 Problem solving formulas in, 134–137 inequalities in, 157–159 tangent ratio in, 313–315 trigonometric ratios in, 327–328 Product, 443 Properties of equality, 117–119 addition property of equality, 118 division property of equality, 118 multiplication property of equality, 118 substitution principle, 118 subtraction property of equality, 118 Properties of inequalities, 146–149 addition property of inequality, Polynomial equation of degree two, 147–148 503 multiplication property of inequality, Polynomial function, second-degree, 148–149 509 Polynomial(s), 169–170 in ascending order, 170 in descending order, 170 division of by a binomial, 200 by a monomial, 198–199 factoring, 447 factoring completely, 461–463 multiplication of, 183–184 by a monomial, 178–179 prime, 447 Positive numbers, 4 absolute value of, 7 Positive slope, 357 Postulate, 246 Power(s), 39, 40, 95–96 division of, 186–187 finding a power of a, 174–176 finding the product of, 173–174 Powers of 10, 192 Precision, 30–31 Price, unit, 212 Prime number, 39, 443 Prime polynomial, 447 over the set of integers, 461 Principal square root, 471–472 finding, of a monomial, 482–483 Prism, 282–283 right, 282–283 Probabilities, 575–659 of (A and B), 596–597 of (A or B), 599–603 of (not A), 605–606 of any event, 592 calculated, 585 combinations and, 639–644 conditional, 619–623 defined, 576 empirical, 576–581 evaluating simple, 590–594 experiments in, 579–581 order property of real numbers, 146–147 transitive property of inequality, 147 Properties of operations, 45–52 addition property of zero, 49–50 additive inverse
, 50 associative of addition, 47–48 of multiplication, 48 closure, 45–47 commutative of addition, 47 of multiplication, 47 distributive, 48–49 multiplication property of one, 50 multiplication property of zero, 52 multiplicative inverses, 51–52 Proportion, 216–220 cross products in, 217 extremes of, 216 general form of, 216 inner terms of, 216 outer terms of, 216 products of the means and extremes in, 217 Pyramid(s), 288 volume of, 288 Pythagoras, 301 Pythagorean Theorem, 227, 301–304 statements of, 302–304 Pythagorean triple, 305 Q Quadrant, 76 Quadratic equation(s), 503 finding roots from a graph, 522–524 roots of, 504, 522 solving, by factoring, 503–507 standard form of, 503 Quadratic function, 509–518 axis of symmetry, 510–511 leading coefficient, 509–511 transformations of, reflection, 517–518 scaling, 517–518 translation, 517–518 turning point, 509 vertex, 509 Quadratic-linear system of equations, 525 solving, algebraically, 529–531 graphically, 525–527 Quadrilateral(s), 78, 272–276 angles of, 273 consecutive angles in, 272 opposite angles in, 272 special, 272–273 Qualitative data, 661 Quantitative data, 661 Quartile, 698–699 first, 698 lower, 698 second, 698 third, 698 upper, 698 R Radical(s), 470, 470–475 addition of like, 487 addition of unlike, 488–489 basic rules for irrational numbers as, 478–481 estimation and, 477–478 index of, 473 irrational numbers and, 476–481 simplifying square-root, 484–485 subtraction of like, 487 subtraction of unlike, 488–489 that are square roots, 471–472 Radical sign, 18, 470 Radicand, 470 Radius of a sphere, 286 Random selection, 586–588 Range, of data, 669 of a function, 341 of a relation, 339 Rate, 212, 227 unit, 212 using ratio to express, 212 Rational approximation, 20, 477–478 Rational expression, 540 Rational number line, 11–12 Rational numbers, 11–15, 470–475 expressing, as decimals, 13–14 expressing decimals as, 15
properties of, 12–13 set of, 11–15 Ratio(s), 208 constant, 222 continued, 209–210 equivalent, 208–209 expression of, in simplest form, 208–209 using, to express a rate, 212 verbal problems involving, 214–215 Ray(s), 247–248 opposite, 248 Real number line, 25 Real numbers, 2, 25–26 completeness property of, 25 ordering, 25–26 order property of, 146–147 set of, 25 using properties of, to multiply signed numbers, 65–67 Reciprocal(s), 51–52 in dividing signed numbers, 69–70 Rectangle, 272, 273 Rectangular solid, 283 Reduced to lowest terms, 541–543 Reflection rule for absolute value functions, 386 for linear functions, 372 for quadratic functions, 517 Regression line, 715–716. See also Linear regression Relation, 339 domain of, 339 range of, 339 Relative error, 228 Relative frequency, 577 cumulative, 578–579 Repeating decimals, 14, 17, 587 Repetition, permutations with, 636– 637 Replacement probability with, 617–618 probability without, 617 Replacement set, 89 of inequalities, 151 Rhombus, 273 Right angle, 248 Right circular cylinder, 283 Right member of an equation, 117 Right prism, 282–283 Right side of an equation, 117 Right triangle, 263, 301 hypotenuse of, 263, 301, 307 legs of, 263, 307 Root(s), 117 calculators and, 473–475 cube, 472 double, 505 of an equation, 504 finding, from a graph, 522–524 square, 470 Roster form, 338 Rounding, 21 Round-off error, 31 S Sample, 662 Sample space(s), 584 subscripts in, 592–594 Sampling, 662 techniques of, 662–663 Scalene triangle, 265 Scaling rules, for absolute value functions, 386 for linear functions, 372 for quadratic functions, 517 Scatter plot, 710–711 Scientific notation, 191–195 changing to ordinary decimal notation, 193–195 Index 737 writing numbers in, 192–193 Second-degree polynomial function, 509 Second quartile, 698 Segment of a line, 247 Set(s), 3, 71 in finding equations, 402 Solids surface area of, 282–284 volumes of, 286–290 Solutions dis
joint, 72 empty, 3, 72 finite, 3, 339–340 infinite, 3, 340–344 of integers, 4–5 of irrational numbers, 17–18 null, 3 operations with, 71–73 intersection of, 71–72, 597 graphing, 153–154 complement of, 73 union of, 72–73 of rational numbers, 11–15 of real numbers, 25 types of, 3 graphing, 154–155 universal, 71 of whole numbers, 3 Set-builder notation, 338–339 Sides, of an angle, 248 classifying triangles according to, 265–266 of polygons, 262 Signed numbers, addition of, with opposite signs, 56–58 with same signs, 54–55 division of, 68–70 multiplication of, 64–67 subtraction of, 59–62 Significant digits, 29–32, 144–145, 195, 280, 284, 289, 304 rules for determining, 29–30 Similar terms, 123, 168 Similar triangles, 307 Simplest form, of a polynomial, 170 expression of ratio in, 208–209 of square-root radical, 485 Simplify a numerical expression, 3 Sine of an angle, 317 Sine ratio, 317–318 applications of, 323–324 finding, on a calculator, 319–320 Single-blind experiment, 663 Singleton event, 585, 590 Slope, 355–360 applications of, 360 graphing linear functions using, 370–373 of parallel lines, 363 of perpendicular lines, 364–365 slope-intercept form, 367–368 types of, negative, 358 positive, 357 undefined, 358 zero, 358 as unit rate of change, 356 writing equations given one point and, 402 Slope-intercept form of a linear equation, 368 extraneous, 565 graphing linear functions using their, 346–350 Solution set(s), 104 graphing, for system of inequalities, 431–434 of inequalities, 151–153 of open sentences, 104–105, 117 Sphere, 286 center of, 286 diameter of, 286 radius of, 286 volume of, 286 Square(s), 18, 273, 470 of monomials, 449 perfect, 449, 471 Square root(s), 18, 470 calculators and, 472 finding the principal, of a monomial, 482–483 of a perfect square, 471 principal, 471–472 radicals that are, 471–472 Square-root radicals division of
, 494–495 multiplication of, 491–492 simplest form of, 485 simplifying, 484–485 Standard deck of cards, 580 Standard form, 503 first-degree equation in, 347 Standard number line, 6 Statistical summary (five statistical summary), 699 Statistics, 660–730 bivariate data, 710–720 correlation, 711–715 causation, 711–715 line of best fit, 715–720 extrapolation, 717 interpolation, 717 regression line, 715–720 central tendency (average), 690–695 and linear transformations, 686 mean, 681 for grouped data, 691–692 for intervals, 694–695 median, 681–682 for grouped data, 691 for intervals, 694 mode, 682–683 for grouped data, 690 for intervals (modal interval), 693 data collection, 661–665 bias, 662 sampling, 662 techniques of, 662–663 defined, 661 experimental design, 663–664 organization of data, 667–672 frequency distribution table, 668 grouped data, 668–670 cumulative frequency, 702–703 percentiles, 701–702 738 Index Statistics cont. quartiles, 698–699 five statistical summary, 699 sampling in, 662–663 visualization of data, types of, box-and-whisker plot, 699–701 cumulative frequency histogram, 703–707 histogram, 675–678 scatter plot, 711–714 stem-and-leaf diagram, 670–672 interpreting graphs of data, 664–665 univariate data, 710 Stem-and-leaf diagram, 670–672 Straight angle, 249 Straight line(s), 246–247 facts about, 246–247 Straightedge, 270 Subscripts, 592 in sample spaces, 592–594 Subsets of the integers, 5 Substitution method for solving a system of linear equations, 422–424 Substitution principle, 118, 423 Subtraction, 3, 38 of algebraic expressions, 168–171 of algebraic fractions, 550–554 distributive property of multiplication over, 168, 178, 183 of like radicals, 487 of signed numbers, 59–62 of unlike radicals, 488–489 verbal phrases involving, 89 Subtraction property of equality, 118 Successor, 2 Sums, probabilities as, 605–606 Supplementary angles, 251 Surface area of cylinders, 283 formulas for,
293 of rectangular solids, 283 of solids, 282–284 Symbol(s) approximately equal to (), 20 factorial (!), 628 grouping, 41–43 of inequality, 7–8 infinity (∞), 151–152 is an element of (∈), 338 is not an element of (βˆ‰), 338 minus (), 61–62 for numbers, 2 operational, 3 for permutations, 631 translating verbal phrases into, 91–93 vertical bar (), 338 System of equations. See Linear system of equations; Quadratic-linear system of equations System of inequalities, graphing the solution set of, 431–434 System of simultaneous equations, 410 T Table(s) frequency distribution, 668 preparing, 668 Tally mark, 668 Tangent of an angle, 308 Tangent ratio(s), 307–311 applications of, 313–317 angle of depression, 313 angle of elevation, 313 in problem solving, 313–315 finding, on a calculator, 308–311 Terminating decimals, 14, 586 Term(s), 95, 122, 168 factors of, 95 like, 123, 168 lowest, 212 multiplication of the sum and difference of two, 450–451 reducing fractions to lowest, 541–543 similar, 123, 168 undefined, 246 unlike, 123, 168 Theorem, 253 Theoretical probability, 584–587 Third quartile, 698 Time series, 715 Total frequency, 668 Transitive property of inequality, 147 Translation rules for absolute value functions, 385–386 for linear functions, 372 for quadratic functions, 517 Transversal, 258 Trapezoid, 272, 275–276 isosceles, 276 Treatment group, 663 Tree diagram, 609–611, 627 Trial, 579 Triangle(s), 262–267 classifying according to angles, 263 according to sides, 265–266 properties of special, 266–267 similar, 307 sides of, 262 sum of the measures of the angles of a, 263–265 types of, acute, 263 equiangular, 263 equilateral, 265–266 isosceles, 265–266 obtuse, 263 right, 263, 301 scalene, 265 vertices of, 262 Trigonometry, 300 cosine ratio in, 318–320, 323–324 problem solving using ratios, 327–328 Pythagorean Theorem in, 301–304 sine ratio in, 317–320, 323–324 tangent
ratio in, 307–317 Trinomial(s), 170 factoring, 457–460 Turning point, 509 Two-valued statistics, 710. See also Bivariate statistics U Undefined term, 246 Unfavorable event, 584 Uniform probability, 586 Union of sets, 72–73 graphing, 154–155 Unit measure, 6 Unit price, 212 Unit rate, 212 of change, 356 Units of measure, changing, 234–236 Univariate statistics, 710 Universal set, 71 Unlike radicals addition of, 488–489 subtraction of, 488–489 Unlike terms, 123, 168 Upper quartile, 698 V Variable(s), 89 dependent, 341 first-degree equations in one, 122 graphing first-degree inequalities in two, 378–381 independent, 341 representing two numbers with the same, 125–126 solving, for in terms of another variable, 142–143 solving equations that have, in both sides, 128–132 writing algebraic expressions involving, 92 Verbal phrases commas in, 90 involving addition, 89 involving division, 90 involving multiplication, 89–90 involving subtraction, 89 translating, into symbols, 91–93 Verbal problems involving ratio, 214–215 using systems of equations in solving, 426–428 Vertex angle of an isosceles triangle, 265 Vertex of a parabola, 509 Vertex of an angle, 248 Vertical angles, 252–254 Vertical bar (|), 338 Vertical number line, 6 Vertices, 78 of a polygon, 262 Volume of cones, 288 formulas for, 293 of pyramids, 286 of solids, 286–290 of spheres, 286 W Whole numbers, 3 set of, 3 With replacement, probability problems, 617–618 Without replacement, probability problems, 617 Words, writing algebraic expressions in, 98–99 X x-axis, 75 lines parallel to, 352 x-coordinate, 76 x-intercept, 366 writing equation given, 407–409 Y y-axis, 75 lines parallel to, 353 y-coordinate, 76 y-intercept, 366 slope and, 367–368 writing equation given, 407–409 Z Zero, 3 addition property of, 49–50 division of, by nonzero number, 69 multiplication property of, 52 Zero exponent, 188–189 Zero slope, 358 Index 739iew Note: Persian mathematician Omar Khayyam would solve algebraic problems geometrically by intersecting graphs rather than solving them algebraically. 35
1.2 Practice - Two-Step Problems Solve each equation. 1) 5 + n 4 = 4 3) 102 = 7r + 4 βˆ’ 8n + 3 = 5) βˆ’ 7) 0 = 6v βˆ’ 8 = x 9) βˆ’ 11 βˆ’ 13) 12 + 3x = 0 βˆ’ 15) 24 = 2n 8 βˆ’ 12 + 2r 17) 2 = βˆ’ 3 + 7 = 10 19) b 77 βˆ’ 6) 8) βˆ’ βˆ’ 4 βˆ’ 2 + x 2) 2 = 2m + 12 βˆ’ βˆ’ 4) 27 = 21 3x βˆ’ b = 8 2 = 4 10) 12) 14) 16) βˆ’ βˆ’ βˆ’ βˆ’ 18) βˆ’ 20) x 5 = a 1 4 βˆ’ 6 = 15 + 3p 5m + 2 = 27 37 = 8 + 3x 8 + n 12 = 7 βˆ’ 8 8 = βˆ’ 1 βˆ’ 21) 152 = 8n + 64 23) βˆ’ 16 = 8a + 64 25) 56 + 8k = 64 27) 29) βˆ’ βˆ’ 2x + 4 = 22 20 = 4p + 4 31) βˆ’ 33 = βˆ’ 40 = 4n 35) βˆ’ 37) 87 = 3 7v βˆ’ x + 1 = 39) βˆ’ 32 βˆ’ 11 βˆ’ 8 + v 2 βˆ’ 3 = βˆ’ 3n = 22) 24) βˆ’ βˆ’ 11 = 2x 26) 4 βˆ’ βˆ’ 28) 67 = 5m βˆ’ βˆ’ 8 29 16 βˆ’ 30) 9 = 8 + x 6 32) m 1 = 4 βˆ’ βˆ’ 80 = 4x 2 28 βˆ’ 34) βˆ’ 36) 33 = 3b + 3 3 = 38) 3x βˆ’ 40) 4 + a 3 = 1 3 βˆ’ 36 1.3 Solving Linear Equations - General Equations Objective: Solve general linear equations with variables on both sides. Often as we are solving linear equations we will need to do some work to set them up into a form we are familiar with solving. This section will focus on manipulating an equation we are asked to solve in such a way that we can use our pattern for solving two-step equations to ultimately arrive at the solution. One such issue that needs to be addressed is parenthesis. Often the parenthesis can get in the way of solving an otherwise easy problem. As you might expect we can get rid of the unwanted parenthesis by using the distributive property. This is shown in the following example. Notice the first step is distributing, then it is solved like any other two-step equation
. Example 59. 4(2x 8x βˆ’ βˆ’ 6) = 16 Distribute 4 through parenthesis 24 = 16 Focus on the subtraction first + 24 + 24 Add 24 to both sides 8x = 40 Now focus on the multiply by 8 8 Divide both sides by 8 8 x = 5 Our Solution! Often after we distribute there will be some like terms on one side of the equation. Example 2 shows distributing to clear the parenthesis and then combining like terms next. Notice we only combine like terms on the same side of the equation. Once we have done this, our next example solves just like any other two-step equation. Example 60. 3(2x 6x βˆ’ βˆ’ 4) + 9 = 15 Distribute the 3 through the parenthesis 12 + 9 = 15 Combine like terms, 6x 12 + 9 Focus on the subtraction first 3 = 15 βˆ’ βˆ’ + 3 + 3 Add 3 to both sides 6x = 18 Now focus on multiply by 6 37 6 6 Divide both sides by 6 x = 3 Our Solution A second type of problem that becomes a two-step equation after a bit of work is one where we see the variable on both sides. This is shown in the following example. Example 61. 4x βˆ’ 6 = 2x + 10 Notice here the x is on both the left and right sides of the equation. This can make it difficult to decide which side to work with. We fix this by moving one of the terms with x to the other side, much like we moved a constant term. It doesn’t matter which term gets moved, 4x or 2x, it would be the author’s suggestion to move the smaller term (to avoid negative coefficients). For this reason we begin this problem by clearing the positive 2x by subtracting 2x from both sides. however, 6 = 2x + 10 Notice the variable on both sides βˆ’ 4x 2x 2x βˆ’ 2x βˆ’ 6 = 10 βˆ’ + 6 + 6 2x = 16 2 2 x = 8 Subtract 2x from both sides Focus on the subtraction first Add 6 to both sides Focus on the multiplication by 2 Divide both sides by 2 Our Solution! The previous example shows the check on this solution. Here the solution is plugged into the x on both the left and right sides before simplifying. Example 62. 4(8) 32 6 = 2
(8) + 10 Multiply 4(8) and 2(8) first βˆ’ 6 = 16 + 10 βˆ’ 26 = 26 Add and Subtract True! The next example illustrates the same process with negative coefficients. Notice first the smaller term with the variable is moved to the other side, this time by adding because the coefficient is negative. 38 Example 63. 3x + 9 = 6x βˆ’ + 3x + 3x 9 = 9x βˆ’ βˆ’ 27 Notice the variable on both sides, Add 3x to both sides Focus on the subtraction by 27 27 3x is smaller βˆ’ + 27 + 27 Add 27 to both sides 36 = 9x 9 9 4 = x Focus on the mutiplication by 9 Divide both sides by 9 Our Solution Linear equations can become particularly intersting when the two processes are combined. In the following problems we have parenthesis and the variable on both sides. Notice in each of the following examples we distribute, then combine like terms, then move the variable to one side of the equation. Example 64. 2(x 2x βˆ’ βˆ’ 5) + 3x = x + 18 Distribute the 2 through parenthesis 10 + 3x = x + 18 Combine like terms 2x + 3x 5x x 4x 10 = x + 18 Notice the variable is on both sides βˆ’ x βˆ’ βˆ’ 10 = 18 βˆ’ + 10 + 10 4x = 28 4 4 x = 7 Subtract x from both sides Focus on the subtraction of 10 Add 10 to both sides Focus on multiplication by 4 Divide both sides by 4 Our Solution Sometimes we may have to distribute more than once to clear several parenthesis. Remember to combine like terms after you distribute! Example 65. 3(4x βˆ’ 12x 5) βˆ’ βˆ’ 15 4(2x + 1) = 5 Distribute 3 and 8x 4 = 5 Combine like terms 12x βˆ’ 4x βˆ’ 19 = 5 βˆ’ Focus on subtraction of 19 βˆ’ + 19 + 19 Add 19 to both sides βˆ’ 4 through parenthesis 8x and 4 15 βˆ’ βˆ’ 4x = 24 Focus on multiplication by 4 39 4 4 Divide both sides by 4 x = 6 Our Solution This leads to a 5-step process to solve any linear equation. While all five steps aren’t always needed, this can serve as a guide to solving equations. 1. Distribute through any parentheses. 2. Combine like terms
on each side of the equation. 3. Get the variables on one side by adding or subtracting 4. Solve the remaining 2-step equation (add or subtract then multiply or divide) 5. Check your answer by plugging it back in for x to find a true statement. The order of these steps is very important. World View Note: The Chinese developed a method for solving equations that involved finding each digit one at a time about 2000 years ago! We can see each of the above five steps worked through our next example. Example 66. 7) + 8x Distribute 4 and 3 through parenthesis 21 + 8x Combine like terms 24 + 9 and 3x + 8x 4(2x 8x 6) + 9 = 3(x βˆ’ 24 + 9 = 3x βˆ’ 8x 8x βˆ’ βˆ’ 15 = 11x 8x βˆ’ 15 = 3x βˆ’ βˆ’ βˆ’ + 21 21 βˆ’ 21 βˆ’ + 21 6 = 3x 3 3 2 = x βˆ’ Notice the variable is on both sides Subtract 8x from both sides Focus on subtraction of 21 Add 21 to both sides Focus on multiplication by 3 Divide both sides by 3 Our Solution Check: 4[2(2) 4[4 βˆ’ βˆ’ 6] + 9 = 3[(2) 6] + 9 = 3[ βˆ’ 5] + 8(2) βˆ’ 7] + 8(2) Plug 2 in for each x. Multiply inside parenthesis Finish parentesis on left, multiply on right 40 4[ βˆ’ βˆ’ 2 15 + 8(2) 15 + 16 Finish multiplication on both sides Add True! When we check our solution of x = 2 we found a true statement, 1 = 1. Therefore, we know our solution x = 2 is the correct solution for the problem. There are two special cases that can come up as we are solving these linear equations. The first is illustrated in the next two examples. Notice we start by distributing and moving the variables all to the same side. Example 67. 3(2x 6x 6x βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 5) = 6x 15 = 6x 6x βˆ’ 15 = 15 βˆ’ 15 Distribute 3 through parenthesis 15 Notice the variable on both sides Subtract 6x from both sides Variable is gone! True! 15 = Here the variable subtracted out completely! We are left with a true statement, 15. If the variables subtract out completely and we are
left with a true βˆ’ statement, this indicates that the equation is always true, no matter what x is. Thus, for our solution we say all real numbers or R. βˆ’ Example 68. 2(3x 6x βˆ’ βˆ’ βˆ’ 5) 10 2x 2x βˆ’ βˆ’ βˆ’ 4x = 2x + 7 Distribute 2 through parenthesis 4x = 2x + 7 Combine like terms 6x 10 = 2x + 7 Notice the variable is on both sides 4x βˆ’ 2x βˆ’ 10 7 Subtract 2x from both sides Variable is gone! False! βˆ’ Again, the variable subtracted out completely! However, this time we are left with a false statement, this indicates that the equation is never true, no matter what x is. Thus, for our solution we say no solution or βˆ…. 41 1.3 Practice - General Linear Equations Solve each equation. 1) 2 3) ( βˆ’ 5( 3a 8) = 1 βˆ’ βˆ’ 4 + 2v) = βˆ’ βˆ’ 5) 66 = 6(6 + 5x) 50 βˆ’ 7) 0 = 8(p 5) βˆ’ βˆ’ 2 + 2(8x 9) 11) βˆ’ βˆ’ 13) βˆ’ 15) 1 βˆ’ 17) 20 βˆ’ 7) = 16 βˆ’ 6 3x βˆ’ βˆ’ 8m + 7 21x + 12 = 1 7m = βˆ’ 12r = 29 βˆ’ 8r βˆ’ 12b + 30 2v βˆ’ βˆ’ 32 2 7b = βˆ’ 24v = 34 βˆ’ 5(2 βˆ’ βˆ’ 19) 21) 23) 25) 29) 31) 33) βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 37) 39) 41) 43) 45) 47) 49) βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 27) 2(4x βˆ’ 4m) = 33 + 5m 4n + 11 = 2(1 8n) + 3n 6v 29 = 5(v + 1) βˆ’ 4v βˆ’ βˆ’ 4x 4) = 20 βˆ’ βˆ’ 1) = 39 5(8a 7a βˆ’ βˆ’ p + 1) + 2(6 + 8p) ( βˆ’ a βˆ’ 57 = βˆ’ βˆ’ βˆ’ 2(m 2) + 7(m 8) = 67 βˆ’ 35) 50 = 8 (7 + 7r) (4r + 6) βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 61 = 2(8n βˆ’ 3( βˆ’ 7(x 5(5r 4) + 4(3r 4) βˆ’ 4) = 8(1 n) βˆ’ 7
v + 3) + 8v = 5v βˆ’ 4(1 βˆ’ βˆ’ 6v) 2) = 4 6(x 1) βˆ’ βˆ’ 6(8k + 4) = βˆ’ βˆ’ 8(6k + 3) βˆ’ 7p) = 8(p 7) βˆ’ 2(1 βˆ’ 2 βˆ’ 2) 2( 3n + 8) = 20 βˆ’ 4 + 3x) = 34 βˆ’ 4) 2 8( βˆ’ 6) 32 = 2 βˆ’ βˆ’ 8) 10) 12) (3 3n βˆ’ βˆ’ βˆ’ 14) 56p βˆ’ 5( 4n + 6) βˆ’ 55 = 8 + 7(k βˆ’ 5) βˆ’ 5n) = 12 27 = βˆ’ 48 = 6p + 2 βˆ’ 27 3n βˆ’ 16) 4 + 3x = 12x + 4 βˆ’ 16n + 12 = 39 7n 18) βˆ’ 20) 17 2x = 35 βˆ’ 8x βˆ’ 7x = 6(2x βˆ’ 25 βˆ’ 7(1 + b) = 5 βˆ’ βˆ’ 8(8r 2) = 3r + 16 1) βˆ’ 5b βˆ’ 19 = βˆ’ 8n βˆ’ 4 + 4k = 4(8k βˆ’ 8) βˆ’ 22) 24) 26) 28) 30) 36) 38) 40) 42) 44) 46(8n 3) + 3n βˆ’ 32) 16 = βˆ’ 34) 7 = 4(n 5(1 6x) + 3(6x + 7) βˆ’ 7) + 5(7n + 7) βˆ’ 8(6 + 6x) + 4( 3 + 6x) = βˆ’ βˆ’ 3) 4 6(x 8) 4(x βˆ’ βˆ’ 4(1 + a) = 2a 2) = βˆ’ 8(5 + 3a) βˆ’ βˆ’ 3) + 5 = 6(x βˆ’ (n + 8) + n = 2 5(x βˆ’ βˆ’ βˆ’ 8n + 2(4n 12 βˆ’ 5) 4) βˆ’ 48) βˆ’ 50) 8( 5(x + 7) = 4( βˆ’ 8n + 4) = 4( 8x 2) βˆ’ 7n + 8) βˆ’ βˆ’ βˆ’ 8(n 7) + 3(3n 3) = 41 76 = 5(1 + 3b) + 3(3b 42 1.4 Solving Linear Equations - Fractions Objective: Solve linear equations with rational coefficients by multiplying by the least common denominator to clear the fractions. Often when solving linear
equations we will need to work with an equation with fraction coefficients. We can solve these problems as we have in the past. This is demonstrated in our next example. Example 69 Focus on subtraction Add 7 2 to both sides Notice we will need to get a common denominator to add 5 6 + 7 common denominator of 6. So we build up the denominator, 7 2 can now add the fractions: 2. Notice we have a = 21 6, and we 3 3 3 4 x βˆ’ 21 6 = 5 6 Same problem, with common denominator 6 + 21 6 + 21 6 Add 21 6 to both sides 3 4 3 4 x = x = 26 6 13 3 Reduce 26 6 to 13 3 Focus on multiplication by 3 4 We can get rid of 3 4 by dividing both sides by 3 same as multiplying by the reciprocal, so we will multiply both sides by 4 3. 4. Dividing by a fraction is the 4 3 3 4 4 3 x = x = 13 3 52 9 Multiply by reciprocal Our solution! While this process does help us arrive at the correct solution, the fractions can make the process quite difficult. This is why we have an alternate method for dealing with fractions - clearing fractions. Clearing fractions is nice as it gets rid of the fractions for the majority of the problem. We can easily clear the fractions 43 by finding the LCD and multiplying each term by the LCD. This is shown in the next example, the same problem as our first example, but this time we will solve by clearing fractions. Example 70 LCD = 12, multiply each term by 12 (12)3 4 x βˆ’ (12)7 2 = (12)5 6 Reduce each 12 with denominators (3)3x βˆ’ (6)7 = (2)5 Multiply out each term 9x Focus on subtraction by 42 42 = 10 βˆ’ + 42 + 42 Add 42 to both sides 9x = 52 9 9 52 9 x = Focus on multiplication by 9 Divide both sides by 9 Our Solution The next example illustrates this as well. Notice the 2 isn’t a fraction in the origional equation, but to solve it we put the 2 over 1 to make it a fraction. Example 71 LCD = 6, multiply each term by 6 (6)2 3 x βˆ’ (6)2 1 = (6)3 2 x + (6)1 6 Reduce 6 with each denominator (
2)2x βˆ’ (6)2 = (3)3x + (1)1 Multiply out each term 4x 4x βˆ’ βˆ’ βˆ’ 12 = 9x + 1 Notice variable on both sides Subtract 4x from both sides Focus on addition of 1 Subtract 1 from both sides Focus on multiplication of 5 Divide both sides by 5 4x 12 = 5x + 1 1 1 βˆ’ 13 = 5x 5 5 13 5 = x Our Solution βˆ’ βˆ’ βˆ’ βˆ’ We can use this same process if there are parenthesis in the problem. We will first distribute the coefficient in front of the parenthesis, then clear the fractions. This is seen in the following example. 44 Example 72. 5 9 x + 4 27 3 2 5 6 (18)18)3 9 (18)5 6 x + = 3 Distribute 3 2 through parenthesis, reducing if possible LCD = 18, multiply each term by 18 Reduce 18 with each denominator (3)5x + (2)2 = (18)3 Multiply out each term βˆ’ 15x + 4 = 54 4 4 βˆ’ 15x = 50 15 10 3. 15 x = Focus on addition of 4 Subtract 4 from both sides Focus on multiplication by 15 Divide both sides by 15. Reduce on right side. Our Solution While the problem can take many different forms, the pattern to clear the fraction is the same, after distributing through any parentheses we multiply each term by the LCD and reduce. This will give us a problem with no fractions that is much easier to solve. The following example again illustrates this process. Example 73 βˆ’ βˆ’ Distribute 1 3, reduce if possible LCD = 4, multiply each term by 4. (4)3 4 x βˆ’ (4)1 2 = (4)1 4 x + (4)2 1 βˆ’ (4)7 2 Reduce 4 with each denominator (1)3x βˆ’ (2)1 = (1)1x + (4)2 3x βˆ’ 2 = x + 8 βˆ’ 2 = x 3x βˆ’ x x 2x (2)7 Multiply out each term 14 Combine like terms 8 6 Notice variable on both sides Subtract x from both sides Focus on subtraction by 2 βˆ’ 2 = 14 βˆ’ βˆ’ 6 βˆ’ βˆ’ βˆ’ + 2 + 2 Add 2 to both sides βˆ’ 2x = 2 x = Focus on multiplication by 2 Divide both sides by 2 4
βˆ’ 2 2 Our Solution βˆ’ World View Note: The Egyptians were among the first to study fractions and linear equations. The most famous mathematical document from Ancient Egypt is the Rhind Papyrus where the unknown variable was called β€œheap” 45 1.4 Practice - Fractions Solve each equation. 1) 3 5(1 + p) = 21 20 6 5 ) 5 4(x βˆ’ βˆ’ m = 113 24 5 4 3) 0 = 5) 3 4 βˆ’ 7) 635 72 = 5 2( βˆ’ βˆ’ 11 4 + x) 9) 2b + 9 5 = 11 5 βˆ’ n + 1) = 3 2 11) 3 2( 7 3 5 4( a βˆ’ 8 3 a + 1) = 19 4 βˆ’ 5 3) 4 3) βˆ’ βˆ’ 4 3 n 13) βˆ’ 15) 55 17) 16 r 19) 21 11 3 + 3 2 2(b 5 3 ) βˆ’ 5 2 x 23) ( βˆ’ βˆ’ 25) 45 16 + 19 16 n βˆ’ 27) 3 2(v + 3 2) = 29) 47 βˆ’ 19 6 x + 1) βˆ’ 4) 3 2 n 8 3 = 29 12 βˆ’ 4 + 3 4 βˆ’ r = 163 32 6) 11 16 9 = βˆ’ 4 3( 5 3 + n) βˆ’ 10) 3 12) 41 14) 1 3( 13 8 βˆ’ 83 24 βˆ’ 2( 2 1 3 x 16) βˆ’ βˆ’ 3(m + 9 18) 2 4) 7 4 k + 1) 10 3 k = βˆ’ 3 4) 7 2 x = βˆ’ 10 3 = βˆ’ 53 18 βˆ’ x + 5 3 (x βˆ’ 7 4) 20) 1 12 = 4 3 βˆ’ n + 2(n + 3 2 ) 3 2 βˆ’ 11 3 r= 22) 7 6 βˆ’ 4 3 n = 24) 26) βˆ’ βˆ’ 149 16 βˆ’ 2 ( 5 7 3 βˆ’ 3 ) = 11 4 βˆ’ a + 25 8 a + 1 7 4 r 5 4( 4 3 βˆ’ r + 1) 1 2 28) 8 3 βˆ’ βˆ’ n + 29 30( 4 βˆ’ n + 2 3) 3 2 3 ( 13 4 x + 1) βˆ’ 46 1.5 Solving Linear Equations - Formulas Objective: Solve linear formulas for a given variable. Solving formulas is much like solving general linear equations. The only difference is we will have several varaibles in the problem and we will be attempting to solve for one speciοΏ½
οΏ½οΏ½c variable. For example, we may have a formula such as A = Ο€r2 + Ο€rs (formula for surface area of a right circular cone) and we may be interested in solving for the varaible s. This means we want to isolate the s so the equation has s on one side, and everything else on the other. So a solution might look like s = A. This second equation gives the same information as the first, they are algebraically equivalent, however, one is solved for the area, while the other is solved for s (slant height of the cone). In this section we will discuss how we can move from the first equation to the second. βˆ’ Ο€s Ο€r2 When solving formulas for a variable we need to focus on the one varaible we are trying to solve for, all the others are treated just like numbers. This is shown in the following example. Two parallel problems are shown, the first is a normal onestep equation, the second is a formula that we are solving for x Example 74. 3x = 12 3 3 x = 4 In both problems, x is multiplied by something wx = z w w To isolate the x we divide by 3 or w. x = Our Solution z w We use the same process to solve 3x = 12 for x as we use to solve w x = z for x. Because we are solving for x we treat all the other variables the same way we would treat numbers. Thus, to get rid of the multiplication we divided by w. This same idea is seen in the following example. Example 75. m + n = p for n Solving for n, treat all other variables like numbers m n = p Subtract m from both sides Our Solution m m βˆ’ βˆ’ βˆ’ As p and m are not like terms, they cannot be combined. For this reason we leave the expression as p m. This same one-step process can be used with grouping symbols. βˆ’ 47 Example 76. y) = b βˆ’ y) (x a(x (x βˆ’ y) βˆ’ for a Solving for a, treat (x y) like a number βˆ’ Divide both sides by (x y) βˆ’ a = x b βˆ’ y Our Solution βˆ’ Because (x y) is in parenthesis, if we are not searching for what is inside the parenthesis, we can keep them together as a group and divide by
that group. However, if we are searching for what is inside the parenthesis, we will have to break up the parenthesis by distributing. The following example is the same formula, but this time we will solve for x. Example 77. βˆ’ βˆ’ y) = b for x Solving for x, we need to distribute to clear parenthesis a(x ax ay = b + ay + ay ax = b + ay a This is a two Add ay to both sides The x is multipied by a Divide both sides by a βˆ’ step equation, ay is subtracted from our x term a b + ay a x = Our Solution Be very careful as we isolate x that we do not try and cancel the a on top and bottom of the fraction. This is not allowed if there is any adding or subtracting in the fraction. There is no reducing possible in this problem, so our final reduced answer remains x = b + ay. The next example is another two-step problem a Example 78. y = mx + b for m Solving for m, focus on addition first b βˆ’ y Subtract b from both sides m is multipied by x. Divide both sides by x b βˆ’ b = mx x b = m Our Solution βˆ’ x y βˆ’ x It is important to note that we know we are done with the problem when the variable we are solving for is isolated or alone on one side of the equation and it does not appear anywhere on the other side of the equation. The next example is also a two-step equation, it is the problem we started with at the beginning of the lesson. 48 Example 79. A = Ο€r2 + Ο€rs for s Solving for s, focus on what is added to the term with s βˆ’ Ο€r2 A βˆ’ βˆ’ Ο€r A Ο€r2 Ο€r2 = Ο€rs Ο€r Ο€r2 βˆ’ Ο€r Subtract Ο€r2 from both sides s is multipied by Ο€r Divide both sides by Ο€r = s Our Solution Again, we cannot reduce the Ο€r in the numerator and denominator because of the subtraction in the problem. Formulas often have fractions in them and can be solved in much the same way we solved with fractions before. First identify the LCD and then multiply each term by the LCD. After we reduce there will be no more fractions in the problem so we can solve like any general
equation from there. Example 80. h = 2m n (n)2m n (n)h = nh = 2m 2 2 nh 2 = m for m To clear the fraction we use LCD = n Multiply each term by n Reduce n with denominators Divide both sides by 2 Our Solution The same pattern can be seen when we have several fractions in our problem. Example 81. (b)a b + + c b = e (b) a b (b)c b a + c = eb c c βˆ’ c a = eb βˆ’ βˆ’ = e for a To clear the fraction we use LCD = b Multiply each term by b Reduce b with denominators Subtract c from both sides Our Solution Depending on the context of the problem we may find a formula that uses the same letter, one capital, one lowercase. These represent different values and we must be careful not to combine a capital variable with a lower case variable. Example 82. a = A βˆ’ b 2 for b Use LCD (2 b) as a group βˆ’ 49 (2 βˆ’ b)a = b)A b (2 βˆ’ 2 βˆ’ Multiply each term by (2 b) βˆ’ βˆ’ βˆ’ (2 2a 2a ab = A a b)a = A ab = A 2a βˆ’ 2a βˆ’ a βˆ’ 2a βˆ’ ) with denominator reduce (2 Distribute through parenthesis Subtract 2a from both sides The b is multipied by Divide both sides by a βˆ’ a βˆ’ Our Solution Notice the A and a were not combined as like terms. This is because a formula will often use a capital letter and lower case letter to represent different variables. Often with formulas there is more than one way to solve for a variable. The next example solves the same problem in a slightly different manner. After clearing the denominator, we divide by a to move it to the other side, rather than distributing. Example 83. a = for b Use LCD = (2 b) as a group βˆ’ Multiply each term by (2 b) βˆ’ Reduce (2 βˆ’ Divide both sides by a b) with denominator Focus on the positive 2 Subtract 2 from both sides 2 Still need to clear the negative (2 βˆ’ b)a = (2 βˆ’ A 2 b βˆ’ b)A (2 βˆ’ b 2 βˆ’ b))( ( βˆ’ βˆ’ b) = ( 1) βˆ’ 1
) Multiply (or divide) each term by 1 βˆ’ Our Solution Both answers to the last two examples are correct, they are just written in a different form because we solved them in different ways. This is very common with formulas, there may be more than one way to solve for a varaible, yet both are equivalent and correct. World View Note: The father of algebra, Persian mathematician Muhammad ibn Musa Khwarizmi, introduced the fundamental idea of blancing by subtracting the same term to the other side of the equation. He called this process al-jabr which later became the world algebra. 50 1.5 Practice - Formulas Solve each of the following equations for the indicated variable. 1) ab = c for b x = b for x 3) f g 5) 3x = a b for x 7) E = mc2 for m 9) V = 4 3 Ο€r3 for Ο€ 11) a + c = b for c 13) c = 4y m + n for y 15) V = Ο€Dn 12 for D 17) P = n(p c) for n 19) T = D βˆ’ d L for D βˆ’ 21) L = Lo(1 + at) for Lo 23) 2m + p = 4m + q for m 25) k m βˆ’ r = q for k 16t2 for v βˆ’ 27) h = vt 29) Q1 = P (Q2 βˆ’ 31) R = kA(T1 + T2) d 2) g = h i for h 4) p = 3y q for y 6) ym b = c d for y 8) DS = ds for D 10) E = mv2 for m 2 12) x f = g for x βˆ’ rs 3 = k for r a 14) βˆ’ 16) F = k(R L) for k βˆ’ 18) S = L + 2B for L 20) I = Ea Eq βˆ’ R for Ea 22) ax + b = c for x 24) q = 6(L p) for L βˆ’ 26) R = aT + b for T 28) S = Ο€rh + Ο€r2 for h Q1) for Q2 for T1 30) L = Ο€(r1 + r2) + 2d for r1 32) P = V1(V2 βˆ’ g V1) for V2 33)
ax + b = c for a 35) lwh = V for w a + b = c a for a 37) 1 39) at βˆ’ bw = s for t 41) ax + bx = c for a 43) x + 5y = 3 for y 45) 3x + 2y = 7 for y 47) 5a 49) 4x βˆ’ βˆ’ 7b = 4 for b 5y = 8 for y 34) rt = d for r 36) V = Ο€r2h a + b = c 38) 1 3 a for b for h 40) at βˆ’ bw = s for w 42) x + 5y = 3 for x 44) 3x + 2y = 7 for x 46) 5a βˆ’ 7b = 4 for a 5y = 8 for x 48) 4x βˆ’ 50) C = 5 9 (F 32) for F βˆ’ 51 1.6 Solving Linear Equations - Absolute Value Objective: Solve linear absolute value equations. When solving equations with absolute value we can end up with more than one possible answer. This is because what is in the absolute value can be either negative or positive and we must account for both possibilities when solving equations. This is illustrated in the following example. Example 84. x | x = 7 or x = | = 7 Absolute value can be positive or negative 7 Our Solution βˆ’ Notice that we have considered two possibilities, both the positive and negative. Either way, the absolute value of our number will be positive 7. World View Note: The first set of rules for working with negatives came from 7th century India. However, in 1758, almost a thousand years later, British mathematician Francis Maseres claimed that negatives β€œDarken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple.” When we have absolute values in our problem it is important to first isolate the absolute value, then remove the absolute value by considering both the positive and negative solutions. Notice in the next two examples, all the numbers outside of the absolute value are moved to the other side first before we remove the absolute value bars and consider both positive and negative solutions. Example 85. x | 5 + 5 = 8 Notice absolute value is not alone Subtract 5 from both sides | βˆ’ x = 3 Absolute value can be positive or negative | x = 3 or x = 3 Our Solution βˆ’ 5 | βˆ’
Example 86. = x | 4 βˆ’ 4 | βˆ’ βˆ’ βˆ’ 20 Notice absolute value is not alone 4 Divide both sides by 4 βˆ’ 52 x = 5 | x = 5 or x = | βˆ’ Absolute value can be positive or negative 5 Our Solution Notice we never combine what is inside the absolute value with what is outside the absolute value. This is very important as it will often change the final result to an incorrect solution. The next example requires two steps to isolate the absolute value. The idea is the same as a two-step equation, add or subtract, then multiply or divide. Example 87. x 5 | 4 = 26 Notice the absolute value is not alone | βˆ’ + 4 + 4 Add 4 to both sides 5 | 5 = 30 Absolute value still not alone Divide both sides by 5 x | 5 = 6 Absolute value can be positive or negative x | x = 6 or x = | 6 Our Solution βˆ’ Again we see the same process, get the absolute value alone first, then consider the positive and negative solutions. Often the absolute value will have more than just a variable in it. In this case we will have to solve the resulting equations when we consider the positive and negative possibilities. This is shown in the next example. Example 88. | 1 = 7 or 2x 2x 1 | βˆ’ 1 = = 7 Absolute value can be positive or negative 7 Two equations to solve βˆ’ βˆ’ 2x βˆ’ Now notice we have two equations to solve, each equation will give us a different solution. Both equations solve like any other two-step equation. 2x 1 = 7 βˆ’ + 1 + 1 2x = 8 2 2 x = 4 2x or 53 βˆ’ 1 = 7 βˆ’ + 1 + 1 2x = 6 2 x = βˆ’ 2 3 βˆ’ Thus, from our previous example we have two solutions, x = 4 or x = 3. βˆ’ Again, it is important to remember that the absolute value must be alone first before we consider the positive and negative possibilities. This is illustrated in below. Example 89. 2 4 2x + 3 | | βˆ’ = βˆ’ 18 4. Notice we cannot combine the 2 and To get the absolute value alone we first need to get rid of the 2 by subtracting, 4 becuase they are then divide by βˆ’ not like terms, the 4 has the absolute value connected to it. Also notice we do not distribute the 4 into the absolute value. This is because the
numbers outside cannot be combined with the numbers inside the absolute value. Thus we get the absolute value alone in the following way: βˆ’ βˆ’ βˆ’ 4 | βˆ’ 2 2 2x + 3 = | βˆ’ | 4 = βˆ’ 2x + 3 4 2x + 3 | βˆ’ = 5 | 2x + 3 = 5 or 2x + 3 = | βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 18 Notice absolute value is not alone Subtract 2 from both sides 2 20 Absolute value still not alone 4 Divide both sides by 4 βˆ’ Absoloute value can be positive or negative 5 Two equations to solve Now we just solve these two remaining equations to find our solutions. βˆ’ 2x + 3 = 5 3 3 βˆ’ 2x = 2 2 2 x = 1 or 5 βˆ’ 3 βˆ’ 8 βˆ’ 2 4 βˆ’ 2x + 3 = 3 βˆ’ 2x = 2 x = 4. βˆ’ We now have our two solutions, x = 1 and x = As we are solving absolute value equations it is important to be aware of special cases. Remember the result of an absolute value must always be positive. Notice what happens in the next example. 54 Example 90. βˆ’ 2x | βˆ’ 5 7 + 7 2x | 5 | βˆ’ = βˆ’ | βˆ’ = 4 Notice absolute value is not alone Subtract 7 from both sides 7 3 Result of absolute value is negative! Notice the absolute value equals a negative number! This is impossible with absolute value. When this occurs we say there is no solution or βˆ…. One other type of absolute value problem is when two absolute values are equal to eachother. We still will consider both the positive and negative result, the difference here will be that we will have to distribute a negative into the second absolute value for the negative possibility. Example 91. 2x = | 7 = 4x + 6 or 2x βˆ’ 7 | 2x βˆ’ | βˆ’ 4x + 6 7 = | βˆ’ Absolute value can be positive or negative (4x + 6) make second part of second equation negative Notice the first equation is the positive possibility and has no significant difference other than the missing absolute value bars. The second equation considers the negative possibility. For this reason we have a negative in front of the expression which will be distributed through the equation on the first step of solving. So we solve both these equations as follows: 2x βˆ’ βˆ’ 2x 7 = 4x + 6 2x 7 =
2x + 6 6 6 βˆ’ βˆ’ βˆ’ βˆ’ 13 = 2x βˆ’ 2 2 13 βˆ’ 2 = x or 6 βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ (4x + 6) 2x 7 = 2x 4x 7 = + 4x + 4x 6x 6 7 = βˆ’ + 7 + 7 6x = 1 6 6 1 6 x = βˆ’ This gives us our two solutions, x = βˆ’ 13 2 or x = 1 6. 55 1.6 Practice - Absolute Value Equations Solve each equation + 8a = 53 | 3k + 8 | 9 + 7x | = 2 = 30 1) 3) 5) 7) | | | | 9) | 11) 13) = 50 | = 24 | | 8 + 6m 6 2x βˆ’ 7 | 3 15) βˆ’ 17) 7 | βˆ’ 7x βˆ’ 3 3r = | = 21 | βˆ’ 19) | βˆ’ 10 | = 3 | βˆ’ 4b βˆ’ 8 3 = 5 21) 8 x + 7 | βˆ’ | 3 + 7m | 25) 3 + 5 23) 5 8 | 2x + 1 = 51 = 63 | 2 βˆ’ | + 10 = 44 27) 29) 31) 33) 35) 6b | | βˆ’ 7 + 8 βˆ’ 5x + 3 | βˆ’ = 1 2x | βˆ’ 2x + 3 = | 4 | | | | 3x 4x βˆ’ 2 βˆ’ 5 | 6x + 3 2 | = | | 7x 3 | βˆ’ = 73 2) 4) 6) 8) | | | | 10) 12) 14) 16) 21 βˆ’ = 7 = 2 n x | | 9n + 8 = 46 | 3 | | | | x = 6 | βˆ’ 5n + 7 = 23 | 9p + 6 = 3 | 2 = 7 | 3n βˆ’ 2 + 2b + 1 = 3 | 18) | βˆ’ 3n 4 βˆ’ 4 | = 2 5 = 11 40 = βˆ’ | 20) 8 22) 3 24) 4 5p + 8 | | βˆ’ 6n + 7 βˆ’ | r + 7 | + 3 = 59 | 26) 5 + 8 10n 2 | βˆ’ | βˆ’ = 101 28) 7 10v | 2 9 = 5 | βˆ’ 30) 8 βˆ’ 3n 3 | βˆ’ 2 + 3x = | 2x 5 βˆ’ 3 3x + 2 2 = = | | | | 5 = 91 | βˆ’ 4 βˆ’ | 3x + 4 2 2x 3 βˆ’ 3 2x | |
| | | 32) 34) 36) | | | 56 1.7 Solving Linear Equations - Variation Objective: Solve variation problems by creating variation equations and finding the variation constant. One application of solving linear equations is variation. Often different events are related by what is called the constant of variation. For example, the time it takes to travel a certain distance is related to how fast you are traveling. The faster you travel, the less time it take to get there. This is one type of variation problem, we will look at three types of variation here. Variation problems have two or three variables and a constant in them. The constant, usually noted with a k, describes the relationship and does not change as the other variables in the problem change. There are two ways to set up a variation problem, the first solves for one of the variables, a second method is to solve for the constant. Here we will use the second method. The greek letter pi (Ο€) is used to represent the ratio of the circumference of a circle to its diameter. World View Note: In the 5th centure, Chinese mathematician Zu Chongzhi calculated the value of Ο€ to seven decimal places (3.1415926). This was the most accurate value of Ο€ for the next 1000 years! If you take any circle and divide the circumference of the circle by the diameter you will always get the same value, about 3.14159... If you have a bigger circumference you will also have a bigger diameter. This relationship is called direct variation or directly proportional. If we see this phrase in the problem we know to divide to find the constant of variation. Example 92. m is varies directly as n β€²β€²Directlyβ€²β€² tells us to divide = k Our formula for the relationship m n In kickboxing, one will find that the longer the board, the easier it is to break. If you multiply the force required to break a board by the length of the board you will also get a constant. Here, we are multiplying the variables, which means as one variable increases, the other variable decreases. This relationship is called indirect variation or inversly proportional. If we see this phrase in the problem we know to multiply to find the constant of variation. Example 93. y is inversely proportional to z β€²β€²Inverselyβ€²β€² tells us to multiply yz = k Our formula for
the relationship 57 The formula for the area of a triangle has three variables in it. If we divide the area by the base times the height we will also get a constant, 1. This relationship 2 is called joint variation or jointly proportional. If we see this phrase in the problem we know to divide the first variable by the product of the other two to find the constant of variation. Example 94. A varies jointly as x and y β€²β€²Jointlyβ€²β€² tells us to divide by the product A xy = k Our formula for the relationship Once we have our formula for the relationship in a variation problem, we use given or known information to calculate the constant of variation. This is shown for each type of variation in the next three examples. Example 95. Example 96. Example 97. w is directly proportional to y and w = 50 when y = 5 w y (50) (5) = k β€²β€²directlyβ€²β€² tells us to divide = k Substitute known values 10 = k Evaluate to find our constant c varies indirectly as d and c = 4.5 when d = 6 cd = k (4.5)(6) = k β€²β€²indirectlyβ€²β€² tells us to multiply Substitute known values 27 = k Evaluate to find our constant x is jointly proportional to y and z and x = 48 when y = 2 and z = 4 x yz (48) (2)(4) = k β€²β€²Jointlyβ€²β€² tells us to divide by the product = k Substitute known values 6 = k Evaluate to find our constant 58 Once we have found the constant of variation we can use it to find other combinations in the same relationship. Each of these problems we solve will have three important steps, none of which should be skipped. 1. Find the formula for the relationship using the type of variation 2. Find the constant of variation using known values 3. Answer the question using the constant of variation The next three examples show how this process is worked out for each type of variation. Example 98. The price of an item varies directly with the sales tax. If a S25 item has a sales tax of S2, what will the tax be on a S40 item? β€²β€²Directlyβ€²β€² tells us to divide price (p) and tax (t) = k p t (25) (2) 12.5 = k Eval
uate to find our constant = k Substitute known values for price and tax 40 t = 12.5 Using our constant, substitute 40 for price to find the tax = 12.5(t) Multiply by LCD = t to clear fraction 40 = 12.5t Reduce the t with the denominator (t)40 t 12.5 12.5 Divide by 12.5 3.2 = t Our solution: Tax is S3.20 Example 99. The speed (or rate) Josiah travels to work is inversely proportional to time it takes to get there. If he travels 35 miles per hour it will take him 2.5 hours to get to work. How long will it take him if he travels 55 miles per hour? rt = k (35)(2.5) = k β€²β€²Inverselyβ€²β€² tells us to multiply the rate and time Substitute known values for rate and time 87.5 = k Evaluate to find our constant 55t = 87.5 Using our constant, substitute 55 for rate to find the time 55 t 55 1.59 Our solution: It takes him 1.59 hours to get to work Divide both sides by 55 β‰ˆ Example 100. 59 The amount of simple interest earned on an investment varies jointly as the principle (amount invested) and the time it is invested. In an account, S150 invested for 2 years earned S12 in interest. How much interest would be earned on a S220 investment for 3 years? I Pt (12) (150)(2) = k β€²β€²Jointlyβ€²β€² divide Interest (I) by product of Principle (P )& time (t) = k Substitute known values for Interest, Principle and time 0.04 = k Evaluate to find our constant I (220)(3) = 0.04 Using constant, substitute 220 for principle and 3 for time I 660 = 0.04 Evaluate denominator = 0.04(660) Multiply by 660 to isolate the variable I = 26.4 Our Solution: The investment earned S26.40 in interest (660)I 660 Sometimes a variation problem will ask us to do something to a variable as we set up the formula for the relationship. For example, Ο€ can be thought of as the ratio of the area and the radius squared. This is still direct variation, we say the area varies directly as the radius square and thus our
variable is squared in our formula. This is shown in the next example. Example 101. The area of a circle is directly proportional to the square of the radius. A circle with a radius of 10 has an area of 314. What will the area be on a circle of radius 4? β€²β€²Directβ€²β€² tells us to divide, be sure we use r2 for the denominator A r2 = k (314) (10)2 = k (314) 100 3.14 = k Divide to find our constant = k Exponents first Substitute known values into our formula A (4)2 = 3.14 Using the constant, use 4 for r, donβ€²t forget the squared! A 16 = 3.14 Evaluate the exponent = 3.14(16) Multiply both sides by 16 A = 50.24 Our Solution: Area is 50.24 (16)A 16 When solving variation problems it is important to take the time to clearly state the variation formula, find the constant, and solve the final equation. 60 1.7 Practice - Variation Write the formula that expresses the relationship described 1. c varies directly as a 2. x is jointly proportional to y and z 3. w varies inversely as x 4. r varies directly as the square of s 5. f varies jointly as x and y 6. j is inversely proportional to the cube of m 7. h is directly proportional to b 8. x is jointly proportional with the square of a and the square root of b 9. a is inversely proportional to b Find the constant of variation and write the formula to express the relationship using that constant 10. a varies directly as b and a = 15 when b = 5 11. p is jointly proportional to q and r and p = 12 when q = 8 and r = 3 12. c varies inversely as d and c = 7 when d = 4 13. t varies directly as the square of u and t = 6 when u = 3 14. e varies jointly as f and g and e = 24 when f = 3 and g = 2 15. w is inversely proportional to the cube of x and w is 54 when x = 3 16. h is directly proportional to j and h = 12 when j = 8 17. a is jointly proportional with the square of x and the square root of y and a = 25 when x = 5 and y = 9 18. m is
inversely proportional to n and m = 1.8 when n = 2.1 Solve each of the following variation problems by setting up a formula to express the relationship, finding the constant, and then answering 61 the question. 19. The electrical current, in amperes, in a circuit varies directly as the voltage. When 15 volts are applied, the current is 5 amperes. What is the current when 18 volts are applied? 20. The current in an electrical conductor varies inversely as the resistance of the conductor. If the current is 12 ampere when the resistance is 240 ohms, what is the current when the resistance is 540 ohms? 21. Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If the distance is 20 cm when the mass is 3 kg, what is the distance when the mass is 5 kg? 22. The volume of a gas varies inversely as the pressure upon it. The volume of a gas is 200 cm3 under a pressure of 32 kg/cm2. What will be its volume under a pressure of 40 kg/cm2? 23. The number of aluminum cans used each year varies directly as the number of people using the cans. If 250 people use 60,000 cans in one year, how many cans are used each year in Dallas, which has a population of 1,008,000? 24. The time required to do a job varies inversely as the number of peopel working. It takes 5hr for 7 bricklayers to build a park well. How long will it take 10 bricklayers to complete the job? 25. According to Fidelity Investment Vision Magazine, the average weekly allowance of children varies directly as their grade level. In a recent year, the average allowance of a 9th-grade student was 9.66 dollars per week. What was the average weekly allowance of a 4th-grade student? 26. The wavelength of a radio wave varies inversely as its frequency. A wave with a frequency of 1200 kilohertz has a length of 300 meters. What is the length of a wave with a frequency of 800 kilohertz? 27. The number of kilograms of water in a human body varies directly as the mass of the body. A 96-kg person contains 64 kg of water. How many kilo grams of water are in a 60-kg person? 28. The time required to drive
a fixed distance varies inversely as the speed. It takes 5 hr at a speed of 80 km/h to drive a fixed distance. How long will it take to drive the same distance at a speed of 70 km/h? 29. The weight of an object on Mars varies directly as its weight on Earth. A person weighs 95lb on Earth weighs 38 lb on Mars. How much would a 100-lb person weigh on Mars? 30. At a constant temperature, the volume of a gas varies inversely as the pres- 62 sure. If the pressure of a certain gas is 40 newtons per square meter when the volume is 600 cubic meters what will the pressure be when the volume is reduced by 240 cubic meters? 31. The time required to empty a tank varies inversely as the rate of pumping. If a pump can empty a tank in 45 min at the rate of 600 kL/min, how long will it take the pump to empty the same tank at the rate of 1000 kL/min? 32. The weight of an object varies inversely as the square of the distance from the center of the earth. At sea level (6400 km from the center of the earth), an astronaut weighs 100 lb. How far above the earth must the astronaut be in order to weigh 64 lb? 33. The stopping distance of a car after the brakes have been applied varies directly as the square of the speed r. If a car, traveling 60 mph can stop in 200 ft, how fast can a car go and still stop in 72 ft? 34. The drag force on a boat varies jointly as the wetted surface area and the square of the velocity of a boat. If a boat going 6.5 mph experiences a drag force of 86 N when the wetted surface area is 41.2 ft2, how fast must a boat with 28.5 ft2 of wetted surface area go in order to experience a drag force of 94N? 35. The intensity of a light from a light bulb varies inversely as the square of the distance from the bulb. Suppose intensity is 90 W/m2 (watts per square meter) when the distance is 5 m. How much further would it be to a point where the intesity is 40 W/m2? 36. The volume of a cone varies jointly as its height, and the square of its radius. If a cone with a height of 8 centimeters and a radius of 2 centimeters has a volume of
33.5 cm3, what is the volume of a cone with a height of 6 centimeters and a radius of 4 centimeters? 37. The intensity of a television signal varies inversely as the square of the dis- tance from the transmitter. If the intensity is 25 W/m2 at a distance of 2 km, how far from the trasmitter are you when the intensity is 2.56 W/m2? 38. The intensity of illumination falling on a surface from a given source of light is inversely proportional to the square of the distance from the source of light. The unit for measuring the intesity of illumination is usually the footcandle. If a given source of light gives an illumination of 1 foot-candle at a distance of 10 feet, what would the illumination be from the same source at a distance of 20 feet? 63 1.8 Linear Equations - Number and Geometry Objective: Solve number and geometry problems by creating and solving a linear equation. Word problems can be tricky. Often it takes a bit of practice to convert the English sentence into a mathematical sentence. This is what we will focus on here with some basic number problems, geometry problems, and parts problems. A few important phrases are described below that can give us clues for how to set up a problem. β€’ β€’ β€’ β€’ A number (or unknown, a value, etc) often becomes our variable Is (or other forms of is: was, will be, are, etc) often represents equals (=) x is 5 becomes x = 5 More than often represents addition and is usually built backwards, writing the second part plus the first Three more than a number becomes x + 3 Less than often represents subtraction and is usually built backwards as well, writing the second part minus the first Four less than a number becomes x 4 βˆ’ Using these key phrases we can take a number problem and set up and equation and solve. Example 102. If 28 less than five times a certain number is 232. What is the number? 5x 5x 28 βˆ’ 28 = 232 Subtraction is built backwards, multiply the unknown by 5 Is translates to equals βˆ’ + 28 + 28 Add 28 to both sides 5x = 260 The variable is multiplied by 5 5 Divide both sides by 5 5 x = 52 The number is 52. This same idea can be extended to a more involved problem as shown in the next example. Example 103. 64 Fifteen more than three times a number is the
same as ten less than six times the number. What is the number 3x + 15 6x 3x + 15 = 6x 3x 3x βˆ’ 15 = 3x βˆ’ βˆ’ βˆ’ First, addition is built backwards 10 Then, subtraction is also built backwards 10 Is between the parts tells us they must be equal Subtract 3x so variable is all on one side 10 Now we have a two βˆ’ βˆ’ + 10 Add 10 to both sides + 10 step equation 3 25 = 3x The variable is multiplied by 3 3 Divide both sides by 3 25 3 = x Our number is 25 3 Another type of number problem involves consecutive numbers. Consecutive numbers are numbers that come one after the other, such as 3, 4, 5. If we are looking for several consecutive numbers it is important to first identify what they look like with variables before we set up the equation. This is shown in the following example. Example 104. The sum of three consecutive integers is 93. What are the integers? First x Make the first number x Second x + 1 To get the next number we go up one or + 1 Third x + 2 Add another 1(2 total) to get the third F + S + T = 93 First (F ) plus Second (S) plus Third (T ) equals 93 (x) + (x + 1) + (x + 2) = 93 Replace F with x, S with x + 1, and T with = 93 Here the parenthesis arenβ€²t needed. 3x + 3 = 93 Combine like terms x + x + x and 2 + 1 3 βˆ’ βˆ’ 3 Add 3 to both sides 3x = 90 The variable is multiplied by 3 3 Divide both sides by 3 x = 30 Our solution for x 3 First 30 Replace x in our origional list with 30 Second (30) + 1 = 31 The numbers are 30, 31, and 32 Third (30) + 2 = 32 Sometimes we will work consective even or odd integers, rather than just consecutive integers. When we had consecutive integers, we only had to add 1 to get to the next number so we had x, x + 1, and x + 2 for our first, second, and third number respectively. With even or odd numbers they are spaced apart by two. So if we want three consecutive even numbers, if the first is x, the next number would be x + 2, then fin
ally add two more to get the third, x + 4. The same is 65 true for consecutive odd numbers, if the first is x, the next will be x + 2, and the third would be x + 4. It is important to note that we are still adding 2 and 4 even when the numbers are odd. This is because the phrase β€œodd” is refering to our x, not to what is added to the numbers. Consider the next two examples. Example 105. The sum of three consecutive even integers is 246. What are the numbers? First x Make the first x Second x + 2 Even numbers, so we add 2 to get the next Third x + 4 Add 2 more (4 total) to get the third F + S + T = 246 Sum means add First (F ) plus Second (S) plus Third (T ) (x) + (x + 2) + (x + 4) = 246 Replace each F, S, and T with what we labeled them x + x + 2 + x + 4 = 246 Here the parenthesis are not needed 3x + 6 = 246 Combine like terms x + x + x and 2 + 4 6 6 Subtract 6 from both sides βˆ’ βˆ’ 3x = 240 The variable is multiplied by 3 3 Divide both sides by 3 3 x = 80 Our solution for x First 80 Replace x in the origional list with 80. Second (80) + 2 = 82 The numbers are 80, 82, and 84. Third ( 80) + 4 = 84 Example 106. Find three consecutive odd integers so that the sum of twice the first, the second and three times the third is 152. First x Make the first x Second x + 2 Odd numbers so we add 2(same as even!) Third x + 4 Add 2 more (4 total) to get the third 2F + S + 3T = 152 Twice the first gives 2F and three times the third gives 3T 2(x) + (x + 2) + 3(x + 4) = 152 Replace F, S, and T with what we labled them 2x + x + 2 + 3x + 12 = 152 Distribute through parenthesis 6x + 14 = 152 Combine like terms 2x + x + 3x and 2 + 14 Subtract 14 from both sides βˆ’ βˆ’ 14 14 6x = 138 Variable is multiplied by 6
6 Divide both sides by 6 6 x = 23 Our solution for x First 23 Replace x with 23 in the original list Second (23) + 2 = 25 The numbers are 23, 25, and 27 Third (23) + 4 = 27 66 When we started with our first, second, and third numbers for both even and odd we had x, x + 2, and x + 4. The numbers added do not change with odd or even, it is our answer for x that will be odd or even. Another example of translating English sentences to mathematical sentences comes from geometry. A well known property of triangles is that all three angles will always add to 180. For example, the first angle may be 50 degrees, the second 30 degrees, and the third 100 degrees. If you add these together, 50 + 30 + 100 = 180. We can use this property to find angles of triangles. World View Note: German mathematician Bernhart Thibaut in 1809 tried to prove that the angles of a triangle add to 180 without using Euclid’s parallel postulate (a point of much debate in math history). He created a proof, but it was later shown to have an error in the proof. Example 107. The second angle of a triangle is double the first. The third angle is 40 less than the first. Find the three angles. First x With nothing given about the first we make that x Second 2x The second is double the first, Third x 40 The third is 40 less than the first F + S + T = 180 All three angles add to 180 βˆ’ (x) + (2x) + (x x + 2x + x 4x 40) = 180 Replace F, S, and T with the labeled values. 40 = 180 Here the parenthesis are not needed. 40 = 180 Combine like terms, x + 2x + x βˆ’ βˆ’ βˆ’ + 40 + 40 Add 40 to both sides 4x = 220 The variable is multiplied by 4 4 x = 55 Our solution for x Divide both sides by 4 4 First 55 Replace x with 55 in the original list of angles Second 2(55) = 110 Our angles are 55, 110, and 15 Third (55) 40 = 15 βˆ’ Another geometry problem involves perimeter or the distance around an object. For example, consider a rectangle has a length of 8 and a width of 3.
There are two lengths and two widths in a rectangle (opposite sides) so we add 8 + 8 + 3 + 3 = 22. As there are two lengths and two widths in a rectangle an alternative to find the perimeter of a rectangle is to use the formula P = 2L + 2W. So for the rectangle of length 8 and width 3 the formula would give, P = 2(8) + 2(3) = 16 + 6 = 22. With problems that we will consider here the formula P = 2L + 2W will be used. Example 108. 67 The perimeter of a rectangle is 44. The length is 5 less than double the width. Find the dimensions. Length x We will make the length x βˆ’ 5 Width is five less than two times the length Width 2x P = 2L + 2W The formula for perimeter of a rectangle 5) Replace P, L, and W with labeled values 10 Distribute through parenthesis 10 Combine like terms 2x + 4x βˆ’ βˆ’ βˆ’ + 10 Add 10 to both sides + 10 (44) = 2(x) + 2(2x 44 = 2x + 4x 44 = 6x 54 = 6x The variable is multiplied by 6 Divide both sides by 6 6 6 9 = x Our solution for x Length 9 Replace x with 9 in the origional list of sides 5 = 13 The dimensions of the rectangle are 9 by 13. Width 2(9) βˆ’ We have seen that it is imortant to start by clearly labeling the variables in a short list before we begin to solve the problem. This is important in all word problems involving variables, not just consective numbers or geometry problems. This is shown in the following example. Example 109. A sofa and a love seat together costs S444. The sofa costs double the love seat. How much do they each cost? Love Seat x With no information about the love seat, this is our x Sofa 2x Sofa is double the love seat, so we multiply by 2 S + L = 444 Together they cost 444, so we add. (x) + (2x) = 444 Replace S and L with labeled values 3x = 444 Parenthesis are not needed, combine like terms x + 2x 3 x = 148 Our solution for x Divide both sides by 3 3 Love Seat 148 Replace x with 148 in the origional list Sofa 2(148) = 296 The love seat costs S148 and the
sofa costs S296. Be careful on problems such as these. Many students see the phrase β€œdouble” and believe that means we only have to divide the 444 by 2 and get S222 for one or both of the prices. As you can see this will not work. By clearly labeling the variables in the original list we know exactly how to set up and solve these problems. 68 1.8 Practice - Number and Geometry Problems Solve. 1. When five is added to three more than a certain number, the result is 19. What is the number? 2. If five is subtracted from three times a certain number, the result is 10. What is the number? 3. When 18 is subtracted from six times a certain number, the result is What is the number? 42. βˆ’ 4. A certain number added twice to itself equals 96. What is the number? 5. A number plus itself, plus twice itself, plus 4 times itself, is equal to What is the number? 104. βˆ’ 6. Sixty more than nine times a number is the same as two less than ten times the number. What is the number? 7. Eleven less than seven times a number is five more than six times the number. Find the number. 8. Fourteen less than eight times a number is three more than four times the number. What is the number? 9. The sum of three consecutive integers is 108. What are the integers? 10. The sum of three consecutive integers is 126. What are the integers? βˆ’ 11. Find three consecutive integers such that the sum of the first, twice the second, and three times the third is 76. βˆ’ 12. The sum of two consecutive even integers is 106. What are the integers? 13. The sum of three consecutive odd integers is 189. What are the integers? 69 14. The sum of three consecutive odd integers is 255. What are the integers? 15. Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 70. 16. The second angle of a triangle is the same size as the first angle. The third angle is 12 degrees larger than the first angle. How large are the angles? 17. Two angles of a triangle are the same size. The third angle is 12 degrees smaller than the first angle. Find
the measure the angles. 18. Two angles of a triangle are the same size. The third angle is 3 times as large as the first. How large are the angles? 19. The third angle of a triangle is the same size as the first. The second angle is 4 times the third. Find the measure of the angles. 20. The second angle of a triangle is 3 times as large as the first angle. The third angle is 30 degrees more than the first angle. Find the measure of the angles. 21. The second angle of a triangle is twice as large as the first. The measure of the third angle is 20 degrees greater than the first. How large are the angles? 22. The second angle of a triangle is three times as large as the first. The measure of the third angle is 40 degrees greater than that of the first angle. How large are the three angles? 23. The second angle of a triangle is five times as large as the first. The measure of the third angle is 12 degrees greater than that of the first angle. How large are the angles? 24. The second angle of a triangle is three times the first, and the third is 12 degrees less than twice the first. Find the measures of the angles. 25. The second angle of a triangle is four times the first and the third is 5 degrees more than twice the first. Find the measures of the angles. 26. The perimeter of a rectangle is 150 cm. The length is 15 cm greater than the width. Find the dimensions. 27. The perimeter of a rectangle is 304 cm. The length is 40 cm longer than the width. Find the length and width. 28. The perimeter of a rectangle is 152 meters. The width is 22 meters less than the length. Find the length and width. 29. The perimeter of a rectangle is 280 meters. The width is 26 meters less than the length. Find the length and width. 70 30. The perimeter of a college basketball court is 96 meters and the length is 14 meters more than the width. What are the dimensions? 31. A mountain cabin on 1 acre of land costs S30,000. If the land cost 4 times as much as the cabin, what was the
cost of each? 32. A horse and a saddle cost S5000. If the horse cost 4 times as much as the saddle, what was the cost of each? 33. A bicycle and a bicycle helmet cost S240. How much did each cost, if the bicycle cost 5 times as much as the helmet? 34. Of 240 stamps that Harry and his sister collected, Harry collected 3 times as many as his sisters. How many did each collect? 35. If Mr. Brown and his son together had S220, and Mr. Brown had 10 times as much as his son, how much money had each? 36. In a room containing 45 students there were twice as many girls as boys. How many of each were there? 37. Aaron had 7 times as many sheep as Beth, and both together had 608. How many sheep had each? 38. A man bought a cow and a calf for S990, paying 8 times as much for the cow as for the calf. What was the cost of each? 39. Jamal and Moshe began a business with a capital of S7500. If Jamal furnished half as much capital as Moshe, how much did each furnish? 40. A lab technician cuts a 12 inch piece of tubing into two pieces in such a way that one piece is 2 times longer than the other. 41. A 6 ft board is cut into two pieces, one twice as long as the other. How long are the pieces? 42. An eight ft board is cut into two pieces. One piece is 2 ft longer than the other. How long are the pieces? 43. An electrician cuts a 30 ft piece of wire into two pieces. One piece is 2 ft longer than the other. How long are the pieces? 44. The total cost for tuition plus room and board at State University is S2,584. Tuition costs S704 more than room and board. What is the tuition fee? 45. The cost of a private pilot course is S1,275. The flight portion costs S625 more than the groung school portion. What is the cost of each? 71 1.9 Solving Linear Equations - Age Problems Objective: Solve age problems by creating and solving a linear equation. An application of linear equations is what are called age problems. When we are solving age problems we generally will be comparing the age of two people both now and in the future (or past). Using the clues given in the problem we will be working to οΏ½
οΏ½nd their current age. There can be a lot of information in these problems and we can easily get lost in all the information. To help us organize and solve our problem we will fill out a three by three table for each problem. An example of the basic structure of the table is below Age Now Change Person 1 Person 2 Table 6. Structure of Age Table Normally where we see β€œPerson 1” and β€œPerson 2” we will use the name of the person we are talking about. We will use this table to set up the following example. Example 110. Adam is 20 years younger than Brian. In two years Brian will be twice as old as Adam. How old are they now? Age Now + 2 Adam Brian We use Adam and Brian for our persons We use + 2 for change because the second phrase is two years in the future Age Now + 2 20 Adam x Brain βˆ’ x Consider the β€²β€²Nowβ€²β€² part, Adam is 20 years youger than Brian. We are given information about Adam, not Brian. So Brian is x now. To show Adam 20. is 20 years younger we subtract 20, Adam is x βˆ’ Age Now Adam x Brian βˆ’ x 20 x + 2 20 + 2 βˆ’ x + 2 Now the + 2 column is filled in. This is done by adding 2 to both Adamβ€²s and Brianβ€²s now column as shown in the table. Age Now + 2 Adam x Brian βˆ’ x 20 x 18 βˆ’ x + 2 Combine like terms in Adamβ€²s future age: 20 + 2 This table is now filled out and we are ready to try and solve. βˆ’ 72 B = 2A (x + 2) = 2(x 18) βˆ’ βˆ’ βˆ’ 36 x + 2 = 2x x x βˆ’ 2 = x 36 + 36 + 36 38 = x βˆ’ Age now Adam 38 Brian βˆ’ 20 = 18 38 βˆ’ Our equation comes from the future statement: Brian will be twice as old as Adam. This means the younger, Adam, needs to be multiplied by 2. Replace B and A with the information in their future cells, Adam (A) is replaced with x 18 and Brian (B) is replaced with (x + 2) This is the equation to solve! Distribute through parenthesis Subtract x from both sides to get variable on one side Need to clear the βˆ’ Add 36 to both sides Our solution for x The first column will help us answer the question. Replace
the xβ€²s with 38 and simplify. Adam is 18 and Brian is 38 36 Solving age problems can be summarized in the following five steps. These five steps are guidelines to help organize the problem we are trying to solve. 1. Fill in the now column. The person we know nothing about is x. 2. Fill in the future/past collumn by adding/subtracting the change to the now column. 3. Make an equation for the relationship in the future. This is independent of the table. 4. Replace variables in equation with information in future cells of table 5. Solve the equation for x, use the solution to answer the question These five steps can be seen illustrated in the following example. Example 111. Carmen is 12 years older than David. Five years ago the sum of their ages was 28. How old are they now? Age Now 5 βˆ’ Carmen David Carmen David 5 βˆ’ Age Now x + 12 x Five years ago is βˆ’ 5 in the change column. Carmen is 12 years older than David. We donβ€²t know about David so he is x, Carmen then is x + 12 Carmen David Age Now x + 12 x 5 βˆ’ x + 12 x βˆ’ βˆ’ 5 5 Subtract 5 from now column to get the change 73 Carmen David Age Now x + 12 x 5 βˆ’ x + 7 x 5 βˆ’ Simplify by combining like terms 12 Our table is ready! 5 βˆ’ (x + 7) + ( = 28 5) = 28 5 = 28 2x + 2 = 28 2 2 βˆ’ 2x = 26 2 2 x = 13 βˆ’ The sum of their ages will be 29. So we add C and D Replace C and D with the change cells. Remove parenthesis Combine like terms x + x and 7 Subtract 2 from both sides Notice x is multiplied by 2 Divide both sides by 2 Our solution for x βˆ’ 5 Age Now Caremen 13 + 12 = 25 David 13 Replace x with 13 to answer the question Carmen is 25 and David is 13 Sometimes we are given the sum of their ages right now. These problems can be tricky. In this case we will write the sum above the now column and make the first person’s age now x. The second person will then turn into the subtraction problem total x. This is shown in the next example. βˆ’ Example 112. The sum of the ages of Nicole and Kristin is 32. In two years Nicole will be three times as
old as Kristin. How old are they now? 32 Age Now + 2 Nicole Kristen x 32 x βˆ’ Nicole Kristen Age Now x + 2 x + 2 32 x βˆ’ x + 2 32 βˆ’ The change is + 2 for two years in the future The total is placed above Age Now The first person is x. The second becomes 32 x βˆ’ Add 2 to each cell fill in the change column Nicole Kristen Age Now + 2 x + 2 x 34 32 x x βˆ’ βˆ’ (x + 2) = 3(34 x + 2 = 102 + 3x N = 3K x) βˆ’ 3x βˆ’ + 3x Combine like terms 32 + 2, our table is done! Nicole is three times as old as Kristin. Replace variables with information in change cells Distribute through parenthesis Add 3x to both sides so variable is only on one side 74 βˆ’ βˆ’ 4x + 2 = 102 2 2 4x = 100 4 4 x = 25 Nicole Kristen 32 Age Now 25 25 = 7 βˆ’ βˆ’ step equation Solve the two Subtract 2 from both sides The variable is multiplied by 4 Divide both sides by 4 Our solution for x Plug 25 in for x in the now column Nicole is 25 and Kristin is 7 A slight variation on age problems is to ask not how old the people are, but rather ask how long until we have some relationship about their ages. In this case we alter our table slightly. In the change column because we don’t know the time to add or subtract we will use a variable, t, and add or subtract this from the now column. This is shown in the next example. Example 113. Louis is 26 years old. Her daughter is 4 years old. In how many years will Louis be double her daughter’s age? Age Now + t Louis Daughter 26 4 As we are given their ages now, these numbers go into the table. The change is unknown, so we write + t for the change Age Now + t Louis Daughter 26 4 26 + t 4 + t Fill in the change column by adding t to each personβ€²s age. Our table is now complete. L = 2D (26 + t) = 2(4 + t) 26 + t = 8 + 2t t t 26 = 8 + t 8 8 βˆ’ 18 = t βˆ’ βˆ’ βˆ’ Louis will be double her daughter Replace variables with information in change cells Distribute through parenthesis Subtract t from both sides Now we have an 8 added
to the t Subtract 8 from both sides In 18 years she will be double her daughterβ€²s age Age problems have several steps to them. However, if we take the time to work through each of the steps carefully, keeping the information organized, the problems can be solved quite nicely. World View Note: The oldest man in the world was Shigechiyo Izumi from Japan who lived to be 120 years, 237 days. However, his exact age has been disputed. 75 1.9 Practice - Age Problems 1. A boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find the present age of each. 2. A father is 4 times as old as his son. In 20 years the father will be twice as old as his son. Find the present age of each. 3. Pat is 20 years older than his son James. In two years Pat will be twice as old as James. How old are they now? 4. Diane is 23 years older than her daughter Amy. In 6 years Diane will be twice as old as Amy. How old are they now? 5. Fred is 4 years older than Barney. Five years ago the sum of their ages was 48. How old are they now? 6. John is four times as old as Martha. Five years ago the sum of their ages was 50. How old are they now? 7. Tim is 5 years older than JoAnn. Six years from now the sum of their ages will be 79. How old are they now? 8. Jack is twice as old as Lacy. In three years the sum of their ages will be 54. How old are they now? 9. The sum of the ages of John and Mary is 32. Four years ago, John was twice as old as Mary. Find the present age of each. 10. The sum of the ages of a father and son is 56. Four years ago the father was 3 times as old as the son. Find the present age of each. 11. The sum of the ages of a china plate and a glass plate is 16 years. Four years ago the china plate was three times the age of the glass plate. Find the present age of each plate. 12. The sum of the ages of a wood plaque and a bronze plaque is 20 years. Four 76 years ago, the bronze plaque was one-half the age of the wood plaque. Find the present age of each plaque. 13. A is now
34 years old, and B is 4 years old. In how many years will A be twice as old as B? 14. A man’s age is 36 and that of his daughter is 3 years. In how many years will the man be 4 times as old as his daughter? 15. An Oriental rug is 52 years old and a Persian rug is 16 years old. How many years ago was the Oriental rug four times as old as the Persian Rug? 16. A log cabin quilt is 24 years old and a friendship quilt is 6 years old. In how may years will the log cabin quilt be three times as old as the friendship quilt? 17. The age of the older of two boys is twice that of the younger; 5 years ago it was three times that of the younger. Find the age of each. 18. A pitcher is 30 years old, and a vase is 22 years old. How many years ago was the pitcher twice as old as the vase? 19. Marge is twice as old as Consuelo. The sum of their ages seven years ago was 13. How old are they now? 20. The sum of Jason and Mandy’s age is 35. Ten years ago Jason was double Mandy’s age. How old are they now? 21. A silver coin is 28 years older than a bronze coin. In 6 years, the silver coin will be twice as old as the bronze coin. Find the present age of each coin. 22. A sofa is 12 years old and a table is 36 years old. In how many years will the table be twice as old as the sofa? 23. A limestone statue is 56 years older than a marble statue. In 12 years, the limestone will be three times as old as the marble statue. Find the present age of the statues. 24. A pewter bowl is 8 years old, and a silver bowl is 22 years old. In how many years will the silver bowl be twice the age of the pewter bowl? 25. Brandon is 9 years older than Ronda. In four years the sum of their ages will be 91. How old are they now? 26. A kerosene lamp is 95 years old, and an electric lamp is 55 years old. How many years ago was the kerosene lamp twice the age of the electric lamp? 27. A father is three times as old as his son, and his daughter is 3 years younger 77 than the son.
If the sum of their ages 3 years ago was 63 years, find the present age of the father. 28. The sum of Clyde and Wendy’s age is 64. In four years, Wendy will be three times as old as Clyde. How old are they now? 29. The sum of the ages of two ships is 12 years. Two years ago, the age of the older ship was three times the age of the newer ship. Find the present age of each ship. 30. Chelsea’s age is double Daniel’s age. Eight years ago the sum of their ages was 32. How old are they now? 31. Ann is eighteen years older than her son. One year ago, she was three times as old as her son. How old are they now? 32. The sum of the ages of Kristen and Ben is 32. Four years ago Kristen was twice as old as Ben. How old are they both now? 33. A mosaic is 74 years older than the engraving. Thirty years ago, the mosaic was three times as old as the engraving. Find the present age of each. 34. The sum of the ages of Elli and Dan is 56. Four years ago Elli was 3 times as old as Dan. How old are they now? 35. A wool tapestry is 32 years older than a linen tapestry. Twenty years ago, the wool tapestry was twice as old as the linen tapestry. Find the present age of each. 36. Carolyn’s age is triple her daughter’s age. In eight years the sum of their ages will be 72. How old are they now? 37. Nicole is 26 years old. Emma is 2 years old. In how many years will Nicole be triple Emma’s age? 38. The sum of the ages of two children is 16 years. Four years ago, the age of the older child was three times the age of the younger child. Find the present age of each child. 39. Mike is 4 years older than Ron. In two years, the sum of their ages will be 84. How old are they now? 40. A marble bust is 25 years old, and a terra-cotta bust is 85 years old. In how many years will the terra-cotta bust be three times as old as the marble bust? 78 1.10 Solving Linear Equations - Distance, Rate and Time Objective: Solve distance
problems by creating and solving a linear equation. An application of linear equations can be found in distance problems. When solving distance problems we will use the relationship rt = d or rate (speed) times time equals distance. For example, if a person were to travel 30 mph for 4 hours. To find the total distance we would multiply rate times time or (30)(4) = 120. This person travel a distance of 120 miles. The problems we will be solving here will be a few more steps than described above. So to keep the information in the problem organized we will use a table. An example of the basic structure of the table is blow: Rate Time Distance Person 1 Person 2 Table 7. Structure of Distance Problem The third column, distance, will always be filled in by multiplying the rate and time columns together. If we are given a total distance of both persons or trips we will put this information below the distance column. We will now use this table to set up and solve the following example 79 Example 114. Two joggers start from opposite ends of an 8 mile course running towards each other. One jogger is running at a rate of 4 mph, and the other is running at a rate of 6 mph. After how long will the joggers meet? Rate Time Distance Jogger 1 Jogger 2 Rate Time Distance Jogger 1 Jogger 2 4 6 Rate Time Distance Jogger 1 Jogger 2 4 6 t t Rate Time Distance Jogger 1 Jogger 2 4 6 t t 4t 6t 8 4t + 6t = 8 10t = 8 10 10 4 5 t = The basic table for the joggers, one and two We are given the rates for each jogger. These are added to the table We only know they both start and end at the same time. We use the variable t for both times The distance column is filled in by multiplying rate by time We have total distance, 8 miles, under distance The distance column gives equation by adding Combine like terms, 4t + 6t Divide both sides by 10 Our solution for t, 4 5 hour (48 minutes) As the example illustrates, once the table is filled in, the equation to solve is very easy to find. This same process can be seen in the following example Example 115. Bob and Fred start from the same point and walk in opposite directions. Bob walks 2 miles per hour faster than Fred. After 3
hours they are 30 miles apart. How fast did each walk? Rate Time Distance Bob Fred 3 3 The basic table with given times filled in Both traveled 3 hours 80 Rate Time Distance Bob r + 2 Fred r 3 3 Bob walks 2 mph faster than Fred We know nothing about Fred, so use r for his rate Bob is r + 2, showing 2 mph faster Bob r + 2 Fred r Rate Time Distance 3r + 6 3r 30 3 3 3r + 6 + 3r = 30 6r + 6 = 30 6 6 βˆ’ 6r = 24 6 6 r = 4 βˆ’ Rate Bob 4 + 2 = 6 Fred 4 Distance column is filled in by multiplying rate by Time. Be sure to distribute the 3(r + 2) for Bob. Total distance is put under distance The distance columns is our equation, by adding Combine like terms 3r + 3r Subtract 6 from both sides The variable is multiplied by 6 Divide both sides by 6 Our solution for r To answer the question completely we plug 4 in for r in the table. Bob traveled 6 miles per hour and Fred traveled 4 mph Some problems will require us to do a bit of work before we can just fill in the cells. One example of this is if we are given a total time, rather than the individual times like we had in the previous example. If we are given total time we will write this above the time column, use t for the first person’s time, and make t, for the second person’s time. This is shown in a subtraction problem, Total the next example βˆ’ Example 116. Two campers left their campsite by canoe and paddled downstream at an average speed of 12 mph. They turned around and paddled back upstream at an average rate of 4 mph. The total trip took 1 hour. After how much time did the campers turn around downstream? Rate Time Distance 12 4 Down Up 1 Rate Time Distance Down Up 12 4 t 1 t βˆ’ Basic table for down and upstream Given rates are filled in Total time is put above time column As we have the total time, in the first time we have t, the second time becomes the subtraction, total t βˆ’ 81 Rate Time Distance Down Up 12 4 t 1 t βˆ’ Distance column is found by multiplying rate by time. Be sure to distribute 4(1 upstream. As they cover the same distance, = is put after the down distance t
) for βˆ’ 4t With equal sign, distance colum is equation 12t = 4 4t βˆ’ 12t = 4 βˆ’ + 4t + 4t 16t = 4 16 16 1 4 t = Add 4t to both sides so variable is only on one side Variable is multiplied by 16 Divide both sides by 16 Our solution, turn around after 1 4 hr (15 min ) Another type of a distance problem where we do some work is when one person catches up with another. Here a slower person has a head start and the faster person is trying to catch up with him or her and we want to know how long it will take the fast person to do this. Our startegy for this problem will be to use t for the faster person’s time, and add amount of time the head start was to get the slower person’s time. This is shown in the next example. Example 117. Mike leaves his house traveling 2 miles per hour. Joy leaves 6 hours later to catch up with him traveling 8 miles per hour. How long will it take her to catch up with him? Rate Time Distance Mike Joy 2 8 Rate Time Distance Mike Joy 2 8 t + 6 t Rate Time Distance t + 6 2t + 12 = Mike Joy 2 8 t 8t βˆ’ 2t + 12 = 8t 2t 2t βˆ’ 12 = 6t 6 6 Basic table for Mike and Joy The given rates are filled in Joy, the faster person, we use t for time Mikeβ€²s time is t + 6 showing his 6 hour head start Distance column is found by multiplying the rate by time. Be sure to distribute the 2(t + 6) for Mike As they cover the same distance, = is put after Mikeβ€²s distance Now the distance column is the equation Subtract 2t from both sides The variable is multiplied by 6 Divide both sides by 6 82 2 = t Our solution for t, she catches him after 2 hours World View Note: The 10,000 race is the longest standard track event. 10,000 meters is approximately 6.2 miles. The current (at the time of printing) world record for this race is held by Ethiopian Kenenisa Bekele with a time of 26 minutes, 17.53 second. That is a rate of 12.7 miles per hour! As these example have shown, using the table can help keep all the given information organized, help fill in the cells, and help find the equation we
will solve. The final example clearly illustrates this. Example 118. On a 130 mile trip a car travled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took 2.5 hours. For how long did the car travel 40 mph? Rate Time Distance 55 Fast Slow 40 Basic table for fast and slow speeds The given rates are filled in 2.5 Rate Time Distance Fast 55 Slow 40 t 2.5 βˆ’ t 2.5 Rate Time Distance Fast 55 Slow 40 t 2.5 βˆ’ 55t t 100 40t βˆ’ 130 55t + 100 βˆ’ 100 40t = 130 15t + 100 = 130 100 15t = 30 15 15 t = 2 βˆ’ βˆ’ Total time is put above the time column As we have total time, the first time we have t The second time is the subtraction problem 2.5 t βˆ’ t) for slow Distance column is found by multiplying rate by time. Be sure to distribute 40(2.5 βˆ’ Total distance is put under distance The distance column gives our equation by adding Combine like terms 55t 40t Subtract 100 from both sides The variable is multiplied by 30 Divide both sides by 15 Our solution for t. βˆ’ Fast Slow 2.5 Time 2 2 = 0.5 βˆ’ To answer the question we plug 2 in for t The car traveled 40 mph for 0.5 hours (30 minutes) 83 1.10 Practice - Distance, Rate, and Time Problems 1. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour at the same time that an automobile at B starts for A at the rate of 25 miles an hour. How long will it be before the automobiles meet? 2. Two automobiles are 276 miles apart and start at the same time to travel toward each other. They travel at rates differing by 5 miles per hour. If they meet after 6 hours, find the rate of each. 3. Two trains travel toward each other from points which are 195 miles apart. They travel at rate of 25 and 40 miles an hour respectively. If they start at the same time, how soon will they meet? 4. A and B start toward each other at the same time from points 150 miles apart. If A went at the rate of 20 miles an hour, at what rate must B travel if they meet in 5 hours? 5. A passenger and
a freight train start toward each other at the same time from two points 300 miles apart. If the rate of the passenger train exceeds the rate of the freight train by 15 miles per hour, and they meet after 4 hours, what must the rate of each be? 6. Two automobiles started at the same time from a point, but traveled in opposite directions. Their rates were 25 and 35 miles per hour respectively. After how many hours were they 180 miles apart? 7. A man having ten hours at his disposal made an excursion, riding out at the rate of 10 miles an hour and returning on foot, at the rate of 3 miles an hour. 84 Find the distance he rode. 8. A man walks at the rate of 4 miles per hour. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 miles per hour, if he must be back home 3 hours from the time he started? 9. A boy rides away from home in an automobile at the rate of 28 miles an hour and walks back at the rate of 4 miles an hour. The round trip requires 2 hours. How far does he ride? 10. A motorboat leaves a harbor and travels at an average speed of 15 mph toward an island. The average speed on the return trip was 10 mph. How far was the island from the harbor if the total trip took 5 hours? 11. A family drove to a resort at an average speed of 30 mph and later returned over the same road at an average speed of 50 mph. Find the distance to the resort if the total driving time was 8 hours. 12. As part of his flight trainging, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 mph, and the average speed returning was 120 mph. Find the distance between the two airports if the total flying time was 7 hours. 13. A, who travels 4 miles an hour starts from a certain place 2 hours in advance of B, who travels 5 miles an hour in the same direction. How many hours must B travel to overtake A? 14. A man travels 5 miles an hour. After traveling for 6 hours another man starts at the same place, following at the rate of 8 miles an hour. When will the second man overtake the first? 15. A motorboat leaves a harbor and travels at an average speed of 8 mph toward a small island. Two hours
later a cabin cruiser leaves the same harbor and travels at an average speed of 16 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cuiser be alongside the motorboat? 16. A long distance runner started on a course running at an average speed of 6 mph. One hour later, a second runner began the same course at an average speed of 8 mph. How long after the second runner started will the second runner overtake the first runner? 17. A car traveling at 48 mph overtakes a cyclist who, riding at 12 mph, has had a 3 hour head start. How far from the starting point does the car overtake the cyclist? 18. A jet plane traveling at 600 mph overtakes a propeller-driven plane which has 85 had a 2 hour head start. The propeller-driven plane is traveling at 200 mph. How far from the starting point does the jet overtake the propeller-driven plane? 19. Two men are traveling in opposite directions at the rate of 20 and 30 miles an hour at the same time and from the same place. In how many hours will they be 300 miles apart? 20. Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average rate of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track. 21. A motorboat leaves a harbor and travels at an average speed of 18 mph to an island. The average speed on the return trip was 12 mph. How far was the island from the harbor if the total trip took 5 h? 22. A motorboat leaves a harbor and travels at an average speed of 9 mph toward a small island. Two hours later a cabin cruiser leaves the same harbor and travels at an average speed of 18 mph toward the same island. In how many hours after the cabin cruiser leaves will the cabin cruiser be alongside the motorboat? 23. A jet plane traveling at 570 mph overtakes a propeller-driven plane that has had a 2 h head start. The propeller-driven plane is traveling at 190 mph. How far from the starting point does the jet overtake the propeller-driven plane? 24. Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 miles per hour more than the rate of the other and they are 168
miles apart at the end of 4 hours, what is the rate of each? 25. As part of flight traning, a student pilot was required to fly to an airport and then return. The average speed on the way to the airport was 100 mph, and the average speed returning was 150 mph. Find the distance between the two airports if the total flight time was 5 h. 26. Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours they are 72 miles apart. Find the rate of each cyclist. 27. A car traveling at 56 mph overtakes a cyclist who, riding at 14 mph, has had a 3 h head start. How far from the starting point does the car overtake the cyclist? 28. Two small planes start from the same point and fly in opposite directions. 86 The first plan is flying 25 mph slower than the second plane. In two hours the planes are 430 miles apart. Find the rate of each plane. 29. A bus traveling at a rate of 60 mph overtakes a car traveling at a rate of 45 mph. If the car had a 1 h head start, how far from the starting point does the bus overtake the car? 30. Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 mph slower than the second plane. In 2 h, the planes are 470 mi apart. Find the rate of each plane. 31. A truck leaves a depot at 11 A.M. and travels at a speed of 45 mph. At noon, a van leaves the same place and travels the same route at a speed of 65 mph. At what time does the van overtake the truck? 32. A family drove to a resort at an average speed of 25 mph and later returned over the same road at an average speed of 40 mph. Find the distance to the resort if the total driving time was 13 h. 33. Three campers left their campsite by canoe and paddled downstream at an average rate of 10 mph. They then turned around and paddled back upstream at an average rate of 5 mph to return to their campsite. How long did it take the campers to canoe downstream if the total trip took 1 hr? 34. A motorcycle breaks down and the rider has to walk the rest of the way to work. The
motorcycle was being driven at 45 mph, and the rider walks at a speed of 6 mph. The distance from home to work is 25 miles, and the total time for the trip was 2 hours. How far did the motorcycle go before if broke down? 35. A student walks and jogs to college each day. The student averages 5 km/hr walking and 9 km/hr jogging. The distance from home to college is 8 km, and the student makes the trip in one hour. How far does the student jog? 36. On a 130 mi trip, a car traveled at an average speed of 55 mph and then reduced its speed to 40 mph for the remainder of the trip. The trip took a total of 2.5 h. For how long did the car travel at 40 mph? 37. On a 220 mi trip, a car traveled at an average speed of 50 mph and then reduced its average speed to 35 mph for the remainder of the trip. The trip took a total of 5 h. How long did the car travel at each speed? 38. An executive drove from home at an average speed of 40 mph to an airport where a helicopter was waiting. The executive boarded the helicopter and flew to the corporate offices at and average speed of 60 mph. The entire distance was 150 mi. The entire trip took 3 h. Find the distance from the airport to the corporate offices. 87 Chapter 2 : Graphing 2.1 Points and Lines......................................................................................89 2.2 Slope........................................................................................................95 2.3 Slope-Intercept Form.............................................................................102 2.4 Point-Slope Form...................................................................................107 2.5 Parallel and Perpendicular Lines...........................................................112 88 2.1 Graphing - Points and Lines Objective: Graph points and lines using xy coordinates. Often, to get an idea of the behavior of an equation we will make a picture that represents the solutions to the equations. A graph is simply a picture of the solutions to an equation. Before we spend much time on making a visual representation of an equation, we first have to understand the basis of graphing. Following is an example of what is called the coordinate plane. -4 -3 -2 -1 2 1 -1 -2 1 2 3 The plane is divided into four sections by a horizontal number line (x-axis) and a vertical number line (y-axis). Where the two lines meet
in the center is called the origin. This center origin is where x = 0 and y = 0. As we move to the right the numbers count up from zero, representing x = 1, 2, 3. To the left the numbers count down from zero, representing x = 3. Similarly, as we move up the number count up from zero, y = 1, 2, 3., and as we move down count down from zero, y = 3. We can put dots on the βˆ’ graph which we will call points. Each point has an β€œaddress” that defines its location. The first number will be the value on the x axis or horizontal number line. This is the distance the point moves left/right from the origin. The second number will represent the value on the y axis or vertical number line. This is the distance the point moves up/down from the origin. The points are given as an ordered pair (x, y). 2, 2, 1, 1, βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ World View Note: Locations on the globe are given in the same manner, each number is a distance from a central point, the origin which is where the prime meridian and the equator. This β€œorigin is just off the western coast of Africa. The following example finds the address or coordinate pair for each of several points on the coordinate plane. Example 119. Give the coordinates of each point. B A C 89 Tracing from the origin, point A is right 1, up 4. This becomes A(1, 4). Point B is left 5, up 3. Left is back5, wards or negative so we have B( βˆ’ 3). C is straight down 2 units. There is no left or right. This means we go right zero so the point is C(0, 2). βˆ’ A(1, 4), B( 5, 3), C(0, βˆ’ βˆ’ 2) Our Solution Just as we can give the coordinates for a set of points, we can take a set of points and plot them on the plane. Example 120. Graph the points A(3, 2), B ( G(0, 0) βˆ’ 2, 1), C (3, 4), D ( 2, βˆ’ βˆ’ βˆ’ 3), E ( βˆ’ 3, 0), F (0, 2), A B Up 1 Left 2 Right 3 Up 2 The first point, A is
at (3, 2) this means x = 3 (right 3) and y = 2 (up 2). Following these instructions, starting from the origin, we get our point. The second point, B ( 2, 1), is left 2 (negative moves backwards), up 1. This is also illustrated on the graph. βˆ’ A Right 3 B Left 2 Down 3 D Down 4 C The third point, C (3, down 4 (negative moves backwards). 4) is right 3, βˆ’ The fourth point, D ( 3) is left 2, down 3 (both negative, both move backwards) βˆ’ βˆ’ 2, The last three points have zeros in them. We still treat these points just like the other points. If there is a zero there is just no movement. B F A Up 2 E Left 3 G D C βˆ’ βˆ’ axis. Next is E ( 3, 0). This is left 3 (negative is backwards), and up zero, right on the x Then is F (0, 2). This is right zero, and up two, right on the y Finally is G (0, 0). This point has no movement. Thus the point is right on the origin. axis. βˆ’ 90 E B D F G A C Our Solution The main purpose of graphs is not to plot random points, but rather to give a picture of the solutions to an equation. We may have an equation such as y = 2x 3. We may be interested in what type of solution are possible in this equation. We can visualize the solution by making a graph of possible x and y combinations that make this equation a true statement. We will have to start by finding possible x and y combinations. We will do this using a table of values. βˆ’ Example 121. Graph y = 2x βˆ’ 3 We make a table of values βˆ’ βˆ’ βˆ’ We will test three values for x. Any three can be used Evaluate each by replacing x with the given value x = 1; y = 2( 3 = x = 0; y = 2(0) x = 1; y = 2(1) 1, ( βˆ’ βˆ’ 5), (0, βˆ’ 3), (1, βˆ’ 1) These then become the points to graph on our equation 91 Plot each point. Once the point are on the graph, con- nect the dots to make a line. The graph is our solution βˆ’ What this line tells us is that any point on the line will work in the equation y = 3
. For example, notice the graph also goes through the point (2, 1). If we use 2x x = 2, we should get y = 1. Sure enough, y = 2(2) 3 = 1, just as the graph suggests. Thus we have the line is a picture of all the solutions for y = 2x 3. We can use this table of values method to draw a graph of any linear equation. 3 = 4 βˆ’ βˆ’ βˆ’ Example 122. Graph 2x βˆ’ 3y = 6 We will use a table of values y x 3 βˆ’ 0 3 We will test three values for x. Any three can be used. 2( 3) βˆ’ 6 βˆ’ + 6 βˆ’ βˆ’ 3y = 6 3y = 6 3, multiply first Substitute each value in for x and solve for y Start with x = + 6 Add 6 to both sides 3y = 12 Divide both sides by 3 y = Solution for y when x = 3, add this to table (0) 2(3) 6 6 βˆ’ 3y = 6 Next x = 0 βˆ’ 3y = 6 Multiplying clears the constant term βˆ’ 3 βˆ’ y = 3 βˆ’ Solution for y when x = 0, add this to table 3 Divide each side by 2 βˆ’ βˆ’ 3y = 6 Next x = 3 3y = 6 Multiply βˆ’ βˆ’ 6 βˆ’ 3y = 0 3 3 βˆ’ y = 0 βˆ’ βˆ’ Subtract 9 from both sides Divide each side by 3 βˆ’ Solution for y when x = 3, add this to table βˆ’ 92 Our completed table. 3, ( βˆ’ βˆ’ 4), (0, 2), (3, 0) Table becomes points to graph Graph points and connect dots Our Solution 93 2.1 Practice - Points and Lines State the coordinates of each point. K E 1) D J I H G C F B Plot each point. 2) L( βˆ’ 5, 5) K(1, 0) J( βˆ’ 3, 4) H( βˆ’ 2) E (3, 4, 2) G(4, -2) 2) D(0, 3) βˆ’ 3, 0) I( βˆ’ F ( 2, βˆ’ βˆ’ C (0, 4) Sketch the graph of each line. 1 4 x 3 βˆ’ 4) y = x 3) y = 5) y = βˆ’ βˆ’ 7) y = βˆ’ 9 βˆ’ 4x + 2 5 βˆ’ 11) y = 4 5 x 3 βˆ’ βˆ’ 13) x + 5y =
βˆ’ 15) 4x + y = 5 15 17) 2x y = 2 βˆ’ 19) x + y = 21) y = βˆ’ 8) y = 5 3 x 10) y = βˆ’ 12) y = 1 x 2 2 βˆ’ 14) 8x y = 5 βˆ’ 16) 3x + 4y = 16 18) 7x + 3y = βˆ’ 20) 3x + 4y = 8 12 22) 9x y = 4 βˆ’ βˆ’ 94 2.2 Graphing - Slope Objective: Find the slope of a line given a graph or two points. As we graph lines, we will want to be able to identify different properties of the lines we graph. One of the most important properties of a line is its slope. Slope is a measure of steepness. A line with a large slope, such as 25, is very steep. A line with a small slope, such as 1 10 is very flat. We will also use slope to describe the direction of the line. A line that goes up from left to right will have a positive slope and a line that goes down from left to right will have a negative slope. As we measure steepness we are interested in how fast the line rises compared to how far the line runs. For this reason we will describe slope as the fraction rise run. Rise would be a vertical change, or a change in the y-values. Run would be a horizontal change, or a change in the x-values. So another way to describe slope would be the fraction change in y change in x. It turns out that if we have a graph we can draw vertical and horiztonal lines from one point to another to make what is called a slope triangle. The sides of the slope triangle give us our slope. The following examples show graphs that we find the slope of using this idea. Example 123. Rise 4 βˆ’ Run 6 To find the slope of this line we will consider the rise, or verticle change and the run or horizontal change. Drawing these lines in makes a slope triangle that we can use to count from one point to the next the graph goes 4, run down 4, right 6. This is rise 4 6. As a fraction it would be, βˆ’ 6. Reduce the fraction to get βˆ’. 2 3 βˆ’ 2 3 βˆ’ Our Solution World View Note: When French mathematicians Rene Descartes and Pierre de Fermat first developed
the coordinate plane and the idea of graphing lines (and other functions) the y-axis was not a verticle line! Example 124. 95 Run 3 Rise 6 To find the slope of this line, the rise is up 6, the run is right 3. Our slope is run or 6 3. then written as a fraction, This fraction reduces to 2. This will be our slope. rise 2 Our Solution There are two special lines that have unique slopes that we need to be aware of. They are illustrated in the following example. Example 125. In this graph there is no rise, but the run is 3 units. This slope becomes 0 line, 3 and all horizontal lines have a zero slope. This 0. = slope This line has a rise of 5, but no run. The = undefined. line, and all vertical lines, have no slope. becomes This 5 0 As you can see there is a big difference between having a zero slope and having no slope or undefined slope. Remember, slope is a measure of steepness. The first slope is not steep at all, in fact it is flat. Therefore it has a zero slope. The second slope can’t get any steeper. It is so steep that there is no number large enough to express how steep it is. This is an undefined slope. We can find the slope of a line through two points without seeing the points on a graph. We can do this using a slope formula. If the rise is the change in y values, we can calculate this by subtracting the y values of a point. Similarly, if run is a change in the x values, we can calculate this by subtracting the x values of a point. In this way we get the following equation for slope. The slope of a line through (x1, y1) and (x2, y2) is y2 x2 βˆ’ βˆ’ y1 x1 96 When mathematicians began working with slope, it was called the modular slope. For this reason we often represent the slope with the variable m. Now we have the following for slope. Slope = m = rise run = change in y change in x = y2 x2 y1 x1 βˆ’ βˆ’ As we subtract the y values and the x values when calculating slope it is important we subtract them in the same order. This process is shown in the
following examples. Example 126. Find the slope between ( 4, 3) and (2, 9) Identify x1, y1, x2, y2 βˆ’ (x1, y1) and (x2, y2) Use slope formula) βˆ’ βˆ’ 12 m = βˆ’ 6 m = βˆ’ Simplify Reduce 2 Our Solution Example 127. Find the slope between (4, 6) and (2, 1) Identify x1, y1, x2, y2 βˆ’ (x1, y1) and (x2, y2) Use slope formula, m = y2 βˆ’ x2 βˆ’ y1 x1 y2 βˆ’ x2 βˆ’ y1 x1 Simplify Reduce, dividing by 1 βˆ’ Our Solution We may come up against a problem that has a zero slope (horiztonal line) or no slope (vertical line) just as with using the graphs. Example 128. Find the slope between ( 4, βˆ’ βˆ’ 1) and ( 4, βˆ’ βˆ’ 5) Identify x1, y1, x2, y2 97 (x1, y1) and (x2, y2) Use slope formula) βˆ’ βˆ’ 4) βˆ’ βˆ’ 4 m = βˆ’ 0 Simplify Canβ€²t divide by zero, undefined m = no slope Our Solution y2 βˆ’ x2 βˆ’ y1 x1 Example 129. Find the slope between (3, 1) and ( 2, 1) Identify x1, y1, x2, y2 βˆ’ (x1, y1) and (x2, y2) Use slope formula βˆ’ Simplify Reduce m = 0 Our Solution y2 βˆ’ x2 βˆ’ y1 x1 Again, there is a big difference between no slope and a zero slope. Zero is an integer and it has a value, the slope of a flat horizontal line. No slope has no value, it is undefined, the slope of a vertical line. Using the slope formula we can also find missing points if we know what the slope is. This is shown in the following two examples. Example 130. Find the value of y between the points (2, y) and (5, 1) with slope βˆ’ 3 βˆ’ We will plug values into slope formula Simplify y1 x1 y m = y2 βˆ’ x2 βˆ’ (3 Multiply both sides by 3 (3) Simplify y Add 1 to both
sides βˆ’ y Divide both sides by 1 1 βˆ’ 8 = y Our Solution 98 Example 131. Find the value of x between the points ( 3, 2) and (x, 6) with slope 2 5 βˆ’ m = 2 5 = y2 βˆ’ x2 βˆ’ 2 6 βˆ’ ( y1 x1 We will plug values into slope formula Simplify βˆ’ = 3x + 3) = 4 Multiply by 5 to clear fraction Multiply both sides by (x + 3) 2 5 (5) 2 5 (x + 3) = 4(5) Simplify 2(x + 3) = 20 Distribute 2x + 6 = 20 6 6 Solve. Subtract 6 from both sides βˆ’ βˆ’ 2x = 14 Divide each side by 2 2 2 x = 7 Our Solution 99 2.2 Practice - Slope Find the slope of each line. 1) 3) 5) 7) 2) 4) 6) 8) 100 9) 10) Find the slope of the line through each pair of points. 11) ( βˆ’ 2, 10), ( 2, βˆ’ 13) ( 15, 10), (16, βˆ’ 15) (10, 18), ( βˆ’ βˆ’ 15) 7) 10) 11, βˆ’ βˆ’ 14), (11, 16, βˆ’ 4, 14), ( βˆ’ 16, 8) βˆ’ 19), (6, 14) 17) ( βˆ’ 19) ( βˆ’ 21) (12, 23) ( βˆ’ 25) ( βˆ’ 27) (7, 29) ( βˆ’ 10), ( βˆ’ βˆ’ 17, 19), (10, 5, 20) 7) βˆ’ 14), ( βˆ’ 5, 7), ( 8, 9) βˆ’ βˆ’ 18, 14) βˆ’ βˆ’ 5, 14) 18) (13, 15), (2, 10) 12) (1, 2), ( 6, 14) βˆ’ βˆ’ 2), (7, 7) 14) (13, 16) ( βˆ’ βˆ’ 3, 6), ( 20, 13) βˆ’ 20) (9, 6), ( 7, 7) βˆ’ βˆ’ βˆ’ 16, 2), (15, 22) ( βˆ’ 24) (8, 11), ( 10) βˆ’ 13) 3, βˆ’ βˆ’ 2), (1, 17) 26) (11, βˆ’ 18, 28) ( βˆ’ 5), (14, 3) βˆ’ βˆ’ 30) (19, 15), (5, 11) Find the value of x or y so that the line through the points has the given slope. 31
) (2, 6) and (x, 2); slope: 4 7 2) and (x, 6); slope: 3, βˆ’ βˆ’ 1, 1); slope: 6 7 8 5 34) ( βˆ’ 36) (x, 33) ( 35) ( βˆ’ βˆ’ 37) (x, 8, y) and ( 7) and ( βˆ’ βˆ’ βˆ’ 9, βˆ’ 9); slope: 2 5 38) (2, βˆ’ 32) (8, y) and ( 2, 4); slope: βˆ’ βˆ’ 2, y) and (2, 4); slope: 1 4 1) and ( 4, 6); slope: βˆ’ βˆ’ 5) and (3, y); slope: 6 1 5 7 10 βˆ’ 39) (x, 5) and (8, 0); slope: 5 6 βˆ’ 40) (6, 2) and (x, 6); slope: 4 5 βˆ’ 101 2.3 Graphing - Slope-Intercept Form Objective: Give the equation of a line with a known slope and y-intercept. When graphing a line we found one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is finding the slope and the y-intercept of the equation. The slope can be represented by m and the yintercept, where it crosses the axis and x = 0, can be represented by (0, b) where b is the value where the graph crosses the vertical y-axis. Any other point on the line can be represented by (x, y). Using this information we will look at the slope formula and solve the formula for y. Example 132. m, (0, b), (x, y) Using the slope formula gives Simplify βˆ’ βˆ’ βˆ’ x b = mx Add b to both sides = m Multiply both sides by x βˆ’ + b + b y = mx + b Our Solution This equation, y = mx + b can be thought of as the equation of any line that as a slope of m and a y-intercept of b. This formula is known as the slope-intercept equation. Slope βˆ’ Intercept Equation: y = m x + b If we know the slope and the y-intercept we can easily find the equation that represents the line. Example 133. Slope = 3 4
, y βˆ’ intercept = 3 Use the slope intercept equation βˆ’ βˆ’ y = mx + b m is the slope, b is the y intercept βˆ’ y = 3 4 x βˆ’ 3 Our Solution We can also find the equation by looking at a graph and finding the slope and yintercept. Example 134. 102 Identify the point where the graph crosses the y-axis (0,3). This means the y-intercept is 3. Idenfity one other point and draw a slope triangle to find the slope. The slope is 2 3 βˆ’ y = mx + b Slope-intercept equation y = 2 3 βˆ’ x + 3 Our Solution We can also move the opposite direction, using the equation identify the slope and y-intercept and graph the equation from this information. However, it will be important for the equation to first be in slope intercept form. If it is not, we will have to solve it for y so we can identify the slope and the y-intercept. Example 135. Write in slope βˆ’ Solve for y Subtract 2x from both sides intercept form: 2x 2x βˆ’ 4y = 4 βˆ’ βˆ’ 4y = 6 2x βˆ’ 2x + 6 Put x term first 4 4 Divide each term by 3 2 Our Solution βˆ’ = Once we have an equation in slope-intercept form we can graph it by first plotting the y-intercept, then using the slope, find a second point and connecting the dots. Example 136. x 1 2 y = mx + b βˆ’ Graph y = 4 Recall the slope intercept formula βˆ’ Idenfity the slope, m, and the y intercept Make the graph Starting with a point at the y-intercept of 4, βˆ’ Then use the slope rise run, so we will rise 1 unit and run 2 units to find the next point. Once we have both points, connect the dots to get our graph. World View Note: Before our current system of graphing, French Mathematician Nicole Oresme, in 1323 sugggested graphing lines that would look more like a 103 bar graph with a constant slope! Example 137. Graph 3x + 4y = 12 Not in slope intercept form Subtract 3x from both sides 3x 3x βˆ’ 4y = 4 y = m = βˆ’ 4 βˆ’ Divide
each term by 4 3x + 12 Put the x term first 4 3 4 y = mx + b 3 4, b = 3 Make the graph x + 3 Recall slope Idenfity m and b βˆ’ βˆ’ βˆ’ intercept equation Starting with a point at the y-intercept of 3, Then use the slope rise run, but its negative so it will go downhill, so we will drop 3 units and run 4 units to find the next point. Once we have both points, connect the dots to get our graph. We want to be very careful not to confuse using slope to find the next point with use a coordinate such as (4, 2) to find an individule point. Coordinates such as (4, 2) start from the origin and move horizontally first, and vertically second. Slope starts from a point on the line that could be anywhere on the graph. The numerator is the vertical change and the denominator is the horizontal change. βˆ’ βˆ’ Lines with zero slope or no slope can make a problem seem very different. Zero slope, or horiztonal line, will simply have a slope of zero which when multiplied by x gives zero. So the equation simply becomes y = b or y is equal to the y-coordinate of the graph. If we have no slope, or a vertical line, the equation can’t be written in slope intercept at all because the slope is undefined. There is no y in these equations. We will simply make x equal to the x-coordinate of the graph. Example 138. Give the equation of the line in the graph. Because we have a vertical line and no slope there is no slope-intercept equation we can use. Rather we make x equal to the x-coordinate of 4 βˆ’ x = 4 βˆ’ Our Solution 104 2.3 Practice - Slope-Intercept Write the slope-intercept form of the equation of each line given the slope and the y-intercept. 1) Slope = 2, y-intercept = 5 3) Slope = 1, y-intercept = βˆ’ 3 4, y-intercept = 4 1 βˆ’ 5) Slope = βˆ’ 7) Slope = 1 3, y-intercept = 1 2) Slope = 4) Slope = βˆ’ βˆ’ 6) Slope = βˆ’ 8) Slope = 2
6, y-intercept = 4 1, y-intercept = βˆ’ 1 4, y-intercept = 3 2 5, y-intercept = 5 Write the slope-intercept form of the equation of each line. 9) 11) 13) 10) 12) 14) 105 15) x + 10y = 37 βˆ’ 17) 2x + y = 1 βˆ’ 3y = 24 19) 7x βˆ’ 21) x = 8 βˆ’ 23) y βˆ’ 25) y βˆ’ 27) y + 5 = 29) y + 1 = (x + 5) 4 = βˆ’ 4 = 4(x βˆ’ 4(x 1 2(x 1) 2) 4) βˆ’ βˆ’ βˆ’ βˆ’ Sketch the graph of each line. 31) y = 1 3 x + 4 33) y = 6 5 x 5 βˆ’ 35) y = 3 2 x 37) x y + 3 = 0 βˆ’ y βˆ’ 3y = βˆ’ βˆ’ 39) 41) 4 + 3x = 0 5x + 9 βˆ’ 10y = 3 16) x βˆ’ 18) 6x 11y = 70 βˆ’ βˆ’ 20) 4x + 7y = 28 22) x 24) y 26) y βˆ’ βˆ’ βˆ’ 7y = 5 = 5 βˆ’ 2 (x 42 2) βˆ’ 3 = βˆ’ 2 3 (x + 3) 4 28) 0 = x βˆ’ 30) y + 2 = 6 5 (x + 5) 32) y = 34 36) y = βˆ’ 38) 4x + 5 = 5y 40) 42) 8 = 6x 3y = 3 2y 3 2 x βˆ’ βˆ’ βˆ’ βˆ’ 106 2.4 Graphing - Point-Slope Form Objective: Give the equation of a line with a known slope and point. The slope-intercept form has the advantage of being simple to remember and use, however, it has one major disadvantage: we must know the y-intercept in order to use it! Generally we do not know the y-intercept, we only know one or more points (that are not the y-intercept). In these cases we can’t use the slope intercept equation, so we will use a different more flexible formula. If we let the slope of an equation be m, and a specific point on the line be (x1, y1), and any other point on the line be (x, y). We can use the slope formula to make a
second equation. Example 139. m, (x1, y1), (x, y) Recall slope formula y2 y1 βˆ’ x2 βˆ’ x1 y y1 βˆ’ x1 x βˆ’ y1 = m(x βˆ’ y βˆ’ = m Plug in values = m Multiply both sides by (x x1) βˆ’ x1) Our Solution If we know the slope, m of an equation and any point on the line (x1, y1) we can easily plug these values into the equation above which will be called the pointslope formula. Point βˆ’ Slope Formula: y y1 = m(x x1) βˆ’ βˆ’ Example 140. Write the equation of the line through the point (3, 4) with a slope of 3 5. βˆ’ y1 = m(x x1) Plug values into point slope formula βˆ’ βˆ’ y y βˆ’ βˆ’ ( 4x (x 3) Simplify signs 3) Our Solution βˆ’ βˆ’ Often, we will prefer final answers be written in slope intercept form. If the direc- 107 tions ask for the answer in slope-intercept form we will simply distribute the slope, then solve for y. Example 141. Write the equation of the line through the point ( slope-intercept form. βˆ’ 6, 2) with a slope of 2 3 in βˆ’ x1) Plug values into point slope formula βˆ’ 6)) Simplify signs βˆ’ βˆ’ y βˆ’ y1 = m(x (x + 6) Distribute slope 2 3 Solve for Our Solution An important thing to observe about the point slope formula is that the operation between the x’s and y’s is subtraction. This means when you simplify the signs you will have the opposite of the numbers in the point. We need to be very careful with signs as we use the point-slope formula. In order to find the equation of a line we will always need to know the slope. If we don’t know the slope to begin with we will have to do some work to find it first before we can get an equation. Example 142. Find the equation of the line through the points ( 2, 5) and (4, 3). βˆ’ βˆ’ m = First we must find the slope y2 βˆ’ x2 βˆ’ 8 = = βˆ’ 6 βˆ’ x1) With slope and either point, use point Plug values in slope formula and evaluate 5 2) y
1 = m(x y1 x1 4 3 slope formula βˆ’ x 4 3 βˆ’ 5 = βˆ’ 2)) ( βˆ’ Simplify signs (x + 2) Our Solution βˆ’ 4 3 Example 143. 108 Find the equation of the line through the points ( intercept form. βˆ’ 3, 4) and ( 1, βˆ’ βˆ’ 2) in slope- m = y1 x1 First we must find the slope y2 βˆ’ x2 βˆ’ 6 = = βˆ’ 2 βˆ’ x1) With slope and either point, point 3)) 3) y1 = m(x 3(x ( Simplify signs 3 Plug values in slope formula and evaluate βˆ’ βˆ’ slope formula βˆ’ (x + 3) Distribute slope βˆ’ y βˆ’ 4 = βˆ’ + 4 βˆ’ 3x 9 Solve for y βˆ’ + 4 Add 4 to both sides y = 3x βˆ’ βˆ’ 5 Our Solution Example 144. Find the equation of the line through the points (6, intercept form. βˆ’ 2) and ( βˆ’ 4, 1) in slope- m = y2 βˆ’ x2 βˆ’ = βˆ’ First we must find the slope y1 x1 3 10 x1) Use slope and either point, use point Plug values into slope formula and evaluate slope formula βˆ’ 6) Simplify signs βˆ’ 6) Distribute slope Solve for y. Subtract 2 from both sides Using 10 5 on right so we have a common denominator Our Solution ) βˆ’ 6 βˆ’ y = 3 10 y1 = m(x βˆ’ ( βˆ’ 2x 3 10 3 (x 10 3 10 βˆ’ βˆ’ x + 2 βˆ’ y = 3 10 βˆ’ βˆ’ x βˆ’ 9 5 10 5 1 5 World View Note: The city of Konigsberg (now Kaliningrad, Russia) had a river that flowed through the city breaking it into several parts. There were 7 bridges that connected the parts of the city. In 1735 Leonhard Euler considered the question of whether it was possible to cross each bridge exactly once and only once. It turned out that this problem was impossible, but the work laid the foundation of what would become graph theory. 109 2.4 Practice - Point-Slope Form Write the point-slope form of the equation of the line through the given point with the given slope. 1) through (2, 3), slope = undefined 2) through (1, 2), slope = undefined 3) through (2, 2), slope = 1 2 4)
through (2, 1), slope = 5) through ( 1, βˆ’ βˆ’ 5), slope = 9 6) through (2, βˆ’ 4, 1), slope = 3 4 8) through (4, βˆ’ 2), slope = 1 2 2 βˆ’ βˆ’ 2 3), slope = βˆ’ 1, 1), slope = 4 3 10) through ( βˆ’ 7) through ( βˆ’ 9) through (0, βˆ’ 11) through (0, 2), slope = βˆ’ 5), slope = βˆ’ 1 4 βˆ’ 13) through ( 15) through ( 5, βˆ’ 3), slope = 1 5 1, 4), slope = 5 4 βˆ’ βˆ’ βˆ’ 12) through (0, 2), slope = 5 4 βˆ’ 14) through ( 1, βˆ’ βˆ’ 4), slope = 16) through (1, 4), slope = βˆ’ βˆ’ 2 3 βˆ’ 3 2 Write the slope-intercept form of the equation of the line through the given point with the given slope. 3 5 3 2 2 5 βˆ’ 17) through: ( βˆ’ 19) through: (5, 21) through: ( βˆ’ 1, 5), slope = 2 βˆ’ 1), slope = βˆ’ βˆ’ 4, 1), slope = 1 2 23) through: (4, 2), slope = 25) through: ( βˆ’ 27) through: (2, 29) through:( βˆ’ βˆ’ 5, βˆ’ 3), slope = βˆ’ 2), slope = 1 βˆ’ 3, 4), slope=undefined 31) through: ( 4, 2), slope = βˆ’ 1 2 βˆ’ 18) through: (2, 2), slope = 20) through: ( βˆ’ 22) through: (4, βˆ’ 2, βˆ’ βˆ’ 2), slope = βˆ’ 3), slope = βˆ’ 2 2 3 βˆ’ 7 4 24) through: ( 2, 0), slope = βˆ’ 26) through: (3, 3), slope = 7 3 5 2 βˆ’ 28) through: ( 4, βˆ’ βˆ’ 3), slope = 0 30) through: ( 2, 5), slope = 2 βˆ’ 32) through: (5, 3), slope = 6 5 βˆ’ 110 Write the point-slope form of the equation of the line through the given points. 33) through: ( 4, 3) and ( βˆ’ 35) through: (5, 1) and ( βˆ’ 3, 1) βˆ’ 3, 0) 37) through: ( 4, βˆ’ βˆ’ 2) and (0, 4) 39) through: (3, 5) and ( 5, 3)
βˆ’ 41) through: (3, 3) and ( 4, 5) βˆ’ βˆ’ 34) through: (1, 3) and ( 3, 3) βˆ’ 36) through: ( 38) through: ( 40) through: ( 42) through: ( βˆ’ βˆ’ βˆ’ βˆ’ 4, 5) and (4, 4) 4, 1) and (4, 4) 4) and ( 5) and ( 1, 1, βˆ’ βˆ’ βˆ’ βˆ’ 5, 0) 4) 5, βˆ’ Write the slope-intercept form of the equation of the line through the given points. 43) through: ( 45) through: ( βˆ’ βˆ’ 5, 1) and ( βˆ’ 5, 5) and (2, 1, 2) βˆ’ 3) βˆ’ 44) through: ( βˆ’ 46) through: (1, 1) and (5, 5, βˆ’ 1) and ( βˆ’ 2) 4) βˆ’ βˆ’ 47) through: (4, 1) and (1, 4) 48) through: (0, 1) and ( βˆ’ 49) through: (0, 2) and (5, 3) 50) through: (0, 2) and (2, 4) 5, βˆ’ 3, 0) 51) through: (0, 3) and ( βˆ’ 1, βˆ’ 1) βˆ’ 52) through: ( 2, 0) and (5, 3) βˆ’ 111 2.5 Graphing - Parallel and Perpendicular Lines Objective: Identify the equation of a line given a parallel or perpendicular line. There is an interesting connection between the slope of lines that are parallel and the slope of lines that are perpendicular (meet at a right angle). This is shown in the following example. Example 145. The above graph has two parallel lines. The slope of the top line is 2 down 2, run 3, or 3. The slope of the bottom line is down 2, run 3 as well, or βˆ’ 2 3. βˆ’ The above graph has two perpendicular lines. The slope of the flatter line or is 2. The slope of the steeper line is down 3, 3 run 2 or run up 2, 3 3 2. βˆ’ World View Note: Greek Mathematician Euclid lived around 300 BC and published a book titled, The Elements. In it is the famous parallel postulate which mathematicians have tried for years to drop from the list of postulates. The attempts have failed, yet all the work done has developed new types of
geometries! As the above graphs illustrate, parallel lines have the same slope and perpendicular lines have opposite (one positive, one negative) reciprocal (flipped fraction) slopes. We can use these properties to make conclusions about parallel and perpendicular lines. Example 146. Find the slope of a line parallel to 5y 2x = 7. βˆ’ 5y 2x = 7 To find the slope we will put equation in slope intercept form βˆ’ βˆ’ + 2x + 2x Add 2x to both sides 5y = 2x + 7 Put x term first 5 Divide each term by The slope is the coefficient of x 112 m = m = 2 5 2 5 Slope of first line. Parallel lines have the same slope Our Solution Example 147. Find the slope of a line perpendicular to 3x 4y = 2 βˆ’ 4y = 2 To find slope we will put equation in slope βˆ’ βˆ’ 3x 3x βˆ’ 4y = 4 βˆ’ βˆ’ 3x Subtract 3x from both sides 3x + 2 Put x term first 4 4 Divide each term by βˆ’ The slope is the coefficient of x intercept form βˆ’ m = m = βˆ’ 3 4 4 3 Slope of first lines. Perpendicular lines have opposite reciprocal slopes Our Solution Once we have a slope, it is possible to find the complete equation of the second line if we know one point on the second line. Example 148. Find the equation of a line through (4, 5) and parallel to 2x 3y = 6. βˆ’ βˆ’ 2x 2x βˆ’ 3y = 3y = 6 We first need slope of parallel line 2x Subtract 2x from each side 2x + 6 Put x term first 3 Divide each term by 3 βˆ’ Identify the slope, the coefficient of x Parallel lines have the same slope We will use this slope and our point (4, 5) βˆ’ y y βˆ’ βˆ’ ( βˆ’ y1 = m(x x1) Plug this information into point slope formula 5) = (x 4) Simplify signs x βˆ’ 4) Our Solution 113 Example 149. Find the equation of the line through (6, slope-intercept form. βˆ’ 9) perpendicular to y = 3 5 βˆ’ x + 4 in y = 3 5 βˆ’ x + 4 Identify
the slope, coefficient of Perpendicular lines have opposite reciprocal slopes We will use this slope and our point (6, 9) βˆ’ y1 = m(x y βˆ’ βˆ’ x1) Plug this information into point slope formula βˆ’ y ( βˆ’ βˆ’ 9x βˆ’ (x βˆ’ 6) Simplify signs 6) Distribute slope βˆ’ βˆ’ 10 Solve for y 9 Subtract 9 from both sides βˆ’ 19 Our Solution Zero slopes and no slopes may seem like opposites (one is a horizontal line, one is a vertical line). Because a horizontal line is perpendicular to a vertical line we can say that no slope and zero slope are actually perpendicular slopes! Example 150. Find the equation of the line through (3, 4) perpendicular to x = 2 βˆ’ x = 2 This equation has no slope, a vertical line no slope Perpendicular line then would have a zero slope βˆ’ m = 0 Use this and our point (3, 4) y βˆ’ y y1 = m(x 4 = 0(x y βˆ’ x1) Plug this information into point 3) Distribute slope Solve for Add 4 to each side y = 4 Our Solution slope formula βˆ’ Being aware that to be perpendicular to a vertical line means we have a horizontal line through a y value of 4, thus we could have jumped from this point right to the solution, y = 4. 114 2.5 Practice - Parallel and Perpendicular Lines Find the slope of a line parallel to each given line. 1) y = 2x + 4 2) y = 3) y = 4x βˆ’ y = 4 5) x βˆ’ 7) 7x + y = 2 βˆ’ 5 10 3 x βˆ’ βˆ’ 5y = 20 6) 6x βˆ’ 8) 3x + 4y = 8 βˆ’ Find the slope of a line perpendicular to each given line. 9) x = 3 11) y = 1 3 x βˆ’ 3y = 13) x βˆ’ βˆ’ 15) x + 2y = 8 1 2 x 1 βˆ’ 10) y = βˆ’ 12) y = 4 5 x 6 14) 3x 16) 8x βˆ’ βˆ’ 3 y = βˆ’ 3y = 9 βˆ’ Write the point-slope form of the equation of the line described. 17) through: (2, 5), parallel to x = 0 18) through: (5, 2), parallel to y = 7 5 x + 4 19) through: (3, 4), parallel to y = 9 2 x 5 βˆ’ 20)
through: (1, 1), parallel to y = βˆ’ 21) through: (2, 3), parallel to βˆ’ 22) through: ( βˆ’ 1, 3), parallel to y = 3x 1 βˆ’ βˆ’ 23) through: (4, 2), parallel to x = 0 24) through: (1, 4), parallel to y = 7 5 x + 2 25) through: (1, 26) through: (1, βˆ’ 5), perpendicular to βˆ’ 2), perpendicular to x + y = 1 βˆ’ x + 2y = 2 βˆ’ 115 27) through: (5, 2), perpendicular to 5x + y = 3 βˆ’ 28) through: (1, 3), perpendicular to x + y = 1 βˆ’ 4x + y = 0 29) through: (4, 2), perpendicular to βˆ’ 5), perpendicular to 3x + 7y = 0 30) through: ( βˆ’ 31) through: (2, 3, 32) through: ( βˆ’ βˆ’ 2) perpendicular to 3y βˆ’ 2, 5). perpendicular to y x = 0 2x = 0 βˆ’ βˆ’ Write the slope-intercept form of the equation of the line described. 33) through: (4, 34) through: ( 35) through: ( 36) through: ( 37) through: ( 3), parallel to y = βˆ’ βˆ’ 5, 2), parallel to y = 3 5 x 2x βˆ’ 3, 1), parallel to y = 4, 0), parallel to ), parallel to y = 4, βˆ’ 1 2 x + 1 βˆ’ βˆ’ βˆ’ βˆ’ 38) through: (2, 3), parallel to y = 5 2 x 1 βˆ’ 39) through: ( 2, βˆ’ βˆ’ 1 2 x 1), parallel to y = βˆ’ 4), parallel to y = 3 5 x 40) through: ( 5, βˆ’ 41) through: (4, 3), perpendicular to βˆ’ 42) through: ( 3, 5), perpendicular to x + 2y = βˆ’ 43) through: (5, 2), perpendicular to x = 0 βˆ’ 4 βˆ’ 44) through: (5, 45) through: ( βˆ’ 46) through: (2, 1), perpendicular to βˆ’ 2, 5), perpendicular to 3), perpendicular to βˆ’ 5x + 2y = 10 x + y = βˆ’ 2x + 5y = 2 10 βˆ’ βˆ’ βˆ’ βˆ’ 47) through: (4, 3), perpendicular to x + 2y = βˆ’ 4, 1), perpendicular to 4x + 3y = βˆ’ 48) through:
( βˆ’ 6 βˆ’ 9 βˆ’ 116 Chapter 3 : Inequalities 3.1 Solve and Graph Inequalities.................................................................118 3.2 Compound Inequalities..........................................................................124 3.3 Absolute Value Inequalities....................................................................128 117 3.1 Inequalities - Solve and Graph Inequalities Objective: Solve, graph, and give interval notation for the solution to linear inequalities. When we have an equation such as x = 4 we have a specific value for our variable. With inequalities we will give a range of values for our variable. To do this we will not use equals, but one of the following symbols: > Greater than > Greater than or equal to < Less than 6 Less than or equal to World View Note: English mathematician Thomas Harriot first used the above symbols in 1631. However, they were not immediately accepted as symbols such as ⊏ and ⊐ were already coined by another English mathematician, William Oughtred. If we have an expression such as x < 4, this means our variable can be any number 2, 0, 3, 3.9 or even 3.999999999 as long as it is smaller smaller than 4 such as βˆ’ 118 than 4. If we have an expression such as x > any number greater than or equal to βˆ’ 2, such as 5, 0, 2, this means our variable can be 1, βˆ’ βˆ’ 1.9999, or even 2. βˆ’ βˆ’ Because we don’t have one set value for our variable, it is often useful to draw a picture of the solutions to the inequality on a number line. We will start from the value in the problem and bold the lower part of the number line if the variable is smaller than the number, and bold the upper part of the number line if the variable is larger. The value itself we will mark with brackets, either ) or ( for less than or greater than respectively, and ] or [ for less than or equal to or greater than or equal to respectively. Once the graph is drawn we can quickly convert the graph into what is called interval notation. Interval notation gives two numbers, the first is the smallest value, the second is the largest value. If there is no largest value, we can use ∞ (infinity). If there is no smallest value, we can use negative infinity. If we use either positive or negative in
finity we will always use a curved bracket for that value. βˆ’ ∞ Example 151. Graph the inequality and give the interval notation x < 2 Start at 2 and shade below Use ) for less than Our Graph Interval Notation ( βˆ’ ∞, 2) Example 152. Graph the inequality and give the interval notation y > 1 βˆ’ Start at 1 and shade above Use [ for greater than or equal βˆ’ Our Graph Interval Notation [ 1, ) ∞ βˆ’ We can also take a graph and find the inequality for it. 119 Example 153. Give the inequality for the graph: Graph starts at 3 and goes up or greater. Curved bracket means just greater than x > 3 Our Solution Example 154. Give the inequality for the graph: Graph starts at equal to βˆ’ 4 and goes down or less. Square bracket means less than or x 6 βˆ’ 4 Our Solution Generally when we are graphing and giving interval notation for an inequality we will have to first solve the inequality for our variable. Solving inequalities is very similar to solving equations with one exception. Consider the following inequality and what happens when various operations are done to it. Notice what happens to the inequality sign as we add, subtract, multiply and divide by both positive and negative numbers to keep the statment a true statement. Subtract 2 from both sides 5 > 1 Add 3 to both sides 8 > 4 6 > 2 Multiply both sides by 3 12 > 6 Divide both sides by 2 6 > 3 Add 1 to both sides 5 > 2 9 > 6 Multiply both sides by 12 Divide both sides by βˆ’ Subtract βˆ’ βˆ’ 6 2 4 from both sides βˆ’ 18 < βˆ’ βˆ’ 3 > 2 Symbol flipped when we multiply or divide by a negative! As the above problem illustrates, we can add, subtract, multiply, or divide on both sides of the inequality. But if we multiply or divide by a negative number, the symbol will need to flip directions. We will keep that in mind as we solve inequalities. Example 155. Solve and give interval notation 2x > 11 5 βˆ’ Subtract 5 from both sides 120 5 βˆ’ 5 βˆ’ 2x > 6 Divide both sides by βˆ’ 2 Divide by a negative 2 βˆ’ x 6 3 Graph, starting at 2 flip symbol! βˆ’ βˆ’ 3, going down with ] for less than or equal to βˆ’ βˆ’ βˆ’ ( 3], βˆ’ βˆ’ ∞ Interval Notation The inequality
we solve can get as complex as the linear equations we solved. We will use all the same patterns to solve these inequalities as we did for solving equations. Just remember that any time we multiply or divide by a negative the symbol switches directions (multiplying or dividing by a positive does not change the symbol!) Example 156. Solve and give interval notation 3(2x 6x βˆ’ βˆ’ 4) + 4x < 4(3x 12 + 4x < 12x 7) + 8 Distribute 28 + 8 Combine like terms βˆ’ βˆ’ 10x 10x βˆ’ βˆ’ 12 < 12x 10x βˆ’ 12 < 2x βˆ’ + 20 βˆ’ 20 Move variable to one side Subtract 10x from both sides 20 Add 20 to both sides βˆ’ + 20 8 < 2x Divide both sides by 2 2 4 < x Be careful with graph, x is larger! 2 (4, ) ∞ Interval Notation It is important to be careful when the inequality is written backwards as in the previous example (4 < x rather than x > 4). Often students draw their graphs the wrong way when this is the case. The inequality symbol opens to the variable, this means the variable is greater than 4. So we must shade above the 4. 121 3.1 Practice - Solve and Graph Inequalities Draw a graph for each inequality and give interval notation. 5 βˆ’ 2 > k 1) n > 3) βˆ’ 5) 5 > x Write an inequality for each graph. 7) 8) 9) 10) 11) 12) 2) n > 4 4) 1 > k 6) βˆ’ 5 < x 122 Solve each inequality, graph each solution, and give interval notation. 13) x 11 > 10 15) 2 + r < 3 17) 8 + n 3 > 6 19) 2 > a 2 βˆ’ 5 14) βˆ’ 16) m 5 2 6 n 13 6 6 5 βˆ’ 18) 11 > 8 + x 2 20) v 9 βˆ’ 4 βˆ’ 6 2 21) 47 > 8 5x βˆ’ βˆ’ 22) 6 + x 12 6 1 βˆ’ 10 > 60 24) 7n βˆ’ 26) 5 > x βˆ’ 5 + 1 2(3 + k) < 23) βˆ’ 25) 18 < 27) 24 > 2( βˆ’ 6(m βˆ’ βˆ’ 44 βˆ’ 8 + p) 6) βˆ’ 6) < 29) r 5(r βˆ’ βˆ’ βˆ’ 31) 24 + 4b < 4(1 + 6b) βˆ’ 18 28) 30) 32) βˆ’ βˆ’
βˆ’ 8(n βˆ’ 60 > 5) > 0 4( βˆ’ βˆ’ 2n) > 8(2 βˆ’ 36 + 6x > 34) βˆ’ βˆ’ 36) 3(n + 3) + 7(8 6x 3) βˆ’ 16 + n βˆ’ 8(x + 2) + 4x 8n) < 5n + 5 + 2 βˆ’ 5p) + 3 > 2(8 βˆ’ βˆ’ 5p) 33) 5v 5 < 5(4v + 1) βˆ’ βˆ’ βˆ’ 35) 4 + 2(a + 5) < 2( a βˆ’ βˆ’ 4) 37) (k βˆ’ βˆ’ 2) > βˆ’ βˆ’ βˆ’ k 20 38) (4 βˆ’ βˆ’ 123 3.2 Inequalities - Compound Inequalities Objective: Solve, graph and give interval notation to the solution of compound inequalities. Several inequalities can be combined together to form what are called compound inequalities. There are three types of compound inequalities which we will investigate in this lesson. The first type of a compound inequality is an OR inequality. For this type of inequality we want a true statment from either one inequality OR the other inequality OR both. When we are graphing these type of inequalities we will graph each individual inequality above the number line, then move them both down together onto the actual number line for our graph that combines them together. When we give interval notation for our solution, if there are two different parts to (union) symbol between two sets of interval notation, the graph we will put a one for each part. βˆͺ Example 157. Solve each inequality, graph the solution, and give interval notation of solution Solve each inequality 4 Add or subtract first 2x βˆ’ + or 4 4 βˆ’ x > 2 Divide 2x > 8 or βˆ’ 2 1 2 βˆ’ x > 4 or x 6 βˆ’ βˆ’ βˆ’ 1 Dividing by negative flips sign 2 Graph the inequalities separatly above number line ( 2], βˆ’ βˆͺ (4, ∞ βˆ’ ∞ ) Interval Notation World View Note: The symbol for infinity was first used by the Romans, although at the time the number was used for 1000. The greeks also used the symbol for 10,000. There are several different results that could result from an OR statement. The graphs could be pointing different directions, as in the graph above, or pointing in the same
direction as in the graph below on the left, or pointing opposite directions, but overlapping as in the graph below on the right. Notice how interval notation works for each of these cases. 124 As the graphs overlap, we take the largest graph for our solution. When the graphs are combined they cover the entire number line. Interval Notation: (, 1) βˆ’ ∞ Interval Notation: (, ∞ βˆ’ ∞ ) or R The second type of compound inequality is an AND inequality. AND inequalities require both statements to be true. If one is false, they both are false. When we graph these inequalities we can follow a similar process, first graph both inequalities above the number line, but this time only where they overlap will be drawn onto the number line for our final graph. When our solution is given in interval notation it will be expressed in a manner very similar to single inequalities (there is a symbol that can be used for AND, the intersection, but we will not use it here). ∩ Example 158. Solve each inequality, graph the solution, and express it interval notation. 2x + 8 > 5x 2x βˆ’ 2x 8 > 3x βˆ’ 7 and 5x 3x βˆ’ 7 and 2x βˆ’ βˆ’ βˆ’ 3x 3 > 3x + 1 Move variables to one side 3 > 1 Add 7 or 3 to both sides + 7 βˆ’ + 3 + 3 βˆ’ + 7 15 > 3x and 2x > 4 Divide 3 2 2 3 5 > x and x > 2 Graph, x is smaller (or equal) than 5, greater than 2 (2, 5] Interval Notation Again, as we graph AND inequalities, only the overlapping parts of the individual graphs makes it to the final number line. As we graph AND inequalities there are also three different types of results we could get. The first is shown in the above 125 example. The second is if the arrows both point the same way, this is shown below on the left. The third is if the arrows point opposite ways but don’t overlap, this is shown below on the right. Notice how interval notation is expressed in each case. In this graph, the overlap is only the smaller graph, so this is what makes it to the final number line. Interval Notation: ( 2), βˆ’ βˆ’ ∞ In this graph there is no overlap of the parts. Because
their is no overlap, no values make it to the final number line. Interval Notation: No Solution or βˆ… The third type of compound inequality is a special type of AND inequality. When our variable (or expression containing the variable) is between two numbers, we can write it as a single math sentence with three parts, such as 5 < x 6 8, to show x is between 5 and 8 (or equal to 8). When solving these type of inequalities, because there are three parts to work with, to stay balanced we will do the same thing to all three parts (rather than just both sides) to isolate the variable in the middle. The graph then is simply the values between the numbers with appropriate brackets on the ends. Example 159. Solve the inequality, graph the solution, and give interval notation. 6 6 2 βˆ’ 4x + 2 < 2 2 2 βˆ’ βˆ’ Subtract 2 from all three parts 8 6 4 βˆ’ βˆ’ βˆ’ βˆ’ 4x < 0 Divide all three parts by βˆ’ Graph x between 0 and 2 βˆ’ 4 βˆ’ 4 Dividing by a negative flips the symbols Flip entire statement so values get larger left to right (0, 2] Interval Notation 126 3.2 Practice - Compound Inequalities Solve each compound inequality, graph its solution, and give interval notation. 1) n 3 6 3 or βˆ’ βˆ’ 5n 6 10 βˆ’ 2) 6m > βˆ’ 24 or m 7 < 12 βˆ’ 3) x + 7 > 12 or 9x < 45 βˆ’ 13 or 6x 6 60 βˆ’ 6 < βˆ’ 5) x 7) v 8 βˆ’ > 1 and v 2 < 1 βˆ’ βˆ’ 9) 8 + b < 3 and 4b < 20 βˆ’ βˆ’ 11) a + 10 > 3 and 8a 6 48 13) 3 6 9 + x 6 7 15) 11 < 8 + k 6 12 17) 19) 21) 23 βˆ’ 16 6 2n 3m 6 11 10 6 22 βˆ’ βˆ’ 5b + 10 6 30 and 7b + 2 6 40 βˆ’ 25) 3x βˆ’ 9 < 2x + 10 and 5 + 7x 6 10x 10 βˆ’ 8v and 7v + 9 6 6 + 10v βˆ’ 27) 8 6v 6 8 βˆ’ βˆ’ 29) 1 + 5k 6 7k 3 or k βˆ’ 31) 2x + 9 > 10x + 1 and 3x βˆ’ 10 > 2k + 10 2 < 7x + 2 βˆ’ βˆ’ 5 < 4
) 10r > 0 or r βˆ’ 6) 9 + n < 2 or 5n > 40 βˆ’ 12 8) βˆ’ 9x < 63 and x 4 < 1 10) βˆ’ 6n 6 12 and n 3 6 2 6 + v > 0 and 2v > 4 12) βˆ’ 14) 0 > x 9 16) βˆ’ 18) 1 6 p 8 > 1 βˆ’ 11 6 n 6 0 9 6 5 βˆ’ βˆ’ 20) 3 + 7r > 59 or 6r 3 > 33 22) 6 8x > βˆ’ βˆ’ 24) n + 10 > 15 or 4n βˆ’ βˆ’ βˆ’ 6 or 2 + 10x > 82 5 < 1 βˆ’ βˆ’ 6 or 10n 26) 4n + 8 < 3n βˆ’ 2a > 2a + 1 or 10a 10r 6 8 + 4r or βˆ’ 28) 5 30) 8 βˆ’ βˆ’ 8 > 9 + 9n βˆ’ 10 > 9a + 9 βˆ’ 6 + 8r < 2 + 8r 32) βˆ’ 9m + 2 < 10 βˆ’ βˆ’ 6m or βˆ’ m + 5 > 10 + 4m 127 3.3 Inequalities - Absolute Value Inequalities Objective: Solve, graph and give interval notation for the solution to inequalities with absolute values. When an inequality has an absolute value we will have to remove the absolute value in order to graph the solution or give interval notation. The way we remove the absolute value depends on the direction of the inequality symbol. Consider < 2. x | | Absolute value is defined as distance from zero. Another way to read this inequality would be the distance from zero is less than 2. So on a number line we will shade all points that are less than 2 units away from zero. This graph looks just like the graphs of the three part compound inequalities! When the absolute value is less than a number we will remove the absolute value by changing the problem to a three part inequality, with the negative value on the 2 < x < 2, as the left and the positive value on the right. So graph above illustrates. < 2 becomes βˆ’ x | | Consider > 2. x | | Absolute value is defined as distance from zero. Another way to read this inequality would be the distance from zero is greater than 2. So on the number line we shade all points that are more than 2 units away from zero. This graph looks just like the graphs of the OR compound inequalities! When the absolute value is greater than a number we will remove the absolute value
by changing the problem to an OR inequality, the first inequality looking just like the problem with no absolute value, the second flipping the inequality symbol and 2, as the graph changing the value to a negative. So above illustrates. > 2 becomes x > 2 or x < x | βˆ’ | World View Note: The phrase β€œabsolute value” comes from German mathematician Karl Weierstrass in 1876, though he used the absolute value symbol for complex numbers. The first known use of the symbol for integers comes from a 1939 128 edition of a college algebra text! For all absolute value inequalities we can also express our answers in interval notation which is done the same way it is done for standard compound inequalities. We can solve absolute value inequalities much like we solved absolute value equations. Our first step will be to isolate the absolute value. Next we will remove the absolute value by making a three part inequality if the absolute value is less than a number, or making an OR inequality if the absolute value is greater than a number. Then we will solve these inequalites. Remember, if we multiply or divide by a negative the inequality symbol will switch directions! Example 160. Solve, graph, and give interval notation for the solution 4x | 5 > 6 OR 4x 5 βˆ’ | 5 6 4x βˆ’ + 5 + 5 6 Solve βˆ’ + 5 + 5 Add 5 to both sides βˆ’ > 6 Absolute value is greater, use OR 4x > 11 OR 4x 6 4 4 x > 11 4 OR x 6 4 1 Divide both sides by 4 βˆ’ 4 1 4 Graph βˆ’ 1 4, βˆ’ 11 4, ∞ βˆͺ βˆ’ ∞ Interval notation Example 161. Solve, graph, and give interval notation for the solution | βˆ’ 16 Add 4 to both sides + 4 12 Divide both sides by 3 Dividing by a negative switches the symbol βˆ’ βˆ’ > 4 Absolute value is greater, use OR βˆ’ 3 | 129 x > 4 OR x 6 4 Graph βˆ’ ( 4], βˆ’ βˆͺ [4, ) ∞ βˆ’ ∞ Interval Notation In the previous example, we cannot combine terms, the βˆ’ adding 4, then divide by 3 has an absolute value attached. So we must first clear the 3. The next example is similar. 3 because they are not like 4 by 4 and βˆ’ βˆ’ βˆ’ βˆ’ Example 162. Solve, graph, and give interval notation for the solution βˆ’
2 βˆ’ 4x + 1 | 2 | βˆ’ 9 9 | > 4x + 1 | 2 4x + 1 3 < 4x + 1 < 3 1 1 βˆ’ | 1 | βˆ’ βˆ’ βˆ’ βˆ’ Subtract 9 from both sides > 3 9 6 Divide both sides by 2 Dividing by negative switches the symbol βˆ’ βˆ’ βˆ’ < 3 Absolute value is less, use three part βˆ’ 2 Solve Subtract 1 from all three parts βˆ’ 4 < 4x < 2 Divide all three parts by 4 4 4 4 1 2 1 < x < βˆ’ Graph 1, 1 2 βˆ’ Interval Notation In the previous example, we cannot distribute the 2 into the absolute value. We can never distribute or combine things outside the absolute value with what is inside the absolute value. Our only way to solve is to first isolate the absolute value by clearing the values around it, then either make a compound inequality (and OR or a three part) to solve. βˆ’ 130 It is important to remember as we are solving these equations, the absolute value is always positive. If we end up with an absolute value is less than a negative number, then we will have no solution because absolute value will always be positive, greater than a negative. Similarly, if absolute value is greater than a negative, this will always happen. Here the answer will be all real numbers. Example 163. Solve, graph, and give interval notation for the solution 6x | βˆ’ 12 + 4 12 4 βˆ’ 6x 4 6x 1 1 | | βˆ’ βˆ’ | | 1 Subtract 12 from both sides < 4 | 12 βˆ’ < 8 Divide both sides by 4 βˆ’ 4 2 Absolute value canβ€²t be less than a negative < βˆ’ No Solution or βˆ… Example 164. Solve, graph, and give interval notation for the solution | | x + 7 6 | βˆ’ 6 17 | βˆ’ βˆ’ > | Subtract 5 from both sides βˆ’ 6 12 Divide both sides by 6 βˆ’ 6 Dividing by a negative flips the symbol 2 Absolute value always greater than negative All Real Numbers or R 131 3.3 Practice - Absolute Value Inequalities Solve each inequality, graph its solution, and give interval notation. 1) 3) 5) 7) 9) | | | | | x | 2x < βˆ’ 3x 2 < 9 βˆ’ 11) 1 + 2 | x | 2x 6 9 > = 3 1 | 5 | βˆ’ βˆ’ 13) 6 15) 17) 19) | |
| > 5 βˆ’ | 3x | x = 3 > = 3 | 3x 5 | βˆ’ > > 3 > = 10 6 15 βˆ’ > 2x | βˆ’ βˆ’ 7 < | > 1 βˆ’ 5 | 21) 4 + 3 x | βˆ’ 23 | 25) 2 3 | βˆ’ βˆ’ 27) 4 29) 3 5 2 | βˆ’ 4x | βˆ’ βˆ’ 31) 5 2 | βˆ’ βˆ’ 4 βˆ’ 2x βˆ’ 3x βˆ’ 2x + 6 6 < > 8 4 βˆ’ βˆ’ | | 33) 4 4 βˆ’ | βˆ’ 10 + x 35) | βˆ’ > 8 | 8 1 26) 28) 30) x 2) | 6 8 | x + 3 4) | 6) x 8) | 10) < 4 | < 12 6 4 8 βˆ’ | x + 3 | 2x + 5 | 3 βˆ’ | > 5 12) 10 x | x βˆ’ 2x 14) 16) 18 | βˆ’ 20) 3 22) 3 24) 4 2 βˆ’ | 2 | βˆ’ βˆ’ 3x βˆ’ 4x 7 > | 3x βˆ’ βˆ’ | βˆ’ βˆ’ βˆ’ | βˆ’ 3x 32) 6 1 3 | βˆ’ βˆ’ 4x | 2x 34) 4 3 βˆ’ βˆ’ | βˆ’ 132 Chapter 4 : Systems of Equations 4.1 Graphing................................................................................................134 4.2 Substitution............................................................................................139 4.3 Addition/Elimination.............................................................................146 4.4 Three Variables......................................................................................151 4.5 Application: Value Problems.................................................................158 4.6 Application: Mixture Problems.............................................................167 133 4.1 Systems of Equations - Graphing Objective: Solve systems of equations by graphing and identifying the point of intersection. βˆ’ We have solved problems like 3x 4 = 11 by adding 4 to both sides and then dividing by 3 (solution is x = 5). We also have methods to solve equations with more than one variable in them. It turns out that to solve for more than one variable we will need the same number of equations as variables. For example, to solve for two variables such as x and y we will need two equations. When we have several equations we are using to solve, we call the equations a system of equations. When solving a system of equations we are looking for a solution that works in both equations. This solution is usually given as an ordered pair (x, y). The following example illustrates a solution working in both equations Example 165. Show (2,1) is the
solution to the system 3x 2, 1) Identify x and y from the orderd pair x = 2, y = 1 Plug these values into each equation 3(2) (1) = 5 First equation βˆ’ 6 βˆ’ 1 = 5 Evaluate 5 = 5 True (2) + (1) = 3 Second equation, evaluate 3 = 3 True As we found a true statement for both equations we know (2,1) is the solution to the system. It is in fact the only combination of numbers that works in both equations. In this lesson we will be working to find this point given the equations. It seems to follow that if we use points to describe the solution, we can use graphs to find the solutions. If the graph of a line is a picture of all the solutions, we can graph two lines on the same coordinate plane to see the solutions of both equations. We are inter- 134 ested in the point that is a solution for both lines, this would be where the lines intersect! If we can find the intersection of the lines we have found the solution that works in both equations. Example 166 = First: m = βˆ’ Second: m = 2 βˆ’ (4,1) To graph we identify slopes and y intercepts βˆ’ Now we can graph both lines on the same plane. To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. Remember a negative slope is downhill! Find the intersection point, (4,1) (4,1) Our Solution Often our equations won’t be in slope-intercept form and we will have to solve both equations for y first so we can idenfity the slope and y-intercept. Example 167. 6x 3y = βˆ’ 2x + 2y = 9 6 βˆ’ βˆ’ Solve each equation for y βˆ’ βˆ’ βˆ’ 6x 6x 3y = 3 3y = 9 βˆ’ 6x βˆ’ 6x 9 βˆ’ 3 3 βˆ’ βˆ’ y = 2x + 3 βˆ’ βˆ’ βˆ’ 6 βˆ’ 2x 2x + 2y = 2x 2y = 2 βˆ’ y = βˆ’ βˆ’ Subtract x terms 6 Put x terms first βˆ’ 2 3 Divide by coefficient of y Identify slope and y βˆ’ 2x 2 x intercepts βˆ’ 135 First: m = 2 1 Second = (-2,-1)
Now we can graph both lines on the same plane 3 βˆ’ To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. Remember a negative slope is downhill! Find the intersection point, ( 2, βˆ’ βˆ’ 1) 2, ( βˆ’ βˆ’ 1) Our Solution As we are graphing our lines, it is possible to have one of two unexpected results. These are shown and discussed in the next two example. Example 168 First: m = 3 2 Second Identify slope and y βˆ’ intercept of each equation Now we can graph both equations on the same plane To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. The two lines do not intersect! They are parallel! If the lines do not intersect we know that there is no point that works in both equations, there is no solution βˆ… No Solution We also could have noticed that both lines had the same slope. Remembering 136 that parallel lines have the same slope we would have known there was no solution even without having to graph the lines. Example 169. 2x 3x βˆ’ βˆ’ 6y = 12 9y = 18 Solve each equation for y 9y = 18 3x Subtract x terms 3x + 18 Put x terms first βˆ’ 9 βˆ’ 1 y = 3 x 9 Divide by coefficient of y 2 Identify the slopes and y βˆ’ βˆ’ intercepts βˆ’ 6y = 12 2x βˆ’ βˆ’ 2x 2x 2x + 12 6 6 βˆ’ βˆ’ 3x 3x βˆ’ 9y = 9 βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 6y = βˆ’ First: m = 1 3 Second βˆ’ Now we can graph both equations together To graph each equation, we start at the y-intercept and use the slope rise run to get the next point and connect the dots. Both equations are the same line! As one line is directly on top of the other line, we can say that the lines β€œintersect” at all the points! Here we say we have infinite solutions Once we had both equations in slope-intercept form we could have noticed that the equations were the same. At this point we could have stated that there are infinite solutions without having to go through the work of graphing the equations. World View Note: The Babylonians were the first to work with systems
of equations with two variables. However, their work with systems was quickly passed by the Greeks who would solve systems of equations with three or four variables and around 300 AD, developed methods for solving systems with any number of unknowns! 137 4.1 Practice - Graphing Solve each equation by graphing. 1) y = y = 3) y = y = 5) y = y = βˆ’ 7) y = 1 3 y = βˆ’ 9 5x βˆ’ βˆ’ 11) x + 3y = βˆ’ 5x + 3y = 3 9 13) x y = 4 2x + y = βˆ’ βˆ’ 15) 2x + 3y = 2x + y = 2 βˆ’ 1 17) 2x + y = 2 x y = 4 βˆ’ 19) 2x + y = βˆ’ x + 3y = 9 2) y = 2x + 2 4 y = x βˆ’ 8) y = 2x 10 12) x + 4y = βˆ’ 2x + y = 4 12 14) 6x + y = x + y = 2 βˆ’ 3 16) 3x + 2y = 2 3x + 2y = βˆ’ 6 18) x + 2y = 6 5x βˆ’ 4y = 16 6 2 20) x y = 3 5x + 2y = 8 βˆ’ 21) 0 = 6x βˆ’ 12 = 6x 9y + 36 βˆ’ 3y 23) 2x 0 = βˆ’ βˆ’ 25) 3 + y = βˆ’ y = 2x βˆ’ βˆ’ x βˆ’ 6x = 1 y 3 βˆ’ 27) 29) 4 βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ y + 7x = 4 y 3 + 7x = 0 βˆ’ 12 + x = 4y 5x = 4y βˆ’ 12 βˆ’ y βˆ’ 22) 24) 2y + = βˆ’ βˆ’ βˆ’ 2y = 2y = βˆ’ βˆ’ x βˆ’ 2x = x 4 βˆ’ 5x + 4 4y βˆ’ 4 βˆ’ βˆ’ 4y 26) 16 = = βˆ’ 5x + βˆ’ 28) 30) 138 4.2 Systems of Equations - Substitution Objective: Solve systems of equations using substitution. When solving a system by graphing has several limitations. First, it requires the graph to be perfectly drawn, if the lines are not straight we may arrive at the wrong answer. Second, graphing is not a great method to use if the answer is really large, over 100 for example, or if the answer is a decimal the that graph will not help us find, 3.2134 for example
. For these reasons we will rarely use graphing to solve our systems. Instead, an algebraic approach will be used. The first algebraic approach is called substitution. We will build the concepts of substitution through several example, then end with a five-step process to solve problems using this method. Example 170. We already know x = 5, substitute this into the other equation x = 5 y = 2x y = 2(5) y = 10 βˆ’ βˆ’ βˆ’ 3 3 Evaluate, multiply first 3 Subtract y = 7 We now also have y (5, 7) Our Solution When we know what one variable equals we can plug that value (or expression) in for the variable in the other equation. It is very important that when we substitute, the substituted value goes in parenthesis. The reason for this is shown in the next example. Example 171. 2x βˆ’ y = 3x 3y = 7 7 βˆ’ 7) = 7 2x βˆ’ 3(3x βˆ’ We know y = 3x βˆ’ 7, substitute this into the other equation Solve this equation, distributing 3 first βˆ’ 139 2x βˆ’ βˆ’ Subtract 21 9x + 21 = 7 Combine like terms 2x 7x + 21 = 7 21 14 Divide by 7 21 7x = 7 βˆ’ βˆ’ 7 9x βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ x = 2 We now have our x, plug into the y = equation to find y βˆ’ y = 3(2) y = 6 y = (2, βˆ’ βˆ’ βˆ’ βˆ’ Subtract 7 Evaluate, multiply first 7 1 We now also have y 1) Our Solution By using the entire expression 3x 7 to replace y in the other equation we were able to reduce the system to a single linear equation which we can easily solve for our first variable. However, the lone variable (a variable without a coefficient) is not always alone on one side of the equation. If this happens we can isolate it by solving for the lone variable. βˆ’ Example 172. βˆ’ 3x + 2y = 1 x 5y = 6 + 5y + 5y x = 6 + 5y 3(6 + 5y) + 2y = 1 Lone variable is x, isolate by adding 5y to both sides. Substitute this into the untouched equation Solve this equation, distributing 3 first 18 +
15y + 2y = 1 Combine like terms 15y + 2y Subtract 18 from both sides 18 + 17y = 1 18 18 17 Divide both sides by 17 17 17y = 17 βˆ’ βˆ’ βˆ’ x = 6 + 5( y = βˆ’ x = 6 1 We have our y, plug this into the x = equation to find x βˆ’ 1) Evaluate, multiply first 5 Subtract x = 1 We now also have x βˆ’ (1, βˆ’ 1) Our Solution The process in the previous example is how we will solve problems using substitu- 140 tion. This process is described and illustrated in the following table which lists the five steps to solving by substitution. Problem 1. Find the lone variable 2. Solve for the lone variable 3. Substitute into the untouched equation 4. Solve 5. Plug into lone variable equation and evaluate Solution 5 βˆ’ 2y = 2 5 4x βˆ’ 2x + y = βˆ’ Second Equation, y 2x + y = 2x βˆ’ y = 4x βˆ’ 4x + 10 + 4x = 2 8x + 10 = 2 10 10 2x βˆ’ 2x βˆ’ 2( βˆ’ 5 βˆ’ βˆ’ 5 2x) = 2 βˆ’ 8x =, ( βˆ’ 8 βˆ’ 8 1 βˆ’ 2) βˆ’ βˆ’ 1) βˆ’ Sometimes we have several lone variables in a problem. In this case we will have the choice on which lone variable we wish to solve for, either will give the same final result. Example 173. βˆ’ Find the lone variable: x or y in first, or x in second. We will chose x in the first Solve for the lone variable, subtract y from both sides Plug into the untouched equation, the second equation x = 5 y = 1 βˆ’ 2y = 1 βˆ’ 5 6 Divide both sides by 2 Solve, parenthesis are not needed here, combine like terms Subtract 5 from both sides 2y = 2 y We have our y! (5 βˆ’ βˆ’ x = 5 (3) Plug into lone variable equation, evaluate βˆ’ x = 2 Now we have our x 141 (2, 3) Our Solution Just as with graphing it is possible to have no solution βˆ… (parallel lines) or infinite solutions (same line) with the substitution method. While we won’t have a parallel line or the same line to look at and conclude if it
is one or the other, the process takes an interesting turn as shown in the following example. Example 174. Solve for the lone variable, subtract 4 from both sides Find the lone variable, y in the first equation βˆ’ βˆ’ y + 4 = 3x 6x = 2y 8 y + 4 = 3x 4 4 βˆ’ y = 3x 4 Plug into untouched equation 6x = 8 8 Combine like terms 6x 6x = 8 Variables are gone!A true statement. 8 = Solve, distribute through parenthesis 4) 8 βˆ’ βˆ’ 6x βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ Infinite solutions Our Solution 2(3x βˆ’ 6x Because we had a true statement, and no variables, we know that anything that works in the first equation, will also work in the second equation. However, we do not always end up with a true statement. Example 175. 6x 3y = 9 βˆ’ βˆ’ 2x + y = 5 βˆ’ 2x + y = 5 βˆ’ + 2x + 2x Find the lone variable, y in the second equation Solve for the lone variable, add 2x to both sides y = 5 + 2x Plug into untouched equation 6x βˆ’ 6x 15 3(5 + 2x) = βˆ’ 6x = βˆ’ 15 βˆ’ βˆ’ βˆ’ βˆ’ Solve, distribute through parenthesis 9 9 Combine like terms 6x 9 Variables are gone!A false statement. 6x βˆ’ No Solution βˆ… Our Solution Because we had a false statement, and no variables, we know that nothing will work in both equations. 142 World View Note: French mathematician Rene Descartes wrote a book which included an appendix on geometry. It was in this book that he suggested using letters from the end of the alphabet for unknown values. This is why often we are solving for the variables x, y, and z. One more question needs to be considered, what if there is no lone variable? If there is no lone variable substitution can still work to solve, we will just have to select one variable to solve for and use fractions as we solve. Example 176. βˆ’ βˆ’ βˆ’ 5x 5x = 5 6y = 14 βˆ’ 2x + 4y = 12 βˆ’ 6y = 5x 14 βˆ’ + 6y + 6y 14 + 6y 5 5 6y 14 x = βˆ’ 5 5 6y 5 12y 5 + 4y = 12 + 4y = 12 + No lone variable,
we will solve for x in the first equation Solve for our variable, add 6y to both sides Divide each term by 5 Plug into untouched equation Solve, distribute through parenthesis Clear fractions by multiplying by 5 + 4y(5) = 12(5) Reduce fractions and multiply βˆ’ 12y + 20y = 60 28 + 8y = 60 28 28 βˆ’ 8y = 32 8 8 βˆ’ Combine like terms Subtract 28 from both sides βˆ’ 12y + 20y Divide both sides by 8 βˆ’ 2 + 14 βˆ’ 5 28 5 βˆ’ 28(5) 5 βˆ’ 12y(5) 5 28 Plug into lone variable equation, multiply + 14 x = βˆ’ 5 14 x = βˆ’ 5 y = 4 We have our y 6(4) 5 24 5 10 5 Add fractions Reduce fraction + x = x = 2 Now we have our x (2, 4) Our Solution Using the fractions does make the problem a bit more tricky. This is why we have another method for solving systems of equations that will be discussed in another lesson. 143 4.2 Practice - Substitution Solve each system by substitution. 1) y = βˆ’ y = 6x 3) y = βˆ’ y = 2x 3x βˆ’ 2x βˆ’ 9 9 βˆ’ 1 2) y = x + 5 2x y = 4 βˆ’ βˆ’ 6x + 3 4) y = βˆ’ y = 6x + 3 5) y = 6x + 4 6) y = 3x + 13 y = 8) y = y = βˆ’ βˆ’ βˆ’ 10) y = 7x y = 2x 22 βˆ’ 2x 5x 9 21 βˆ’ βˆ’ 24 βˆ’ 3x + 16 βˆ’ x + 3y = 12 βˆ’ y = 6x + 21 24 βˆ’ 12) y = 3x βˆ’ βˆ’ 7) y = 3x + 2 5 y = 3x + 8 βˆ’ 9) y = 2x y = βˆ’ 11) y = 6x 3x βˆ’ 13) y = 3x βˆ’ 15) y = 3 βˆ’ 2x + 9 6 βˆ’ 3y = βˆ’ 6 βˆ’ 6y = 30 14) 6x y = 4y = βˆ’ 6x + 2 βˆ’ βˆ’ 8 7 5 βˆ’ 3x + 4y = 17 βˆ’ 16) 7x + 2y = y = 5x + 5 βˆ’ 17) 2x + 2y = 18 βˆ’ y = 7x + 15 19) y = 8x + 19 x + 6y = 16 βˆ’ βˆ’ 21) 7x βˆ’ y =