data/Gr8A_Mathematics_Learner_Eng.json

{
  "source_file": "Gr8A_Mathematics_Learner_Eng.txt",
  "title": "Grade 8A Mathematics",
  "table_of_contents": [
    {
      "section_id": "1",
      "title": "Whole numbers"
    },
    {
      "section_id": "2",
      "title": "Integers"
    },
    {
      "section_id": "3",
      "title": "Exponents"
    },
    {
      "section_id": "4",
      "title": "Numeric and geometric patterns"
    },
    {
      "section_id": "5",
      "title": "Functions and relationships"
    },
    {
      "section_id": "6",
      "title": "Algebraic expressions 1"
    },
    {
      "section_id": "7",
      "title": "Algebraic equations 1"
    },
    {
      "section_id": "8",
      "title": "Term 1: Revision and assessment"
    },
    {
      "section_id": "8",
      "title": "Algebraic expressions 2"
    },
    {
      "section_id": "9",
      "title": "Algebraic equations 2"
    },
    {
      "section_id": "10",
      "title": "Construction of geometric figures"
    },
    {
      "section_id": "11",
      "title": "Geometry of 2D shapes"
    },
    {
      "section_id": "12",
      "title": "Geometry of straight lines"
    },
    {
      "section_id": "14",
      "title": "Term 2: Revision and assessment"
    }
  ],
  "front_matter": "MATHEMATICS\nGrade 8\nBook 1\nCAPS\nLearner Book\nDeveloped and funded as an ongoing project by the Sasol Inzalo \nFoundation in partnership with the Ukuqonda Institute.\nMaths_LB_gr8_bookA.indb   1\n2014/07/03   10:53:13 AM\n\nPublished by The Ukuqonda Institute\n9 Neale Street, Rietondale 0084\nRegistered as a Title 21 company, registration number 2006/026363/08\nPublic Benefit Organisation, PBO Nr. 930035134\nWebsite: http://www.ukuqonda.org.za\nFirst published in 2014\n© 2014. Copyright in the work is vested in the publisher.\nCopyright in the text remains vested in the contributors.\nISBN: 978-1-920705-26-8\nThis book was developed with the participation of the Department of Basic  \nEducation of South Africa with funding from the Sasol Inzalo Foundation.\nContributors:\nPiet Human, Erna Lampen, Marthinus de Jager, Louise Keegan, Paul van Koersveld,  \nNathi Makae, Enoch Masemola, Therine van Niekerk, Alwyn Olivier, Cerenus Pfeiffer,  \nRenate Röhrs, Dirk Wessels, Herholdt Bezuidenhout\nIllustrations and computer graphics:\nLeonora van Staden, Lisa Steyn Illustration\nZhandre Stark, Lebone Publishing Services\nComputer graphics for chapter frontispieces: Piet Human\nCover illustration: Leonora van Staden\nText design: Mike Schramm\nLayout and typesetting: Lebone Publishing Services\nPrinted by: [printer name and address]\nMaths_LB_gr8_bookA.indb   2\n2014/07/03   10:53:13 AM\n\nCOPYRIGHT NOTICE\nYour freedom to legally copy this book\nThis book is published under a Creative Commons Attribution-NonCommercial 4.0  \nUnported License (CC BY-NC).\nYou are allowed and encouraged to freely copy this book. You can photocopy, print  \nand distribute it as often as you like. You may download it onto any electronic device, \ndistribute it via email, and upload it to your website, at no charge. You may also adapt \nthe text and illustrations, provided you acknowledge the copyright holders (‘attribute \nthe original work’).\nRestrictions: You may not make copies of this book for a profit-seeking purpose.  \nThis holds for printed, electronic and web-based copies of this book,  \nand any part of this book.\nFor more information about the Creative Commons Attribution-NonCommercial 4.0 \nUnported (CC BY-NC 4.0) license, see http://creativecommons.org/ \nlicenses/by-nc/4.0/\nAll reasonable efforts have been made to ensure that materials included are not already \ncopyrighted to other entities, or in a small number of cases, to acknowledge copyright \nholders. In some cases this may not have been possible. The publishers welcome the \nopportunity for redress with any unacknowledged copyright holders.\nExcept where otherwise noted, this work is licensed under  \nhttp://creativecommons.org/licenses/by-nc/4.0/\nMaths_English_TG_gr7_prelim.indd   3\n2014/08/26   02:53:16 PM\n\nTable of contents\nTerm 1\nChapter 1:\nWhole numbers....................................................................... 1\t\nChapter 2:\nIntegers..................................................................................... 29\t\nChapter 3:\nExponents................................................................................. 51\t\nChapter 4:\nNumeric and geometric patterns .......................................... 77\t\nChapter 5:\nFunctions and relationships....................................................93\nChapter 6:\nAlgebraic expressions 1...........................................................105\nChapter 7:\nAlgebraic equations 1..............................................................119\t\nTerm 1: Revision and assessment...........................................127\nMaths_LB_gr8_bookA.indb   4\n2014/07/03   10:53:13 AM\n\nTerm 2\nChapter 8:\nAlgebraic expressions 2........................................................... 145\t\nChapter 9:\nAlgebraic equations 2.............................................................. 165\t\nChapter 10: \nConstruction of geometric figures.........................................173\t\nChapter 11:\nGeometry of 2D shapes..........................................................191\t\nChapter 12:\nGeometry of straight lines......................................................211\t\nTerm 2: Revision and assessment...........................................231\nMaths_LB_gr8_bookA.indb   5\n2014/07/03   10:53:13 AM\n\nMaths_LB_gr8_bookA.indb   6\n2014/07/03   10:53:13 AM",
  "chapters": [
    {
      "title": "Whole numbers",
      "content": "1\t Whole numbers\n1.1\t\nProperties of whole numbers\nthe commutative property of addition and multiplication\n1.\t Which of the following calculations would you choose  \nto calculate the number of yellow beads in this pattern?  \nDo not do any calculations now, just make a choice.\n\t\n\t\n(a)\t 7 + 7 + 7 + 7 + 7\n\t\n(b)\t 10 + 10 + 10 + 10 + 10 + 10 + 10\n\t\n(c)\t 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5\n\t\n(d)\t 5 + 5 + 5 + 5 + 5 + 5 + 5 \n\t\n(e)\t 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7 + 7\n\t\n(f)\t 10 + 10 + 10 + 10 + 10\n\t\nMy choice: \n2.\t (a)\t How many red beads are there in the pattern, and how many yellow beads?\n\t\n(b)\t How many beads are there in the pattern in total? \n3.\t (a)\t Which expression describes what you did to calculate the total number of beads:\n\t\n\t\n70 + 50 or 50 + 70? \n\t\n(b)\t Does it make a difference? \n\t\n(c)\t Which expression describes what you did to calculate the number of red beads: \n\t\n\t\n7 × 10 or 10 × 7? \n\t\n(d)\t Does it make a difference? \nWe say: addition and multiplication are \ncommutative. The numbers can be swopped \naround and their order does not change the answer. \nThis does not work for subtraction and division, \nhowever.\n4.\t Calculate each of the following:\n\t\n5 × 8  \n\t\n10 × 8  \n\t\n12 × 8  \n \t\n8 × 12  \n\t\n6 × 8  \n\t\n3 × 7     \n\t\n6 × 7    \n \t\n7 × 6    \nMaths_LB_gr8_bookA.indb   3\n2014/07/03   10:53:13 AM\n\n4\t\nMATHEMATICS Grade 8: Term 1\nthe associative property of addition and multiplication\nLebogang and Nathi both have to calculate 25 × 24.\nLebogang calculates 25 × 4 and then multiplies by 6.\nNathi calculates 25 × 6 and then multiplies by 4.\n1.\t Will they get the same answer or not?   \nIf three or more numbers have to be multiplied, \nit does not matter which two of the numbers are \nmultiplied first. \nThis is called the associative property of \nmultiplication. We also say multiplication is \nassociative.\n2.\t Do the following calculations. Do not use a calculator now.\n\t\n(a)\t 4 + 7 + 5 + 6  \n\t\n\t\n\t\n\t\n(b)\t 7 + 6 + 5 + 4  \n\t\n(c)\t 6 + 5 + 7 + 4  \n\t\n\t\n\t\n\t\n(d)\t 7 + 5 + 4 + 6  \n3.\t (a)\t Is addition associative?   \n\t\n(b)\t Illustrate your answer with an example. \n4.\t Find the value of each expression by working in the easiest possible way. \n\t\n(a)\t 2 × 17 × 5  \t \t\n\t\n(b)\t 4 × 7 × 5\t\n\t\n\t\n\t\n(c)\t 75 + 37 + 25\t\n\t\n(d)\t 60 + 87 + 40 + 13 \t\n5.\t What must you add to each of the following numbers to get 100?\n\t\n82\t \t\n\t\n44\t \t\n\t\n56\t \t\n\t\n78\t \t\n\t\n24\t \t\n\t\n89\t \t\n\t\n77\n6.\t What must you multiply each of these numbers by to get 1 000?\n\t\n250\t\t\n\t\n125\t\n\t\n25\t \t\n\t\n500\t\n\t\n\t\n200\t\n\t\n\t\n50\n7.\t Calculate each of the following. Note that you can make the work very easy by being \nsmart in deciding how to group the operations. \n\t\n(a)\t 82 + 54 + 18 + 46 + 237\t\t\n\t\n\t\n\t\n(b)\t 24 + 89 + 44 + 76 + 56 + 11 \n\t\n(c)\t 25 × (86 × 4)\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 32 × 125\nMaths_LB_gr8_bookA.indb   4\n2014/07/03   10:53:14 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n5\nmore conventions and the distributive property\nThe distributive property is a useful property because it allows us to do this:\n3 × (2 + 4)\n3 × 2 + 3 × 4\nBoth answers are 18. Notice that we have to use brackets in the first example to show \nthat the addition operation must be done first. Otherwise, we would have done the \nmultiplication first. For example, the expression 3 × 2 + 4 means “multiply 3 by 2; then \nadd 4”. It does not mean “add 2 and 4; then multiply by 3”.\nThe expression 4 + 3 × 2 also means “multiply 3 by 2; then add 4”.\nIf you wish to specify that addition or subtraction \nshould be done first, that part of the expression \nshould be enclosed in brackets.\nThe distributive property can be used to break up a difficult multiplication into smaller \nparts. For example, it  can be used to make it easier to calculate 6 × 204:\n\t\n\t\n6 × 204 can be rewritten as 6 × (200 + 4)\t \t\n(Remember the brackets!)\n\t\n= 6 × 200 + 6 × 4 \n\t\n= 1 200 + 24\n\t\n= 1 224\nMultiplication can also be distributed over subtraction, for example to calculate 7 × 96:\n\t\n7 × 96 = 7 × (100 − 4)\n\t\n= 7 × 100 − 7 × 4 \n\t\n= 700 − 28\n\t\n= 672\n1. \t Here are some calculations with answers. Rewrite them with brackets to make all  \nthe answers correct.\n\t\n(a) \t 8 + 6 × 5 = 70 \t \t\n\t\n\t\n\t\n\t\n\t\n(b) \t8 + 6 × 5 = 38\n\t\n(c)\t 5 + 8 × 6 − 2 = 52\t\n\t\n\t\n\t\n\t\n\t\n(d) \t5 + 8 × 6 − 2 = 76\n\t\n(e) \t 5 + 8 × 6 − 2 = 51 \t \t\n\t\n\t\n\t\n\t\n(f) \t 5 + 8 × 6 − 2 = 37\nMaths_LB_gr8_bookA.indb   5\n2014/07/03   10:53:14 AM\n\n6\t\nMATHEMATICS Grade 8: Term 1\n2.\t Calculate the following:\n\t\n(a)\t 100 × (10 + 7)\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t 100 × 10 + 100 × 7\n\t\n(c)\t 100 × (10 − 7)\t \t\n\t\n\t\n\t\n\t\n\t\n(d)\t 100 × 10 − 100 × 7\n3.\t Complete the table.\n×\n8\n5\n4\n9\n7\n3\n6\n2\n10\n11\n12\n7\n3\n27\n6\n9\n5\n8\n6\n4\n28\n2\n10\n80\n110\n12\n11\n4.\t Use the various mathematical conventions for numerical expressions to make these \ncalculations easier. Show all your working.    \n\t\n(a)\t 18 × 50\t\n(b)\t 125 × 28\t\n(c)\t 39 × 220\n\t\n(d)\t 443 + 2 100 + 557\t\n(e)\t 318 + 650 + 322\t\n(f)\t 522 + 3 003 + 78\nTwo more properties of numbers are:\n• The additive property of 0: when we add zero to \nany number, the answer is that number. \n• The multiplicative property of 1: when we \nmultiply any number by 1, the answer is that number.\nMaths_LB_gr8_bookA.indb   6\n2014/07/03   10:53:14 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n7\n1.2\t Calculations with whole numbers\nestimating, approximating and rounding\n1.\t Try to give answers that you trust to these questions, without doing any calculations \nwith the given numbers.  \n\t\n(a) \t Is 8 × 117 more than 2 000 or less than 2 000? \n than 2 000\n \n \n \n\t\n(b)\t Is 27 × 88 more than 3 000 or less than 3 000? \n than 3 000\n \n \n \n\t\n(c)\t Is 18 × 117 more than 3 000 or less than 3 000? \n than 3 000\n \n \n \n\t\n(d)\t Is 47 × 79 more than 3 000 or less than 3 000? \n than 3 000\n \n \n \nWhat you have done when you tried to give answers  \nto questions 1(a) to (d), is called estimation. To  \nestimate is to try to get close to an answer without  \nactually doing the required calculations with the given numbers.\n2.\t Look at question 1 again. \n\t\n(a)\t The numbers 1 000, 2 000, 3 000, 4 000, 5 000, 6 000, 7 000, 8 000, 9 000 and \n    10 000 are all multiples of a thousand. In each case, write down the multiple of  \n    1 000 that you think is closest to the answer. Write it on the short dotted \n    line. The numbers you write down are called estimates.\n\t\n(b)\t In some cases you may think that you may achieve a better estimate by  \n    adding 500 to your estimate, or subtracting 500 from it. If so, you may add or \n    subtract 500. \n\t\n(c)\t If you wish, you may write what you believe is an even better estimate by  \n    adding or subtracting some hundreds. \n3.\t (a)\t Use a calculator to find the exact answers for  \n    the calculations in question 1, or look up the  \n    answers in one of the tables on page 2.  \n    Calculate the error in your last approximation \n    of each of the answers in question 1.\n\t\n(b)\t What was your smallest error? \nAn estimate may also be \ncalled an approximation.\nThe difference between \nan estimate and the actual \nanswer is called the error. \nMaths_LB_gr8_bookA.indb   7\n2014/07/03   10:53:14 AM\n\n8\t\nMATHEMATICS Grade 8: Term 1\n4.\t Think again about what you did in question 2. In 2(a) you tried to approximate the \nanswers to the nearest 1 000. In 2(c) you tried to approximate the answers to the \nnearest 100. Describe what you tried to achieve in question 2(b).  \n5.\t Estimate the answers for each of the following products and sums. Try to approximate \nthe answers for the products to the nearest thousand, and for the sums to the nearest \nhundred. Use the first line in each question to do this.\n\t\n(a)\t 84 × 178\t\n\t\n(b)\t 677 + 638\t\n\t\n(c)\t 124 × 93\t\n\t\n(d)\t 885 + 473\t\n\t\n(e)\t 79 × 84\t\t\n\t\n(f)\t 921 + 367\t\n\t\n(g)\t 56 × 348\t\n\t\n(h)\t764 + 829\t\n6.\t Use a calculator to find the exact answers for the calculations in question 5, or \nlook up the answers in the tables on page 2. Calculate the error in each of your \napproximations. Use the second line in each question to do this.\nCalculating with “easy” numbers that are close to given numbers is a good way to obtain \napproximate answers, for example:\n• To approximate 764 + 829 one may calculate 800 + 800 to get the approximate answer \n1 600, with an error of 7.\n• To approximate 84 × 178 one may calculate 80 × 200 to get the approximate answer  \n16 000, with an error of 1 048.\n7.\t Calculate with “easy” numbers close to the given numbers to produce approximate \nanswers for each product below. Do not use a calculator. When you have made \nyour approximations, look up the precise answers in the top table on page 2.\n\t\n(a)\t 78 × 46 \n\t\n(b)\t 67 × 88 \n\t\n(c)\t 34 × 276 \n\t\n(d)\t 78 × 178 \n\t\nMaths_LB_gr8_bookA.indb   8\n2014/07/03   10:53:14 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n9\nrounding off and compensating\n1.\t (a)\t Approximate the answer for 386 + 3 435, by  \n    rounding both numbers off to the nearest \n    hundred, and adding the rounded numbers.\n\t\n(b)\t Because you rounded 386 up to 400, you introduced an error of 14 in your  \n    approximate answer. What error did you introduce by rounding 3 435 down to  \n    3 400?  \n\t\n(c)\t What was the combined (total) error introduced by rounding both numbers off  \n    before calculating?\n\t\n(d)\t Use your knowledge of the total error to correct your approximate answer, so  \n    that you have the correct answer for 386 + 3 435. \nWhat you have done in question 1 to find the correct answer for 386 + 3 435 is called \nrounding off and compensating. By rounding the numbers off you introduced \nerrors. You then compensated for the errors by making adjustments to your answer.\n2.\t Round off and compensate to calculate each of the following accurately: \n\t\n(a)\t 473 + 638   \n\t\n\t\n\t\n(b)\t 677 + 921  \nSubtraction can also be done in this way. For example, to calculate R5 362 − R2 687, you \nmay round R2 687 up to R3 000. The calculation can proceed as follows:\n• Rounding R2 687 up to R3 000 can be done in two steps: 2 687 + 13 = 2 700, and  \n2 700 + 300 = 3 000. In total, 313 is added.\n• 313 can now be added to 5 362 too: R5 362 + 313 = 5 675.\n• Instead of calculating R5 362 − R2 687, which is a bit difficult, you may calculate  \nR5 675 − R3 000. This is easy: R5 675 − R3 000 = R2 675.\nThis means that R5 362 − R2 687 = R2 675, because\nR5 362 − R2 687 = (R5 362 + R313) − (R2 687 + R313).\nThe word compensate \nmeans to do things that will \nremove damage.\nMaths_LB_gr8_bookA.indb   9\n2014/07/03   10:53:14 AM\n\n10\t\nMATHEMATICS Grade 8: Term 1\nadding numbers in parts written in columns\nNumbers can be added by thinking of their parts as we say the numbers. \nFor example, we say 4 994 as four thousand nine hundred and ninety-four.  \nThis can be written in expanded notation as 4 000 + 900 + 90 + 4.\nSimilarly, we can think of 31 837 as 30 000 + 1 000 + 800 + 30 + 7.\n31 837 + 4 994 can be calculated by working with the various kinds \nof parts separately. To make this easy, the numbers can be written  \nbelow each other so that the units are below the units, the tens  \nbelow the tens and so on, as shown on the right. \nWe write only this:\nIn your mind you can see this:\n31 837\n30 000\n1 000\n800\n30\n7\n4 994\n4 000\n900\n90\n4\nThe numbers in each column can be added to get a new set of numbers.\n31 837\n30 000\n1 000\n800\n30\n7\n  4 994\n4 000\n900\n90\n4\n11\n11\n120\n120\n1 700\n1 700\n5 000\n5 000\n30 000\n30 000\n36 831\nIt is easy to add the new set of numbers to get the answer. \nThe work may start with the 10 000s or any other parts. Starting with the units as  \nshown above makes it possible to do more of the work mentally, and write less, as  \nshown below.\n31 837\nTo achieve this, only the units digit 1 of the 11 is written in \nthe first step. The 10 of the 11 is remembered and added to \nthe 30 and 90 of the tens column, to get 130. \n  4 994\n36 831\nWe say the 10 is carried from the units column to the tens column. The same is done \nwhen the tens parts are added to get 130: only the digit “3” is written (in the tens \ncolumn, so it means 30), and the 100 is carried to the next step.\n1.\t Calculate each of the following without using a calculator:\n\t\n(a)\t 4 638 + 2 667  \t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t 748 + 7 246  \n31 837\n4 994\nMaths_LB_gr8_bookA.indb   10\n2014/07/03   10:53:14 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n11\n2.\t Impilo Enterprises plans a new computerised training facility in their existing \nbuilding. The training manager has to keep the total expenditure budget under R1 \nmillion. This is what she has written so far:\n\t\nArchitects and builders\n\t\nPainting and carpeting\n\t\nSecurity doors and blinds\n\t\nData projector\n\t\n25 new secretary chairs\n\t\n24 desks for work stations\n\t\n1 desk for presenter\n\t\n25 new computers\n\t\n12 colour laser printers\nR102 700\nR  42 600\nR  52 000\nR     4 800\nR  50 400\nR123 000\nR  28 000\nR300 000\nR  38 980\n\t\nWork out the total cost of all the items the training manager has budgeted for.\n3.\t Calculate each of the following without using a calculator:\n\t\n(a)\t 7 828 + 6 284   \t\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 7 826 + 888 + 367\n\t\n(c)\t 657 + 32 890 + 6 542   \t \t\n\t\n\t\n\t\n(d)\t 6 666 + 3 333 + 1   \n\t\nMaths_LB_gr8_bookA.indb   11\n2014/07/03   10:53:14 AM\n\n12\t\nMATHEMATICS Grade 8: Term 1\nmethods of subtraction\nThere are many ways to find the difference between two numbers. For example, to find \nthe difference between 267 and 859 one may think of the numbers as they may be \nwritten on a number line.\n100\n200\n300\n400\n500\n600\n700\n800\n900\n1 000\n267\n859\nWe may think of the distance between 267 and 859 as three steps: from 267 to 300, from \n300 to 800, and from 800 to 859. How big are each of these three steps?\n100\n200\n300\n400\n500\n600\n700\n800\n900\n1 000\n267\n859\n267 + 33  300\n 800 + 59 = 859\n+ 500\nThe above shows that 859 − 267 is 33 + 500 + 59.\n1.\t Calculate 33 + 500 + 59 to find the answer for 859 − 267.\n2.\t Calculate each of the following. You may think of working out the distance between \nthe two numbers as shown above, or use any other method you prefer.  \nDo not use a calculator now.   \n\t\n(a)\t 823 − 456\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 1 714 − 829\n\t\n(c)\t 3 045 − 2 572\t \t\n\t\n\t\n\t\n\t\n\t\n(d)\t 5 131 − 367\nYou can use the tables of sums on page 2 to check your answers for question 2.\nLike addition, subtraction can also be done by working with the different parts in  \nwhich we say numbers. For example, 8 764 − 2 352 can be calculated as follows:\n8 thousand − 2 thousand = 6 thousand\n    7 hundred − 3 hundred = 4 hundred\n                       6 tens − 5 tens = 1 ten\n                   4 units − 2 units = 2 units\nSo, 8 764 − 2 352 = 6 412\nMaths_LB_gr8_bookA.indb   12\n2014/07/03   10:53:15 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n13\nSubtraction by parts is more difficult in some cases, for example 6 213 − 2 758:\n\t\n 6 000 − 2 000 = 4 000. This step is easy, but the following steps cause problems:\n\t\n        200 − 700 = ?\n\t\n             10 − 50 = ?\n\t\n\t\n    3 − 8 = ?\nFortunately, the parts and sequence of work may be  \nrearranged to overcome these problems, as shown  \nbelow:\ninstead of \nwe may do\n3 − 8 =\n?\n13 − 8 =\n“borrow” 10 from below\n10 − 50 =\n?\n100 − 50 =\n“borrow” 100 from below\n200 − 700 =\n?\n1 100 − 700 =\n“borrow” 1 000 from below\n6 000 − 2 000 =\n?\n5 000 − 2 000 =\nThis reasoning can also be set out in columns:\n             instead of             \nwe may do             \nbut write only this\n6 000\n200\n10\n3\n5 000\n1 100\n100\n13\n6\n2\n1\n3\n2 000\n700\n50\n8\n2 000\n700\n50\n8\n2\n7\n5\n8\n3 000\n400\n50\n5\n3\n4\n5\n5\n3.\t (a)\t Complete the above calculations and find the answer for 6 213 − 2 758.\n\t\n(b)\t Use the borrowing technique to calculate 823 − 376 and 6 431 − 4 968.\n4.\t Check your answers in question 3(b) by doing addition. \nOne way to overcome these \nproblems is to work with \nnegative numbers: \n200 − 700 = (−500) \n10 − 50 = (−40) \n3 − 8 = (−5) \n4 000 − 500  3 500 − 45 = \nMaths_LB_gr8_bookA.indb   13\n2014/07/03   10:53:15 AM\n\n14\t\nMATHEMATICS Grade 8: Term 1\nWith some practice, you can learn to subtract using borrowing  \nwithout writing all the steps. It is convenient to work in  \ncolumns, as shown on the right for calculating  \n6 213 − 2 758.\nIn fact, by doing more work mentally, you may learn to save  \nmore paper by writing even less as shown below.\n6 213\n2 758\n3 455\nDo not use a calculator when you do question 5, because the purpose of this work is for \nyou to come to understand methods of subtraction. What you will learn here will later \nhelp you to understand algebra better.    \n5.\t Calculate each of the following: \n\t\n(a)\t 7 342 − 3 877\t\n(b)\t 8 653 − 1 856\t\n(c)\t 5 671 − 4 528\nYou may use a calculator to do questions 6 and 7.\n6.\t Estimate the difference between the two car prices in each case to the nearest R1 000 \nor closer. Then calculate the difference.\n\t\n(a)\t R102 365 and R98 128\t\t\n\t\n\t\n\t\n(b)\t R63 378 and R96 889\n7.\t First estimate the answers to the nearest 100 000 or 10 000 or 1 000. Then calculate.\n\t\n(a)\t 238 769 −141 453\t\n(b)\t 856 333 − 739 878\t\n(c)\t 65 244 − 39 427\n6 213\n2 758\n5\n50\n400\n3 000\n3 455\nMaths_LB_gr8_bookA.indb   14\n2014/07/03   10:53:15 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n15\na method of multiplication\n7 × 4 59I can be calculated in parts, as shown here:\n\t\n7 × 4 000\t = \t 28 000\n\t\n7 × 500\t = \t 3 500\n\t\n7 × 90\t = \t 630\n\t\n7 × 8\t =   56\nThe four partial products can now be added to get the  \nanswer, which is 32 186. It is convenient to write the  \nwork in vertical columns for units, tens, hundreds and  \nso on, as shown on the right.\nThe answer can be produced with less writing, by “carrying”  \nparts of the partial answers to the next column, when working  \nfrom right to left in the columns. Only the 6 of the product 7 × 8  \nis written down instead of 56. The 50 is kept in mind, and added  \nto the 630 obtained when 7 × 90 is calculated in the next step.\n1.\t Calculate each of the following. Do not use a calculator now.\n\t\n(a)\t 27 × 649    \t\n(b)\t 75 × 1 756    \t\n(c)\t 348 × 93\n    \t \t\n2.\t Use your calculator to check your answers for question 1. Redo the questions for \nwhich you had the wrong answers.\n3.\t Calculate each of the following. Do not use a calculator now.\n\t\n(a)\t 67 × 276\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 84 × 178\t\n\t\n\t\n4.\t Use the product table on page 2 or a calculator to check your answers for question 3. \nRedo the questions for which you had the wrong answers.\n4\n5\n9\n8\n7\n5\n6\n6\n3\n0\n3\n5\n0\n0\n2\n8\n0\n0\n0\n3\n2\n1\n8\n6\n4\n5\n9\n8\n7\n3\n2\n1\n8\n6\nMaths_LB_gr8_bookA.indb   15\n2014/07/03   10:53:15 AM\n\n16\t\nMATHEMATICS Grade 8: Term 1\nlong division\n1.\t The municipal head gardener wants to buy young trees to plant along the main  \nstreet of the town. The young trees cost R27 each, and an amount of R9 400 has  \nbeen budgeted for trees. He needs 324 trees. Do you think he has enough money?\n2.\t (a)\t How much will 300 trees cost? \n\t\n(b)\t How much money will be left if 300 trees are bought? \n\t\n(c)\t How much money will be left if 20 more trees are bought? \nThe municipal gardener wants to work out exactly how many trees, at R27 each, he can \nbuy with the budgeted amount of R9 400. His thinking and writing are described below.\nStep 1\nWhat he writes:\nWhat he thinks:\n27\n9 400\nI want to find out how many chunks of 27 there are in 9 400.\nStep 2\nWhat he writes:\nWhat he thinks:\n300\nI think there are at least 300 chunks of 27 in 9 400.\n27\n9 400\n8 100\n300 × 27 = 8 100. I need to know how much is left over.\n1 300\nI want to find out how many chunks of 27 there are in 1 300.\nStep 3    (He has to rub out the one “0” of the 300 on top, to make space.)\nWhat he writes:\nWhat he thinks:\n340\nI think there are at least 40 chunks of 27 in 1 300.\n27\n9 400\n8 100\n1 300\n1 080\n40 × 27 = 1 080. I need to know how much is left over.\n220\nI want to find out how many chunks of 27 there are in 220. \nPerhaps I can buy some extra trees.\nStep 4    (He rubs out another “0”.)\nWhat he writes:\nWhat he thinks: \n348\nI think there are at least 8 chunks of 27 in 220.\n27\n9 400\n8 100\n1 300\n1 080\n220\n216\n8 × 27 = 216\n4\nSo, I can buy 348 young trees and will have R4 left.\nMaths_LB_gr8_bookA.indb   16\n2014/07/03   10:53:15 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n17\nDo not use a calculator to do questions 3 and 4. The purpose of this work is for you \nto develop a good understanding of how division can be done. Check all your answers by \ndoing multiplication.\n3.\t (a) Graham bought 64 goats, all at the \t\n\t\nsame price. He paid R5 440 in total. \t\n\t\nWhat was the price for each goat? \t\n\t\nYour first step can be to work out \t\n\t\nhow much he would have paid if he \t\n\t\npaid R10 per goat, but you can start \t\n\t\nwith a bigger step if you wish.\n(b)\tMary has R2 850 and she wants to \nbuy candles for her sister’s wedding \nreception. The candles cost R48 each. \nHow many candles can she buy?\n4.\t Calculate each of the following, without using a calculator:\n\t\n(a)\t 7 234 ÷ 48\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 3 267 ÷ 24\n\t\n(c)\t 9 500 ÷ 364\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 8 347 ÷ 24\nMaths_LB_gr8_bookA.indb   17\n2014/07/03   10:53:15 AM\n\n18\t\nMATHEMATICS Grade 8: Term 1\n1.3\t Multiples, factors and prime factors\nmultiples and factors\n1.\t The numbers 6; 12; 18; 24; ... are multiples of 6. \n\t\nThe numbers 7; 14; 21; 28; ... are multiples of 7. \n\t\n(a)\t What is the 100th number in each sequence above? \n\t\n(b)\t Is 198 a number in the first sequence? \n\t\n(c)\t Is 175 a number in the second sequence?\nOf which numbers is 20 a multiple?\n20 = 1 × 20 = 2 × 10 = 4 × 5 = 5 × 4 = 10 × 2 = 20 × 1 \nFactors come in pairs. The following pairs are factors  \nof 20: \n1\n2\n4\n5\n10\n20\n2.\t A rectangle has an area of 30 cm. What are the possible lengths of the sides of the \nrectangle in centimetres if the lengths of the sides are natural numbers? \n3.\t Are 4; 8; 12 and 16 factors of 48? Simon says that all multiples of 4 smaller than 48 are \nfactors of 48. Is he right? \n4.\t We have defined factors in terms of the product of two numbers. What happens if we \nhave a product of three or more numbers, for example 210 = 2 × 3 × 5 × 7? \n\t\n(a)\t Explain why 2; 3; 5 and 7 are factors of 210.\n\t\n(b)\t Are 2 × 3; 3 × 5; 5 × 7; 2 × 5 and 2 × 7 factors of 210? \n\t\n(c)\t Are 2 × 3 × 5; 3 × 5 × 7 and 2 × 5 × 7 factors of 210? \n5.\t Is 20 a factor of 60? What factors of 20 are also factors of 60?\nIf n is a natural number, 6n \nrepresents the multiples of 6.\n20 is a multiple of 1; 2; 4; 5; \n10 and 20 and all of these \nnumbers are factors of 20.\nMaths_LB_gr8_bookA.indb   18\n2014/07/03   10:53:15 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n19\nprime numbers and composite numbers\n1.\t Express each of the following numbers as a  \nproduct of as many factors as possible, including  \nrepeated factors. Do not use 1 as a factor. \n\t\n(a)\t 66\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 67\n\t\n(c)\t 68\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 69\n\t\n(e)\t 70\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 71\n\t\n(g)\t 72\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(h)\t73\n2.\t Which of the numbers in question 1 cannot be expressed as a product  \nof two whole numbers, except as the product 1 × the number itself? \nA number that cannot be expressed as a product of \ntwo whole numbers, except as the product \n1 × the number itself, is called a prime number.\n3.\t Which of the numbers in question 1 are prime? \nComposite numbers are natural numbers with \nmore than two different factors. The sequence of \ncomposite numbers is 4; 6; 8; 9; 10; 12; ... \n4.\t Are the statements below true or false? If you answer “false”, explain why.\n\t\n(a)\t All prime numbers are odd numbers. \n\t\n(b)\t All composite numbers are even numbers. \n\t\n(c)\t 1 is a prime number. \n\t\n(d)\t If a natural number is not prime, then it is composite. \nThe number 36 can be formed \nas 2 × 2 × 3 × 3. Because 2 and \n3 are used twice, they are called \nrepeated factors of 36.\nMaths_LB_gr8_bookA.indb   19\n2014/07/03   10:53:15 AM\n\n20\t\nMATHEMATICS Grade 8: Term 1\n\t\n(e)\t 2 is a composite number.\n\t\n(f)\t 785 is a prime number.\n\t\n(g)\t A prime number can only end in 1; 3; 7 or 9.\n\t\n(h)\t Every composite number is divisible by at least one prime number.\n5.\t We can find out whether a given number is prime by systematically checking \nwhether the primes 2; 3; 5; 7; 11; 13; … are factors of the given number or not. \n\t\nTo find possible factors of 131, we need to consider only the primes 2; 3; 5; 7 and 11. \nWhy not 13; 17; 19; …?\n6.\t Determine whether the following numbers are prime or composite. If the number is \ncomposite, write down at least two factors of the number (besides 1 and the number \nitself).\n\t\n(a)\t 221\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 713\nprime factorisation\nTo find all the factors of a number you can write the  \nnumber as the product of prime factors, first by writing it  \nas the product of two convenient (composite) factors  \nand then by splitting these factors into smaller factors  \nuntil all factors are prime. Then you take all the possible  \ncombinations of the products of the prime factors. \nExample: Find the factors of 84. \nWrite 84 as the product of prime factors by starting with different known factors: \n\t\n84\t = 4 × 21 \t\nor\t\n84 \t= 7 × 12\t\nor\t\n84 \t= 2 × 42\n\t\n\t = 2 × 2 × 3 × 7 \t\n\t\n\t\n= 7 × 3 × 4\t\n\t\n\t\n= 2 × 6 × 7\n\t\n\t\t\n\t\n\t\n= 7 × 3 × 2 × 2\t\n\t\n\t\n= 2 × 2 × 3 × 7\nEvery composite number can \nbe expressed as the product \nof prime factors and this can \nhappen in only one way.\nMaths_LB_gr8_bookA.indb   20\n2014/07/03   10:53:15 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n21\nA more systematic way of finding the prime factors of a number would be to start \nwith the prime numbers and try the consecutive prime numbers 2; 3; 5; 7; ... as possible \nfactors. The work may be set out as shown below.\n2\n1 430\n3\n2 457\n5\n715\n3\n819\n11\n143\n3\n273\n13\n13\n7\n91\n1\n13\n13\n1\n\t\n1 430 = 2 × 5 × 11 × 13\t\n2 457 = 3 × 3 × 3 × 7 × 13\nWe can use exponents to write the products of prime factors more compactly as  \nproducts of powers of prime factors.\n\t\n2 457\t =    3 × 3 × 3 × 7 × 13\t\n=  33 × 7 × 13\n\t\n      72\t =     2 × 2 × 2 × 3 × 3\t\n=  23 × 32\n\t\n1 500\t =  2 × 2 × 3 × 5 × 5 × 5\t =  22 × 3 × 53\n1.\t Express the following numbers as the product of powers of primes:\n\t\n(a)\t 792 = \n\t\n\t\n(b)\t 444 = \n2.\t Find the prime factors of the numbers below.\n2\n28\n32\n124\n36\n42\n345\n182\n14\nMaths_LB_gr8_bookA.indb   21\n2014/07/03   10:53:15 AM\n\n22\t\nMATHEMATICS Grade 8: Term 1\ncommon multiples and factors\n1.\t Is 4 × 5 a multiple of 4? \n\t    Is 4 × 5 a multiple of 5? \n2.\t Comment on the following statement:  \nThe product of numbers is a multiple of each of the numbers in the product.\nWe use common multiples when fractions with different denominators are added. \nTo add 2\n3  + 3\n4 , the common denominator is 3 × 4, so the sum becomes 8\n12  + 9\n12 .\nIn the same way, we could use 6 × 8 = 48 as a common denominator to add 1\n6  + 3\n8 , but \n24 is the lowest common multiple (LCM) of 6 and 8. \nPrime factorisation makes it easy to find the lowest common multiple or highest \ncommon factor.\nWhen we simplify a fraction, we divide the same  \nnumber into the numerator and the denominator. For  \nthe simplest fraction, use the highest common  \nfactor (HCF) to divide into both numerator and  \ndenominator.\n\t\n\t\nSo 36\n144  = \n2\n2\n3\n3\n2\n2\n2\n2\n3\n3\n×\n×\n×\n×\n×\n×\n×\n×\n = 1\n4\nUse prime factorisation to determine the LCM and HCF of 32, 48 and 84 in a systematic \nway:\t \t\n\t\n32 = 2 × 2 × 2 × 2 × 2\t= 25\n\t\n\t\n\t\n48 = 2 × 2 × 2 × 2 × 3\t= 24 × 3\n\t\n\t\n\t\n84 = 2 × 2 × 3 × 7\t\n= 22 × 3 × 7\nThe LCM is a multiple, so all of the factors of all the numbers must divide into it.\nAll of the factors that are present in the three numbers must also be factors of the \nLCM, even if it is a factor of only one of the numbers. But because it has to be the lowest \ncommon multiple, no unnecessary factors are in the LCM.\nThe highest power of each factor is in the LCM, because then all of the other factors \ncan divide into it. In 32, 48 and 84, the highest power of 2 is 25, the highest power of 3 is \n3 and the highest power of 7 is 7. \nLCM = 25 × 3 × 7 = 672\nThe HCF is a common factor. Therefore, for a factor to be in the HCF, it must be a factor \nof all of the numbers. 2 is the only number that appears as a factor of all three numbers. \nThe lowest power of 2 is 22, so the HCF is 22.\nThe HCF is divided into the  \nnumerator and the denominator \nto write the fraction in its  \nsimplest form.\nMaths_LB_gr8_bookA.indb   22\n2014/07/03   10:53:17 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n23\n3.\t Determine the LCM and the HCF of the numbers in each case.\n\t\n(a)\t 24; 28; 42\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 17; 21; 35\n\t\n(c)\t 75; 120; 200\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 18; 30; 45\ninvestigate prime numbers\nYou may use a calculator for this investigation.\n1.\t Find all the prime numbers between 110 and 130.\n\t\n\t\n2.\t Find all the prime numbers between 210 and 230.\n\t\n\t\n\t\n\t\n3.\t Find the biggest prime number smaller than 1 000.\n\t\nMaths_LB_gr8_bookA.indb   23\n2014/07/03   10:53:17 AM\n\n24\t\nMATHEMATICS Grade 8: Term 1\n1.4\t Solving problems\nrate and ratio\nYou may use a calculator for the work in this section.\n1.\t Tree plantations in the Western Cape are to be cut down in favour of natural \nvegetation. There are roughly 3 000 000 trees on plantations in the area and it is \npossible to cut them down at a rate of 15 000 trees per day with the labour available. \nHow many working days will it take before all the trees will be cut down?\nInstead of saying “… per day”, people often say “at a \nrate of … per day”. Speed is a way to describe the rate \nof movement.\n \n2.\t A car travels a distance of 180 km in 2 hours on a straight road. How many kilometres \ncan it travel in 3 hours at the same speed? \n3.\t Thobeka wants to order a book that costs $56,67. The rand-dollar exchange rate is \nR7,90 to a dollar. What is the price of the book in rands?\n4.\t In pattern A below, there are 5 red beads for every 4 yellow beads.\nPattern A\nPattern B\nPattern C\n\t\nDescribe patterns B and C in the same way.\nThe word per is often used \nto describe a rate and can \nmean for every, for, in each, \nin, out of, or every. \nMaths_LB_gr8_bookA.indb   24\n2014/07/03   10:53:17 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n25\n5.\t Complete the table to show how many screws are produced by two machines in \ndifferent periods of time.\nNumber of hours\n1\n2\n3\n5\n8\nNumber of screws at machine A\n1 800\nNumber of screws at machine B\n2 700\n\t\n\t\n(a)\t How much faster is machine B than machine A? \n\t\n(b)\t How many screws will machine B produce in the same time that it takes  \n    machine A to make 100 screws?\nThe patterns in question 4 can be described like this: In pattern A, the ratio of yellow \nbeads to red beads is 4 to 5. This is written as 4 : 5. In pattern B, the ratio between yellow \nbeads and red beads is 3 : 6. In pattern C the ratio is 2 : 7. \nIn question 5, machine A produces 2 screws for every 3 screws that machine B  \nproduces. This can be described by saying that the ratio between the production speeds \nof machines A and B is 2 : 3.\n6.\t Nathi, Paul and Tim worked in Mr Setati’s garden. Nathi worked for 5 hours, Paul for \n4 hours and Tim for 3 hours. Mr Setati gave the boys R600 for their work. How should \nthey divide the R600 among the three of them? \nA ratio is a comparison of two (or more) quantities.  \nThe number of hours that Nathi, Paul and Tim  \nworked are in the ratio 5 : 4 : 3. To be fair, the money  \nshould also be shared in that ratio. That means that Nathi should receive 5 parts, Paul  \n4 parts and Tim 3 parts of the money. There were 12 parts, which means Nathi should \nreceive 5\n12  of the total amount, Paul should get 4\n12  and Tim should get 3\n12.\n7.\t Ntabi uses 3 packets of jelly to make a pudding for 8 people. How many packets of \njelly does she need to make a pudding for 16 people? And for 12 people?\nWe use ratios to show how \nmany times more, or less, \none quantity is than another.\nMaths_LB_gr8_bookA.indb   25\n2014/07/03   10:53:18 AM\n\n26\t\nMATHEMATICS Grade 8: Term 1\n8.\t Which rectangle is more like a square: a 3 × 5 rectangle or a 6 × 8 rectangle? Explain.\nTo increase 40 in the ratio 2 : 3 means that the 40 \nrepresents two parts and must be increased so that \nthe new number represents 3 parts. If 40 represents \ntwo parts, 20 represents 1 part. The increased number \nwill therefore be 20 × 3 = 60.\n9.\t (a)\t Increase 56 in the ratio 2 : 3.\n\t\n(b)\t Decrease 72 in the ratio 4 : 3.\n10.\t(a)\t Divide 840 in the ratio 3 : 4.\n\t\n(b)\t Divide 360 in the ratio 1 : 2 : 3.\n11.\tSome data about the performance of different athletes during a walking event is \ngiven below. Investigate the data to find out who walks fastest and who walks slowest. \nArrange the athletes from the fastest walker to the slowest walker. \n\t\n(a)\t First make estimates to do the investigation.\n\t\n(b)\t Then use your calculator to do the investigation.\n\t\nAthlete\nA\nB\nC\nD\nE\nF\nDistance walked in m\n2 480\n4 283\n3 729\n6 209\n3 112\n5 638\nTime taken in minutes\n17\n43\n28\n53\n24\n45\nRemember that if you  \nmultiply by 1, the number \ndoes not change. If you \nmultiply by a number \ngreater than 1, the number \nincreases. If you multiply by \na number smaller than 1, the \nnumber decreases.\nMaths_LB_gr8_bookA.indb   26\n2014/07/03   10:53:18 AM\n\n\t\nCHAPTER 1: WHOLE NUMBERS\t\n27\nprofit, loss, discount and interest\n1.\t (a)\t How much is 1 eighth of R800? \n\t\n(b)\t How much is 1 hundredth of R800? \n\t\n(b)\t How much is 7 hundredths of R800? \nRashid is a furniture dealer. He buys a couch for R2 420. He displays the couch in \nhis showroom with the price marked as R3 200. Rashid offers a discount of R320 to \ncustomers who pay cash.\nThe amount for which a dealer buys an article from \na producer or manufacturer is called the cost price. \nThe price marked on the article is called the marked \nprice and the price of the article after discount is the \nselling price.\n2.\t (a)\t What is the cost price of the couch in Rashid’s furniture shop? \n\t\n(b)\t What is the marked price? \n\t\n(c)\t What is the selling price for a customer who pays cash? \n\t\n(d)\t How much is 10 hundredths of R3 200? \nThe discount on an article is always less than the  \nmarked price of the article. In fact, it is only a fraction  \nof the marked price. The discount of R320 that Rashid  \noffers on the couch is 10 hundredths of the marked price. \nAnother word for hundredths is percentage, and  \nthe symbol for percentage is %. So we can say that  \nRashid offers a discount of 10%.\nA percentage is a number of hundredths.\n18% is 18 hundredths, and 25% is 25 hundredths.\nA discount of 6% on an article can be calculated in two steps:\nStep 1: Calculate 1 hundredth of the marked price (divide by 100).\nStep 2: Calculate 6 hundredths of the marked price (multiply by 6).\n3.\t Calculate a discount of 6% on each of the following marked prices of articles:\n\t\n(a)\t R3 600\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t R9 360\n% is a symbol for hundredths.  \n8% means 8 hundredths and \n15% means 15 hundredths.\nThe symbol % is just a\nvariation of the 100  that is \nused in the common fraction \nnotation for hundredths.\n8% is 8\n100 .\nMaths_LB_gr8_bookA.indb   27\n2014/07/03   10:53:18 AM\n\n28\t\nMATHEMATICS Grade 8: Term 1\n4.\t (a)\t How much is 1 hundredth of R700? \n\t\n(b)\t A customer pays cash for a coat marked at R700.  \n\t\nHe is given R63 discount. How many hundredths of R700 is this? \n\t\n(c)\t What is the percentage discount? \n5.\t A client buys a blouse marked at R300 and she is given R36 discount for paying cash. \nWork as in question 4 to determine what percentage discount she was given.\nYou may use a calculator to do questions 6, 7 and 8.\n6.\t A dealer buys an article for R7 500 and makes the price 30% higher. The article is sold \nat a 20% discount.\n\t\n(a)\t What is the selling price of the article?\n\t\n(b)\t What is the dealer’s percentage profit?\nWhen a person borrows money from a bank or some other institution, he or she \nnormally has to pay for the use of the money. This is called interest.\n7.\t Sam borrows R7 000 from a bank at 14% interest for one year. How much does he \nhave to pay back to the bank at the end of the period?\n8.\t Jabu invests R5 600 for one year at 8% interest. \n\t\n(a)\t What will the value of his investment be at the end of that year?\n\t\n(b)\t At the end of the year Jabu does not withdraw the investment or the interest \n    earned, but reinvests it for another year. How much will it be worth at the end  \n\t\nof the second year?\n\t\n(c)\t What will the value of Jabu’s investment be after five years?\t\nMaths_LB_gr8_bookA.indb   28\n2014/07/03   10:53:19 AM\n\nChapter 2\nIntegers\n\t\nCHAPTER 2: INTEGERS\t\n29\nIn this chapter you will work with whole numbers smaller than 0. These numbers are  \ncalled negative numbers. The whole numbers larger than 0, 0 itself and the negative  \nwhole numbers together are called the integers. Mathematicians have agreed that  \nnegative numbers should have certain properties that would make them useful for various \npurposes. You will learn about these properties and how they make it possible to do  \ncalculations with negative numbers. \n2.1\t What is beyond 0?.................................................................................................... 31\n2.2\t Adding and subtracting with integers....................................................................... 35\n2.3\t Multiplying and dividing with integers...................................................................... 40\n2.4\t Squares, cubes and roots with integers..................................................................... 47\nDo what you can.\n  5 − 0 = ?\n5 − 7 = ?\n5 + 5 = ?\n  5 − 1 = ?\n5 − 6 = ?\n5 + 4 = ?\n  5 − 2 = ?\n5 − 5 = ?\n5 + 3 = ?\n  5 − 3 = ?\n5 − 4 = ?\n5 + 2 = ?\n  5 − 4 = ?\n5 − 3 = ?\n5 + 1 = ?\n  5 − 5 = ?\n5 − 2 = ?\n5 + 0 = ?\n  5 − 6 = ?\n5 − 1 = ?\n5 + ? = ?\n  5 − 7 = ?\n5 − 0 = ?\n5 + ? = ?\n  5 − 8 = ?\n5 − ? = ?\n5 + ? = ?\n  5 − 9 = ?\n5 −? = ?\n5 + ? = ?\n5 − 10 = ?\n5 − ? = ?\n5 + ? = ?\nMaths_LB_gr8_bookA.indb   29\n2014/07/03   10:53:19 AM\n\n30\t\nMATHEMATICS Grade 8: Term 1\nChoose a good plan to complete this table.\n  Plan A: Look at where the number 4 appears in the table. All the 4s lie on a diagonal,  \n  going down from left to right. Complete the other diagonals in the same way.\n  Plan B: Look at the number 13 in the table. It is on the right, in the fourth row from \n  the top. It can be obtained by adding the two numbers indicated by arrows. Complete \n  all the cells by adding numbers from the yellow column and blue row in this way.\n  Plan C: In each row, add 1 to go right and subtract 1 to go left.\n  Plan D: In each column, add 1 to go up and subtract 1 to go down.\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n4\n7\n4\n6\n–3\n–2\n–1\n0\n4\n5\n13\n4\n–5\n–4\n3\n4\n2\n4\n1\n4\n–8\n–7\n–6\n–5\n–4\n–3\n–2\n–1\n0\n1\n2\n3\n4\n5\n6\n7\n8\n–1\n4\n–2\n4\n–3\n4\n–4\n4\n–5\n–6\n–7\n0\n–8\n0\nMaths_LB_gr8_bookA.indb   30\n2014/07/03   10:53:19 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n31",
      "chapter_id": "1"
    },
    {
      "title": "Integers",
      "content": "2\t Integers\n2.1\t What is beyond 0?\nwhy people decided to have negative numbers\nOn the right, you can see how Jimmy prefers to work  \nwhen doing calculations such as 542 + 253.\nHe tries to calculate 542 − 253 in a similar way:\n\t\n500 − 200\t =\t 300\n\t\n     40 − 50\t =\t ?\nJimmy clearly has a problem. He reasons as follows:\nI can subtract 40 from 40; that gives 0. But then there is still 10 that I have to subtract.\nHe decides to deal with the 10 that he still has to subtract later, and continues:\n\t\n500 − 200\t =\t 300\n\t\n40 − 50\t =\t 0, but there is still 10 that I have to subtract.\n\t\n2 − 3\t =\t 0, but there is still 1 that I have to subtract.\n1.\t (a)\t What must Jimmy still subtract, and what will his final answer be?\n\t\n(b)\t When Jimmy did another subtraction problem, he ended up with this writing at \n    one stage:\n\t\n\t\n\t\n\t\n600   and   (−)50   and   (−)7\n\t\n\t\nWhat do you think is Jimmy’s final answer for this subtraction problem?\nAbout 500 years ago, some mathematicians proposed that a “negative number” may be \nused to describe the result in a situation such as in Jimmy’s subtraction problem above, \nwhere a number is subtracted from a number smaller than itself.\nFor example, we may say 10 − 20 = (−10)\nThis proposal was soon accepted by other  \nmathematicians, and it is now used all over the world.\n500 + 200 = 700\n40 + 50 = 90\n2 + 3 = 5\n700 + 90 + 5 = 795\nMathematicians are people \nwho do mathematics for a \nliving. Mathematics is their \nprofession, like health care is \nthe profession of nurses and \nmedical doctors.\nMaths_LB_gr8_bookA.indb   31\n2014/07/03   10:53:19 AM\n\n32\t\nMATHEMATICS Grade 8: Term 1\n2.\t Calculate each of the following:\n\t\n(a)\t 16 − 20 \t\n\t\n\t\n\t\n(b)\t 16 − 30  \t\n\t\n(c)\t 16 − 40 \t\n\t\n\t\n\t\n(d)\t 16 − 60  \t\n\t\n(e)\t 16 − 200 \t\n\t\n\t\n\t\n(f)\t 5 − 1 000\t\n3.\t Some numbers are shown on the lines below. Fill in the missing numbers.\n–9\n–5\n–4\n–2\n–1\n0\n1\n2\n3\n4\n9\n–10\n–5\n0\n5\n10\n15\nThe following statement is true if the number is 5: \n\t\n\t\n15 − (a certain number) = 10 \nA few centuries ago, some mathematicians decided  \nthey wanted to have numbers that will also make  \nsentences like the following true:\n\t\n\t\n15 + (a certain number) = 10 \nBut to go from 15 to 10 you have to subtract 5.\nThe number we need to make the sentence 15 + (a certain number) = 10 true must have \nthe following strange property:\nIf you add this number, it should have the same effect as to subtract 5.\nNow the mathematicians of a few centuries ago really wanted to have numbers for which \nsuch strange sentences would be true. So they thought:\n\t\n\t\nLet us decide, and agree amongst ourselves, that the number we call negative 5 \t\n\t\n\t\nwill have the property that if you add it to another number, the effect will be the \t\n\t\n\t\nsame as when you subtract the natural number 5.\nThis means that the mathematicians agreed that 15 + (−5) is equal to 15 − 5.\nStated differently, instead of adding negative 5 to a number, you may subtract 5.\nAdding a negative number has the same effect as \nsubtracting a natural number. \nFor example: 20 + (−15) = 20 − 15 = 5\n4.\t Calculate each of the following:\n\t\n(a)\t 500 + (−300) \t \t\n\t\n\t\n(b)\t 100 + (−20) + (−40) \t\n\t\n(c)\t 500 + (−200) + (−100) \t\n\t\n(d)\t 100 + (−60) \t\n\t\n\t\nThe numbers 1; 2; 3; 4 etc. \nare called the natural  \nnumbers. The natural \nnumbers, 0 and the negative \nwhole numbers together are \ncalled the integers.\nMaths_LB_gr8_bookA.indb   32\n2014/07/03   10:53:19 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n33\n5.\t Make a suggestion of what the answer for (−20) + (−40) should be. Give reasons for \nyour suggestion.\n6.\t Continue the lists of numbers below to complete the table.\n(a)\n(b)\n(c)\n(d)\n(e)\n(f)\n(g)\n10\n100\n3\n−3\n−20\n150\n0\n9\n90\n6\n−6\n−18\n125\n−5\n8\n80\n9\n−9\n−16\n100\n−10\n7\n70\n12\n−12\n−14\n75\n−15\n6\n60\n15\n−15\n50\n−20\n5\n50\n−25\n4\n40\n3\n30\n2\n20\n1\n10\n−1\n−10\nThe following statement is true if the number is 5: \n\t\n\t\n15 + (a certain number) = 20 \nWhat properties should a number have so that it makes the following statement true?\n\t\n\t\n15 − (a certain number) = 20 \nTo go from 15 to 20 you have to add 5. The number we need to make the sentence  \n\t\n\t\n15 − (a certain number) = 20 true must have the following property:\nIf you subtract this number, it should have the same effect as to add 5.\nLet us agree that 15 − (−5) is equal to 15 + 5. \nStated differently, instead of subtracting negative 5 from a number, you may add 5.\nMaths_LB_gr8_bookA.indb   33\n2014/07/03   10:53:19 AM\n\n34\t\nMATHEMATICS Grade 8: Term 1\nSubtracting a negative number has the same effect as \nadding a natural number. \nFor example: 20 − (−15) = 20 + 15 = 35\n7.\t Calculate.\n\t\n(a)\t 30 − (−10)\t \t\n\t\n\t\n\t\n(b)\t 30 + 10\t\t\n\t\n(c)\t 30 + (−10)\t \t\n\t\n\t\n\t\n(d)\t 30 − 10\t\t\n\t\n(e)\t 30 − (−30)\t \t\n\t\n\t\n\t\n(f)\t 30 + 30\t\t\n\t\n(g)\t 30 + (−30)\t \t\n\t\n\t\n\t\n(h)\t30 − 30\t\t\nYou probably agree that \n5 + (−5) = 0        10 + (−10) = 0    and    20 + (−20) = 0\nWe may say that for each “positive” number there is a corresponding or opposite \nnegative number. Two positive and negative numbers that correspond, for example 3 \nand (−3), are called additive inverses. They wipe each other out when you add them.\nWhen you add any number to its additive \ninverse, the answer is 0 (the additive \nproperty of 0). For example, 120 + (−120) = 0.\n8.\t Write the additive inverse of each of the  \n\t\nfollowing numbers:\n\t\n(a)\t 24\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t −24\t\n\t\n(c)\t −103\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 2 348\t\nThe idea of additive inverses may be used to explain why 8 + (−5) is equal to 3:\n8 + (−5) = 3 + 5 + (−5) = 3 + 0 = 3\n9.\t Use the idea of additive inverses to explain why each of these statements is true:\n\t\n(a)\t 43 + (−30) = 13\t\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 150 + (−80) = 70\nSTATEMENTS THAT ARE TRUE FOR MANY DIFFERENT NUMBERS\nFor how many different pairs of numbers can the following statement be true, if only natural \n(positive) numbers are allowed?  \na number + another number = 10\nFor how many different pairs of numbers can the statement be true if negative numbers are also \nallowed?  \nWhat may each of the  \nfollowing be equal to?\n  (−8) + 5 \n  (−5) + (−8) \nMaths_LB_gr8_bookA.indb   34\n2014/07/03   10:53:19 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n35\n2.2\t Adding and subtracting with integers\nadding can make less and subtraction can make more\n1.\t Calculate each of the following:\n\t\n(a)\t 10 + 4 + (−4)\t\n\t\n\t\n\t\n(b)\t 10 + (−4) + 4\t\n\t\n(c)\t 3 + 8 + (−8)\t\t\n\t\n\t\n\t\n(d)\t 3 + (−8) + 8\t\t\nNatural numbers can be arranged in any order to add  \nand subtract them. This is also the case for integers.\n2.\t Calculate each of the following:\n\t\n(a)\t 18 + 12 \t\n\t\n(b)\t 12 + 18\t\n\t\n(c)\t 2 + 4 + 6 \t\n\t\n(d)\t 6 + 4 + 2 \t\n\t\n(e)\t 2 + 6 + 4 \t\n\t\n(f)\t 4 + 2 + 6 \t\n\t\n(g)\t 4 + 6 + 2 \t\n\t\n(h)\t 6 + 2 + 4 \t\n\t\n(i)\t 6 + (−2) + 4 \t\n\t\n(j)\t 4 + 6 + (−2) \t\n\t\n(k)\t 4 + (−2) + 6 \t\n\t\n(l)\t (−2) + 4 + 6 \t\n\t\n(m)\t6 + 4 + (−2) \t\n\t\n(n)\t (−2) + 6 + 4 \t\n\t\n(o)\t (−6) + 4 + 2 \t\n3.\t Calculate each of the following:\n\t\n(a)\t (−5) + 10\t\n\t\n\t\n\t\n\t\n(b)\t 10 + (−5)\t\n\t\n\t\n\t\n(c)\t (−8) + 20\t\n\t\n\t\n\t\n\t\n(d)\t 20 − 8\t \t\n\t\n\t\n\t\n(e)\t 30 + (−10)\t \t\n\t\n\t\n\t\n(f)\t 30 + (−20)\t \t\n\t\n\t\n(g)\t 30 + (−30)\t \t\n\t\n\t\n\t\n(h)\t10 + (−5) + (−3)\t\n\t\n(i)\t (−5) + 7 + (−3) + 5\t\n\t\n\t\n(j)\t (−5) + 2 + (−7) + 4\t\n4.\t In each case, find the number that makes the  \nstatement true. Give your answer by writing a  \nclosed number sentence.\n\t\n\t\n(a)\t 20 + (an unknown number) = 50\n\t\n(b)\t 50 + (an unknown number) = 20\n\t\n(c)\t 20 + (an unknown number) = 10\nThe numbers 1; 2; 3; 4; etc. \nthat we use to count, are \ncalled natural numbers.\nStatements like these are also \ncalled number sentences.\nAn incomplete number  \nsentence, where some numbers \nare not known at first, is  \nsometimes called an open \nnumber sentence:\n8 − (a number) = 10\nA closed number sentence \nis where all the numbers are \nknown:\n8 + 2 = 10\nMaths_LB_gr8_bookA.indb   35\n2014/07/03   10:53:19 AM\n\n36\t\nMATHEMATICS Grade 8: Term 1\n\t\n(d)\t (an unknown number) + (−25) = 50\n\t\n(e)\t (an unknown number) + (−25) = −50\n5.\t Use the idea of additive inverses to explain why each of the following statements  \nis true:\n\t\n(a)\t 43 + (−50) = −7\t\t\n\t\n(b)\t 60 + (−85) = −25\t\n6.\t Complete the table as far as you can.\n(a)\n(b)\n(c)\n5 − 8 = \n5 + 8 = \n8 − 3 = \n5 − 7 = \n5 + 7 = \n7 − 3 = \n5 − 6 = \n5 + 6 = \n6 − 3 = \n5 − 5 = \n5 + 5 = \n5 − 3 = \n5 − 4 = \n5 + 4 = \n4 − 3 = \n5 − 3 = \n5 + 3 = \n3 − 3 = \n5 − 2 = \n5 + 2 = \n2 − 3 = \n5 − 1 = \n5 + 1 = \n1 − 3 = \n5 − 0 = \n5 + 0 = \n0 − 3 = \n5 − (−1) = \n5 + (−1) = \n(−1) − 3 = \n5 − (−2) = \n5 + (−2) = \n(−2) − 3 = \n5 − (−3) = \n5 + (−3) = \n(−3) − 3 = \n5 − (−4) = \n5 + (−4) = \n(−4) − 3 = \n5 − (−5) = \n5 + (−5) = \n(−5) − 3 = \n5 − (−6) = \n5 + (−6) = \n(−6) − 3 = \nMaths_LB_gr8_bookA.indb   36\n2014/07/03   10:53:19 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n37\n7.\t Calculate.\n\t\n(a)\t 80 + (−60)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 500 + (−200) + (−200)\n8.\t (a)\t Is 100 + (−20) + (−20) = 60, or does it equal something else? \n\t\n(b)\t What do you think (−20) + (−20) is equal to? \n9.\t Calculate.\n\t\n(a)\t 20 − 20\t\t\n\t\n\t\n\t\n\t\n(b)\t 50 – 20\t\t\n\t\n\t\n\t\n(c)\t (−20) − (−20)\t\n\t\n\t\n\t\n(d)\t (−50) − (−20)\t\n\t\n10.\tCalculate.\n\t\n(a)\t 20 − (−10)\t \t\n\t\n\t\n\t\n(b)\t 100 − (−100)\t\n\t\n\t\n(c)\t 20 + (−10)\t \t\n\t\n\t\n\t\n(d)\t 100 + (−100)\t\n\t\n\t\n(e)\t (−20) − (−10)\t\n\t\n\t\n\t\n(f)\t (−100) − (−100)\t\n\t\n(g)\t (−20) + (−10)\t\n\t\n\t\n\t\n(h)\t(−100) + (−100)\t\n11.\tComplete the table as far as you can.\n(a)\n(b)\n(c)\n5 − (−8) =\n(−5) + 8 = \n8 − (−3)  = \n5 − (−7) = \n(−5) + 7 = \n7 − (−3)  = \n5 − (−6) = \n(−5) + 6 = \n6 − (−3) = \n5 − (−5) = \n(−5) + 5 = \n5 − (−3) = \n5 − (−4) = \n(−5) + 4 = \n4 − (−3) = \n5 − (−3) = \n(−5) + 3 = \n3 − (−3) = \n5 − (−2) = \n(−5) + 2 = \n2 − (−3) = \n5 − (−1) = \n(−5) + 1 = \n1 − (−3) = \n5 − 0 = \n(−5) + 0 = \n0 − (−3) = \n5 − 1 = \n(−5) + (−1) = \n(−1) − (−3) = \n5 − 2 = \n(−5) + (−2) = \n(−2) − (−3) = \n5 − 3 = \n(−5) + (−3) = \n(−3) − (−3) = \n5 − 4 = \n(−5) + (−4) = \n(−4) − (−3) = \n5 − 5 = \n(−5) + (−5) = −\n(−5) − (−3) = \nMaths_LB_gr8_bookA.indb   37\n2014/07/03   10:53:20 AM\n\n38\t\nMATHEMATICS Grade 8: Term 1\n12.\tIn each case, state whether the statement is true or false and give a numerical \nexample to demonstrate your answer.\n\t\n(a)\t Subtracting a positive number from a negative number has the same effect as \n    adding the additive inverse of the positive number.\n\t\n(b)\t Adding a negative number to a positive number has the same effect as adding  \n    the additive inverse of the negative number.\n\t\n(c)\t Subtracting a negative number from a positive number has the same effect as  \n    subtracting the additive inverse of the negative number.\n\t\n(d)\t Adding a negative number to a positive number has the same effect as subtracting \n    the additive inverse of the negative number.\n\t\n(e)\t Adding a positive number to a negative number has the same effect as adding the \n    additive inverse of the positive number.\n\t\n(f)\t Adding a positive number to a negative number has the same effect as subtracting \n    the additive inverse of the positive number.\n\t\n(g)\t Subtracting a positive number from a negative number has the same effect as  \n    subtracting the additive inverse of the positive number.\n\t\n(h)\t Subtracting a negative number from a positive number has the same effect as \n    adding the additive inverse of the negative number.\nMaths_LB_gr8_bookA.indb   38\n2014/07/03   10:53:20 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n39\ncomparing integers and solving problems\n1.\t Fill <, > or = into the block to make the relationship between the numbers true:\n\t\n(a)\t −103 \n  −99\n\t\n(b)\t −699 \n  −701\n\t\n(c)\t 30 \n  −30\n\t\n(d)\t 10 − 7 \n  −(10 − 7)\n\t\n(e)\t −121 \n  −200\n\t\n(f)\t 12 − 5 \n  −(12 + 5)\n\t\n(g)\t −199 \n  −110\n2.\t At 5 a.m. in Bloemfontein the temperature was −5 °C. At 1 p.m., it was 19 °C. By how \nmany degrees did the temperature rise?\n3.\t A diver swims 150 m below the surface of the sea. She moves 75 m towards the \nsurface. How far below the surface is she now?\n4.\t One trench in the ocean is 800 m deep and another is 2 200 m deep. What is the \ndifference in their depths?\n5.\t An island has a mountain which is 1 200 m high. The surrounding ocean has a depth \nof 860 m. What is the difference in height?\n6.\t On a winter’s day in Upington the temperature rose by 19 °C. If the minimum \ntemperature was −4 °C, what was the maximum temperature?\nMaths_LB_gr8_bookA.indb   39\n2014/07/03   10:53:20 AM\n\n40\t\nMATHEMATICS Grade 8: Term 1\n2.3\t Multiplying and dividing with integers\nmultiplication with integers\n1.\t Calculate.\n\t\n(a)\t −5 + −5 + −5 + −5 + −5 + −5 + −5 + −5 + −5 + −5\t\t\n\t\n(b)\t −10 + −10 + −10 + −10 + −10 \t\n\t\n\t\n\t\n\t\n\t\n\t\n(c)\t −6 + −6 + −6 + −6 + −6 + −6 + −6 + −6 \t \t\n\t\n\t\n\t\n(d)\t −8 + −8 + −8 + −8 + −8 + −8 \t\n\t\n\t\n\t\n\t\n\t\n\t\n(e)\t −20 + −20 + −20 + −20 + −20 + −20 + −20 \t \t\n\t\n2.\t In each case, show whether you agree (✓) or disagree (✗) with the given statement.\n\t\n(a)\t 10 × (−5) = 50\t \t\n\t\n\t\n\t\n(b)\t 8 × (−6) = (−8) × 6\t\n\t\n(c)\t (−5) × 10 = 5 × (−10)\t\n\t\n\t\n(d)\t 6 × (−8) = −48\t \t\n\t\n(e)\t (−5) × 10 = 10 × (−5) \t\n\t\n\t\n(f)\t 8 × (−6) = 48\t\n\t\n\t\n(g)\t 4 × 12 = −48\t\n\t\n\t\n\t\n\t\n(h)\t(−4) × 12 = −48\t\t\nMultiplication of integers is commutative:\n(−20) × 5 = 5 × (−20)\n3.\t Is addition of integers commutative?  \nDemonstrate your answer with three different examples.\n4.\t Calculate.\n\t\n(a)\t 20 × (−10) \t \t\n\t\n\t\n(b)\t (−5) × 4 \t\n\t\n\t\n(c)\t (−20) × 10 \t \t\n\t\n\t\n(d)\t 4 × (−25) \t \t\n\t\n(e)\t 29 × (−20) \t\t\n\t\n\t\n(f)\t (−29) × (−2) \t\n5.\t Calculate.\n\t\n(a)\t 10 × 50 + 10 × (−30)\t\n\t\n\t\n\t\n\t\n(b)\t 50 + (−30)\n\t\n(c)\t 10 × (50 + (−30))\t\n\t\n\t\n\t\n\t\n\t\n(d)\t (−50) + (−30)\n\t\n(e)\t 10 × (−50) + 10 × (−30)\t\t\n\t\n\t\n\t\n(f)\t 10 × ((−50) + (−30))\nMaths_LB_gr8_bookA.indb   40\n2014/07/03   10:53:20 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n41\nThe product of two positive numbers is a positive \nnumber, for example 5 × 6 = 30.\nThe product of a positive number and a negative \nnumber is a negative number, for example  \n5 × (−6) = −30.\nThe product of a negative number and a positive \nnumber is a negative number, for example  \n(−5) × 6 = −30.\n6.\t (a)\t Four numerical expressions are given below. Underline the expressions that you \n    would expect to have the same answers. Do not do the calculations.\n\t\n\t\n14 × (23 + 58)      23 × (14 + 58)      14 × 23 + 14 × 58      14 × 23 + 58\n\t\n(b)\t What property of operations is demonstrated by the fact that two of the above \n    expressions have the same value?\n7.\t Consider your answers for question 6.\n\t\n(a)\t Does multiplication distribute over addition in the case of integers? \n\t\n(b)\t Illustrate your answer with two examples.\n8.\t Three numerical expressions are given below. Underline the expressions that you \nwould expect to have the same answers. Do not do the calculations.\n\t\n10 × ((−50) − (−30))\t\n\t\n10 × (−50) − (−30)\t\n\t\n10 × (−50) − 10 × (−30)\n9.\t Do the three sets of calculations given in question 8.\nYour work in questions 5, 8 and 9 demonstrates that multiplication with a positive \nnumber distributes over addition and subtraction of integers. For example: \n\t\n10 × (5 + (−3)) = 10 × 2 = 20 and 10 × 5 + 10 × (−3) = 50 + (−30) = 20\n\t\n10 × (5 − (−3)) = 10 × 8 = 80 and 10 × 5 − 10 × (−3) = 50 − (−30) = 80\n10.\tCalculate: (−10) × (5 + (−3)) \nMaths_LB_gr8_bookA.indb   41\n2014/07/03   10:53:20 AM\n\n42\t\nMATHEMATICS Grade 8: Term 1\nNow consider the question of whether multiplication with a negative number  \ndistributes over addition and subtraction of integers. For example, would  \n(−10) × 5 + (−10) × (−3) also have the answer −20, as does (−10) × (5 + (−3))?\t\n11.\tWhat must (−10) × (−3) be equal to, if we want (−10) × 5 + (−10) × (−3) to be equal  \nto −20?\nIn order to ensure that multiplication distributes over addition and subtraction in the \nsystem of integers, we have to agree that \n\t\n(a negative number) × (a negative number) is a positive number, \nfor example (−10) × (−3) = 30.\n12.\tCalculate.\n\t\n(a)\t (−10) × (−5)\t\n\t\n\t\n(b)\t (−10) × 5\t\n\t\n(c)\t 10 × 5  \t\t\n\t\n\t\n\t\n(d)\t 10 × (−5)\t\n\t\n(e)\t (−20) × (−10) + (−20) × (−6)\t\n\t\n\t\n(f)\t (−20) × ((−10) + (−6))\n\t\n(g)\t (−20) × (−10) − (−20) × (−6)\t\n\t\n\t\n(h)\t(−20) × ((−10) − (−6))\nHere is a summary of the properties of integers that make it possible to do \ncalculations with integers:\n• When a number is added to its additive inverse, the \nresult is 0, for example (+12) + (−12) = 0.\n• Adding an integer has the same effect as \nsubtracting its additive inverse. For example, \n3 + (−10) can be calculated by doing 3 − 10,  \nand the answer is −7.\n• Subtracting an integer has the same effect as \nadding its additive inverse. For example, 3 − (−10) \ncan be calculated by doing 3 + 10, and the answer  \nis 13.\n• The product of a positive and a negative integer is \nnegative, for example (−15) × 6 = −90.\n• The product of a negative and a negative integer is \npositive, for example (−15) × (−6) = 90.\nMaths_LB_gr8_bookA.indb   42\n2014/07/03   10:53:20 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n43\ndivision with integers\n1.\t (a)\t Calculate 25 × 8. \n\t\n(b)\t How much is 200 ÷ 25? \n\t\n(c)\t How much is 200 ÷ 8? \nDivision is the inverse of multiplication. Hence, \nif two numbers and the value of their product are \nknown, the answers to two division problems are  \nalso known.\n2.\t Calculate.\n\t\n(a)\t 25 × (−8)\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t (−125) × 8\n3.\t Use the work you have done for question 2 to write the answers for the following \ndivision questions:\n\t\n(a)\t (−1 000) ÷ (−125)\t\n\t\n\t\n(b)\t (−1 000) ÷ 8\t\n\t\n(c)\t (−200) ÷ 25\t\n\t\n\t\n\t\n(d)\t (−200) ÷ 8\t \t\n4.\t Can you also work out the answers for the following division questions by using the \nwork you have done for question 2?\n\t\n(a)\t 1 000 ÷ (−125)\t \t\n\t\n\t\n(b)  (−1 000) ÷ (−8)\t\n\t\n(c)\t (−100) ÷ (−25)\t \t\n\t\n\t\n(d)  100 ÷ (−25)\t\nWhen two numbers are multiplied, for example 30 × 4 = 120, the word “product” can  \nbe used in various ways to describe the situation:\n• An expression that specifies multiplication only, such as 30 × 4, is called a product  \nor a product expression. \n• The answer obtained is also called the product of the two numbers. For example,  \n120 is called the product of 30 and 4.\nAn expression that specifies division only, such as 30 ÷ 5, is called a quotient or a \nquotient expression. The answer obtained is also called the quotient of the two \nnumbers. For example, 6 is called the quotient of 30 and 5.\n5.\t In each case, state whether you agree or disagree with the statement, and give an \nexample to illustrate your answer.\n\t\n(a)\t The quotient of a positive and a negative integer is negative.\nMaths_LB_gr8_bookA.indb   43\n2014/07/03   10:53:20 AM\n\n44\t\nMATHEMATICS Grade 8: Term 1\n\t\n(b)\t The quotient of a positive and a positive integer is negative.\n\t\n(c)\t The quotient of a negative and a negative integer is negative.\n\t\n(d)\t The quotient of a negative and a negative integer is positive.\n6.\t Do the necessary calculations to enable you to provide the values of the quotients.\n\t\n(a)\t (−500) ÷ (−20)\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t (−144) ÷ 6\n\t\n(c)\t 1 440 ÷ (−60)\t \t\n\t\n\t\n\t\n\t\n\t\n(d)\t (−1 440) ÷ (−6)\n\t\n(e)\t −14 400 ÷ 600\t \t\n\t\n\t\n\t\n\t\n\t\n(f)\t 500 ÷ (−20)\nthe associative properties of operations with integers\nMultiplication of whole numbers is associative. This means that in a product with \nseveral factors, the factors can be placed in any sequence, and the calculations can be \nperformed in any sequence. For example, the following sequences of calculations will all \nproduce the same answer:\nA.\t 2 × 3, the answer of 2 × 3 multiplied by 5, the new answer multiplied by 10\nB.\t 2 × 5, the answer of 2 × 5 multiplied by 10, the new answer multiplied by 3\nC.\t 10 × 5, the answer of 10 × 5 multiplied by 3, the new answer multiplied by 2\nD.\t 3 × 5, the answer of 3 × 5 multiplied by 2, the new answer multiplied by 10\n1.\t Do the four sets of calculations given in A to D to check whether they really produce \nthe same answers.\nA.\nB.\nC.\nD.\nMaths_LB_gr8_bookA.indb   44\n2014/07/03   10:53:20 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n45\n2.\t (a)\t If the numbers 3 and 10 in the calculation sequences A, B, C and D are replaced \n    with −3 and −10, do you think the four answers will still be the same? \n\t\n(b)\t Investigate, to check your expectation.\nMultiplication with integers is associative.\nThe calculation sequence A can be represented in symbols in only two ways:\n• 2 × 3 × 5 × 10. The convention to work from left to right unless otherwise indicated \nwith brackets ensures that this representation corresponds to A.\n• \t5 × (2 × 3) × 10, where brackets are used to indicate that 2 × 3 should be calculated first. \nWhen brackets are used, there are different possibilities to describe the same sequence.\n3.\t Express the calculation sequences B, C and D given on page 44 symbolically, without \nusing brackets.\n4.\t Investigate, in the same way that you did for multiplication in question 2, whether \naddition with integers is associative. Use sequences of four integers.\n5.\t (a)\t Calculate: 80 − 30 + 40 − 20\t\n\t\n\t\n\t\n(b)\t Calculate: 80 + (−30) + 40 + (−20)\t \t\n\t\n(c)\t Calculate: 30 − 80 + 20 − 40\t\n\t\n\t\n\t\n(d)\t Calculate: (−30) + 80 + (− 20) + 40\t\t\n\t\n(e)\t Calculate: 20 + 30 − 40 − 80\t\n\t\n\t\nMaths_LB_gr8_bookA.indb   45\n2014/07/03   10:53:20 AM\n\n46\t\nMATHEMATICS Grade 8: Term 1\nmixed calculations with integers\n1.\t Calculate.\n\t\n(a)\t −3 × 4 + (−7) × 9\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −20(−4 − 7)\n\t\n(c)\t 20 × (−5) − 30 × 7\t \t\n\t\n\t\n\t\n\t\n(d)\t −9(20 − 15)\n\t\n(e)\t −8 × (−6) − 8 × 3\t\n\t\n\t\n\t\n\t\n\t\n(f)\t (−26 − 13) ÷ (−3)\n\t\n(g)\t −15 × (−2) + (−15) ÷ (−3)\t\n\t\n\t\n\t\n(h)\t−15(2 − 3)\n\t\n(i)\t (−5 + −3) × 7\t\n\t\n\t\n\t\n\t\n\t\n\t\n(j)\t −5 × (−3 + 7) + 20 ÷ (−4)\n2.\t Calculate.\n\t\n(a)\t 20 × (−15 + 6) − 5 × (−2 − 8) − 3 × (−3 − 8)\t\n\t\n(b)\t 40 × (7 + 12 − 9) + 25 ÷ (−5) − 5 ÷ 5\n\t\n(c)\t −50(20 − 25) + 30(−10 + 7) − 20(−16 + 12)\t \t\nMaths_LB_gr8_bookA.indb   46\n2014/07/03   10:53:20 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n47\n  (d)\t\n−5 × (−3 + 12 − 9)\n\t\n\t\n(e)\t −4 × (30 − 50) + 7 × (40 − 70) − 10 × (60 − 100)\n\t\n\t\n(f)\t −3 × (−14 + 6) × (−13 + 7) × (−20 + 5)\n\t\n(g)\t 20 × (−5) + 10 × (−3) + (−5) × (−6) − (3 × 5)\n\t\n(h)\t −5(−20 − 5) + 10(−7 − 3) − 20(−15 − 5) + 30(−40 − 35)\n\t\n(i)\t (−50 + 15 − 75) ÷ (−11) + (6 − 30 + 12) ÷ (−6)\n2.4\t Squares, cubes and roots with integers\nsquares and cubes of integers\n1.\t Calculate.\n\t\n(a)\t 20 × 20 \n\t\n\t\n\t\n\t\n(b)\t 20 × (−20) \n2.\t Write the answers for each of the following:\n\t\n(a)\t (−20) × 20 \n\t\t\n\t\n\t\n(b)\t (−20) × (−20) \nMaths_LB_gr8_bookA.indb   47\n2014/07/03   10:53:20 AM\n\n48\t\nMATHEMATICS Grade 8: Term 1\n3.\t Complete the table.\nx\n1\n−1\n2\n−2\n5\n−5\n10\n−10\nx2 which is x × x\nx3\n4.\t In each case, state for which values of x, in the table in question 3, the given \nstatement is true.\n\t\n(a)\t x3 is a negative number \n\t\n(b)\t x2 is a negative number \n\t\n(c)\t x2 > x3 \n\t\n(d)\t x2 < x3 \n5.\t Complete the table.\nx\n3\n−3\n4\n−4\n6\n−6\n7\n−7\nx2\nx3\n6.\t Ben thinks of a number. He adds 5 to it, and his answer is 12. \n\t\n(a)\t What number did he think of? \n\t\n(b)\t Is there another number that would also give 12 when 5 is added to it? \n7.\t Lebo also thinks of a number. She multiplies the number by itself and gets 25. \n\t\n(a)\t What number did she think of? \n\t\n(b)\t Is there more than one number that will give 25 when multiplied by itself? \n8.\t Mary thinks of a number and calculates (the number) × (the number) × (the number). \nHer answer is 27.\n\t\nWhat number did Mary think of? \n102 is 100 and (−10)2 is also 100.\nBoth 10 and (−10) are called square roots of 100.  \n10 may be called the positive square root of 100, \nand (−10) may be called the negative square root \nof 100.\nMaths_LB_gr8_bookA.indb   48\n2014/07/03   10:53:20 AM\n\n\t\nCHAPTER 2: INTEGERS\t\n49\n9.\t Write the positive square root and the negative square root of each number.\n\t\n(a)\t 64  \t\t\n\t\n\t\n\t\n(b)\t 9  \t \t\n10.\tComplete the table.\nNumber\n1\n4\n9\n16\n25\n36\n49\n64\nPositive square root\n3\n8\nNegative square root\n−3\n−8\n11.\tComplete the tables.\n(a)\nx\n1\n2\n3\n4\n5\n6\n7\n8\nx3\n(b)\nx\n−1\n−2\n−3\n−4\n−5\n−6\n−7\n−8\nx3\n33 is 27 and (−5)3 is −125.\n3 is called the cube root of 27, because 33 = 27.\n−5 is called the cube root of −125 because (−5)3 = −125.\n12.\tComplete the table.\nNumber\n−1\n8\n−27\n−64\n−125\n−216\n1 000\nCube root\n−3\n10\nThe symbol \n is used to indicate “root”.\n−125\n3\n represents the cube root of −125. That means \n−125\n3\n = −5.\n36\n2\n represents the positive square root of 36, and −36\n2\n represents the negative square \nroot. The “2” that indicates “square” is normally omitted, so 36  = 6 and −36 = −6.\n13.\tComplete the table.\n−8\n3\n121\n−64\n3\n−\n64\n64\n−1\n3\n−1\n−216\n3\nMaths_LB_gr8_bookA.indb   49\n2014/07/03   10:53:22 AM\n\nWorksheet\nINTEGERS\n1.\t Use the numbers −8, −5 and −3 to demonstrate each of the following:\n\t\n(a)\tMultiplication with integers distributes over addition.\n\t\n(b)\tMultiplication with integers distributes over subtraction.\n\t\n(c)\tMultiplication with integers is associative.\n\t\n(d)\tAddition with integers is associative.\n2.\t Calculate each of the following without using a calculator:\n\t\n(a)\t5 × (−2)3\t\n(b)\t 3 × (−5)2\t\n\t\n(c)\t2 × (−5)3\t\n(d)\t 10 × (−3)2\n3.\t Use a calculator to calculate each of the following:\n\t\n(a)\t24 × (−53) + (−27) × (−34) − (−55) × 76    \n\t\n(b)\t64 × (27 − 85) − 29 × (−47 + 12)   \n4.\t Use a calculator to calculate each of the following:\n\t\n(a)\t−24 × 53 + 27 × 34 + 55 × 76 \n\t\n(b)\t64 × (−58) + 29 × (47 − 12) \nIf you don’t get the same answers in questions 3 and 4, you have made mistakes.\nMaths_LB_gr8_bookA.indb   50\n2014/07/03   10:53:22 AM\n\nChapter 3\nExponents\n\t\nCHAPTER 3: EXPONENTS\t\n51\nIn this chapter, you will revise work you have done on squares, cubes, square roots and \ncube roots. You will learn about laws of exponents that will enable you to do calculations \nusing numbers written in exponential form.\nVery large numbers are written in scientific notation. Scientific notation is a convenient \nway of writing very large numbers as a product of a number between 1 and 10 and a \npower of 10.\n3.1\t Revision.................................................................................................................... 53\n3.2\t Working with integers............................................................................................... 58\n3.3\t Laws of exponents.................................................................................................... 60\n3.4\t Calculations.............................................................................................................. 70\n3.5\t Squares, cubes and roots of rational numbers........................................................... 71\n3.6\t Scientific notation..................................................................................................... 74\nMaths_LB_gr8_bookA.indb   51\n2014/07/03   10:53:22 AM\n\n52\t\nMATHEMATICS Grade 8: Term 1\nMaths_LB_gr8_bookA.indb   52\n2014/07/03   10:53:23 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n53",
      "chapter_id": "2"
    },
    {
      "title": "Exponents",
      "content": "3\t Exponents\n3.1\t Revision\nexponential notation\n1.\t Calculate.\n\t\n(a)\t 2 × 2 × 2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 2 × 2 × 2 × 2 × 2 × 2\n\t\n(c)  3 × 3 × 3\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 3 × 3 × 3 × 3 × 3 × 3\nInstead of writing 3 × 3 × 3 × 3 × 3 × 3 we can write 36.\nWe read this as “3 to the power of 6”. The number 3 is \nthe base, and 6 is the exponent. \nWhen we write 3 × 3 × 3 × 3 × 3 × 3 as 36, we are using \nexponential notation.\n2.\t Write each of the following in exponential form:\n\t\n(a)\t 2 × 2 × 2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 2 × 2 × 2 × 2 × 2 × 2\n\t\n(c)\t 3 × 3 × 3\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 3 × 3 × 3 × 3 × 3 × 3\n3.\t Calculate.\n\t\n(a)\t 52\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 25\n\t\n(c)\t 102\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 152\n\t\n(e)\t 34\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 43\n\t\n(g)\t 23\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(h)\t32\nMaths_LB_gr8_bookA.indb   53\n2014/07/03   10:53:23 AM\n\n54\t\nMATHEMATICS Grade 8: Term 1\nsquares\nTo square a number is to multiply it by itself. The \nsquare of 8 is 64 because 8 × 8 equals 64.\nWe write 8 × 8 as 82 in exponential form.\nWe read 82 as eight squared.\n1.\t Complete the table.\nNumber\nSquare the \nnumber\nExponential  \nform\nSquare\n(a)\n1\n(b)\n2\n(c)\n3\n(d)\n4\n(e)\n5\n(f)\n6\n(g)\n7\n(h)\n8\n8 × 8\n82\n64\n(i)\n9\n(j)\n10\n(k)\n11\n(l)\n12\n2.\t Calculate the following:\n\t\n(a)\t 32 × 42\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 22 × 32\n\t\n(c)\t 22 × 52\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 22 × 42\n3.\t Complete the following statements to make them true:\n\t\n(a)\t 32 × 42 = \n2\t\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 22 × 32 = \n2\n\t\n(c)\t 22 × 52 = \n2\t\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 22 × 42 = \n2\nMaths_LB_gr8_bookA.indb   54\n2014/07/03   10:53:23 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n55\ncubes\nTo cube a number is to multiply it by itself and then \nby itself again. The cube of 3 is 27 because 3 × 3 × 3 \nequals 27.\nWe write 3 × 3 × 3 as 33 in exponential form.\nWe read 33 as three cubed.\n1.\t Complete the table.\nNumber\nCube the  \nnumber\nExponential  \nform\nCube\n(a)\n1\n(b)\n2\n(c)\n3\n3 × 3 × 3\n33\n27\n(d)\n4\n(e)\n5\n(f)\n6\n(g)\n7\n(h)\n8\n(i)\n9\n(j)\n10\n2.\t Calculate the following:\n\t\n(a)\t 23 × 33\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 23 × 53\n\t\n(c)\t 23 × 43\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 13 × 93\n3.\t Which of the following statements are true? If a statement is false, rewrite it as a  \ntrue statement.\n\t\n(a)\t 23 × 33 = 63\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 23 × 53 = 73\n\t\n(c)\t 23 × 43 = 83\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 13 × 93 = 103\nMaths_LB_gr8_bookA.indb   55\n2014/07/03   10:53:23 AM\n\n56\t\nMATHEMATICS Grade 8: Term 1\nsquare and cube roots\nTo find the square root of a number we ask the \nquestion: Which number was multiplied by itself to \nget a square?\nThe square root of 16 is 4 because 4 × 4 = 16.\nThe question: Which number was multiplied \nby itself to get 16? is written mathematically as 16.\nThe answer to this question is written as 16 = 4.\n1.\t Complete the table.\nNumber\nSquare of the \nnumber\nSquare root of  \nthe square of  \nthe number\nReason\n(a)\n1\n(b)\n2\n(c)\n3\n(d)\n4\n16\n4\n4 × 4 = 16\n(e)\n5\n(f)\n6\n(g)\n7\n(h)\n8\n(i)\n9\n(j)\n10\n(k)\n11\n(l)\n12\n2.\t Calculate the following. Justify your answer.\n\t\n(a)\t\n144 \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n100\n\t\n(c)\t\n81 \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t\n64\nMaths_LB_gr8_bookA.indb   56\n2014/07/03   10:53:24 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n57\nTo find the cube root of a number we ask the \nquestion: Which number was multiplied by itself and \nagain by itself to get a cube? \nThe cube root of 64 is 4 because 4 × 4 × 4 = 64. \nThe question: Which number was multiplied by \nitself and again by itself (or cubed) to get 64? \nis written mathematically as \n64\n3\n. The answer to this \nquestion is written as \n64\n3\n = 4.\n3.\t Complete the table.\nNumber\nCube of the \nnumber\nCube root of the \ncube of the number\nReason\n(a)\n1\n(b)\n2\n(c)\n3\n(d)\n4\n64\n4\n4 × 4 × 4 = 64\n(e)\n5\n(f)\n6\n(g)\n7\n(h)\n8\n(i)\n9\n(j)\n10\n4.\t Calculate the following and give reasons for your answers:\n\t\n(a)\t\n216\n3\n\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n8\n3\n\t\n(c)\t\n125\n3\n\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t\n27\n3\n\t\n(e)\t\n64\n3\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t\n1 000\n3  \nMaths_LB_gr8_bookA.indb   57\n2014/07/03   10:53:26 AM\n\n58\t\nMATHEMATICS Grade 8: Term 1\n3.2\t Working with integers\nrepresenting integers in exponential form\n1.\t Calculate the following, without using a calculator:\n\t\n(a)\t −2 × −2 × −2\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −2 × −2 × −2 × −2\n\t\n(c)\t −5 × −5\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t −5 × −5 × −5\n\t\n(e)\t −1 × −1 × −1 × −1 \t \t\n\t\n\t\n\t\n\t\n(f)\t −1 × −1 × −1 \n2.\t Calculate the following:\n\t\n(a)\t −22\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t (−2)2\n\t\n(c)\t (−5)3\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t −53\n3.\t Use your calculator to calculate the answers to question 2. \n\t\n(a)\t Are your answers to question 2(a) and (b) different or the same as those of the \t \t\n\t\ncalculator?\n\t\n(b)\t If your answers are different to those of the calculator, try to explain how the  \n\t\ncalculator did the calculations differently from you.\nThe calculator “understands” −52 and (−5)2 as two \ndifferent numbers.\nIt understands −52 as −5 × 5 = −25 and \n(−5)2 as −5 × −5 = 25\nMaths_LB_gr8_bookA.indb   58\n2014/07/03   10:53:26 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n59\n4.\t Write the following in exponential form:\n\t\n(a)\t −2 × −2 × −2 \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −2 × −2 × −2 × −2\n\t\n(c)\t −5 × −5\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t −5 × −5 × −5\n\t\n(e)\t −1 × −1 × −1 × −1 \t \t\n\t\n\t\n\t\n\t\n(f)\t −1 × −1 × −1 \n5.\t Calculate the following:\n\t\n(a)\t (−3)2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t (−3)3\n\t\n(c)\t (−2)4\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t (−2)6\n\t\n(e)\t (−2)5\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t (−3)4\n6.\t Say whether the sign of the answer is negative or positive. Explain why.\n\t\n(a)\t (−3)6\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t (−5)11\n\t\n(c)\t (−4)20\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t (−7)5\n7.\t Say whether the following statements are true or false. If a statement is false rewrite it \nas a correct statement.\n\t\n(a)\t (−3)2 = −9\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −32 = 9\n\t\n(c)\t (−52) = −52\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t (−1)3 = −13\n\t\n(e)\t (−6)3 = −18\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t (−2)6 = 26\nMaths_LB_gr8_bookA.indb   59\n2014/07/03   10:53:26 AM\n\n60\t\nMATHEMATICS Grade 8: Term 1\n3.3\t Laws of exponents\nproduct of powers\n1.\t A product of 2s is given below. Describe it using exponential notation, that is, write  \nit as a power of 2.\n\t\n2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 \n2.\t Express each of the following as a product of the powers of 2, as indicated by the \nbrackets.\n\t\n(a)\t (2 × 2 × 2) × (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2)\n\t\n(b)\t (2 × 2 × 2 × 2 × 2) × (2 × 2 × 2 × 2 × 2) × (2 × 2)\n\t\n(c)\t (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2) × (2 × 2)\n\t\n(d)\t (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)\n\t\n(e)\t (2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2) × (2 × 2)\n3.\t Complete the following statements so that they are true. You may want to refer to \nyour answers to question 2 (a) to (e) to help you. \n\t\n(a)\t 23 × \n = 212\n\t\n(b)\t 25 × \n × 22 =212\n\t\n(c)\t 22 × 22 × 22 × 22 × 22 × 22 = \n\t\n(d)\t 28 × \n = 212\n\t\n(e)\t 23 × 23 × 23 × \n = 212\n\t\n(f)\t 26 × \n = 212\n\t\n(g)\t 22 × 210 = \n \nMaths_LB_gr8_bookA.indb   60\n2014/07/03   10:53:26 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n61\nSuppose we are asked to simplify: 32 × 34.\nThe solution is: 32 × 34 \t= 9 × 81\n\t\n= 729\n\t\n= 36\nWe can explain this solution in the following manner:\n32 × 34 = 3 × 3 × 3 × 3 × 3 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 36\n\t\n2 factors\t 4 factors\t\n6 factors\n4.\t Complete the table.\nProduct of \npowers\nRepeated \nfactor\nTotal number of times \nthe factor is repeated\nSimplified form\n(a)\n27 × 23\n(b)\n52 × 54\n(c)\n41 × 45\n(d)\n63 × 62\n(e)\n28 × 22\n(f)\n53 × 53\n(g)\n42 × 44\n(h)\n21 × 29\nWhen you multiply two or more powers that have \nthe same base, the answer has the same base, but its \nexponent is equal to the sum of the exponents of the \nnumbers you are multiplying.\nWe can express this symbolically as am × an = am + n, \nwhere m and n are natural numbers and a is not zero.\n5.\t What is wrong with these statements? Correct each one.\n\t\n(a)\t 23 × 24 = 212\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 10 × 102 × 103 = 101 × 2 × 3 = 106\n\t\n(c)\t 32 × 33 = 36\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 53 × 52 = 15 × 10\nThe base (3) is a repeated \nfactor. The exponents (2 and \n4) tell us the number of times \neach factor is repeated.\nMaths_LB_gr8_bookA.indb   61\n2014/07/03   10:53:26 AM\n\n62\t\nMATHEMATICS Grade 8: Term 1\n6.\t Express each of the following numbers as a single power of 10.\n\t\nExample: 1 000 000 as a power of 10 is 106.\n\t\n(a)\t 100\t\n(b)\t 1 000\t\n(c)\t 10 000\n\t\n(d)\t 102 × 103 × 104\t\n(e)\t 100 × 1 000 × 10 000\t\n(f)\t 1 000 000 000\n7.\t Write each of the following products in exponential form:\n\t\n(a)\t x × x × x × x × x × x × x × x × x = \n\t\n(b)\t (x × x) × (x × x × x) × (x × x × x × x)\n\t\n(c)\t (x × x × x × x) × (x × x) × (x × x) × x\n\t\n(d)\t (x × x × x × x × x × x) × (x × x × x)\n\t\n(e)\t (x × x × x) × (y × y × y)\n\t\n(f)\t (a × a) × (b × b)\n8.\t Complete the table.\nProduct of \npowers\nRepeated \nfactor\nTotal number of times  \nthe factor is repeated\nSimplified \nform\n(a)\nx7 × x3\n(b)\nx2 × x4\n(c)\nx1 × x5\n(d)\nx3 × x2\n(e)\nx8 × x2\n(f)\nx3 × x3\n(g)\nx1 × x9\nMaths_LB_gr8_bookA.indb   62\n2014/07/03   10:53:26 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n63\nraising a power to a power\n1.\t Complete the table of powers of 2.\nx\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n2x\n2\n4\n21\n22\n23\nx\n12\n13\n14\n15\n16\n17\n18\n2x\n2.\t Complete the table of powers of 3.\nx\n1\n2\n3\n4\n5\n6\n7\n8\n9\n3x\n31\n32\n33\nx\n10\n11\n12\n13\n14\n3x\n3.\t Complete the table. You can read the values from the tables you made in questions 1 \nand 2.\nProduct of powers\nRepeated \nfactor\nPower of  \npower \nnotation\nTotal \nnumber of \nrepetitions\nSimplified \nform\nValue\n24  × 24 × 24\n2\n(24)3\n12\n212\n4 096\n32 × 32 × 32 × 32\n23 × 23 × 23 × 23 × 23\n34 × 34 × 34\n26 × 26 × 26\n4.\t Use your table of powers of 2 to find the answers for the following:\n\t\n(a)\t 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2   =   \n  =   \n\t\n(b)\t (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2) × (2 × 2 × 2 × 2)   =   \n   =   \n\t\n(c)\t 163   =   \n   =   \n  =   \nMaths_LB_gr8_bookA.indb   63\n2014/07/03   10:53:26 AM\n\n64\t\nMATHEMATICS Grade 8: Term 1\n5.\t Use your table of powers of 2 to find the answers for the following:\n\t\n(a)\t Is 163 = 212? \t\n\t\n\t \t\n\t\n\t\n(b)\t Is 24 × 24 × 24 = 212 \t \t\n\t\n(c)\t Is 24 × 23 = 212? \t\t\n\t \t\n\t\n\t\n(d)\t Is (24)3 = 24 × 24 × 24?  \t\n\t\n(e)\t Is (24)3 = 212? \t\n\t\n\t \t\n\t\n\t\n(f)\t Is (24)3 = 24 + 3? \t \t\n\t\n\t\n(g)\t Is (24)3 = 24 × 3? \t \t\n\t \t\n\t\n\t\n(h)\tIs (22)5 = 22 + 5? \t \t\n\t\n6.\t (a)\t Express 85 as a power of 2. It may help to first express 8 as a power of 2.\n\t\n(b)\t Can (23) × (23) × (23) × (23) × (23) be expressed as (23)5? \n\t\n(c)\t Is (23)5 = 23 + 5 or is (23)5 = 23 × 5? \n7.\t (a)\t Express 43 as a power of 2. \n\t\n(b)\t Calculate 22 × 22 × 22 and express your answer as a single power of 2.\n\t\n(c)\t Can (22) × (22) × (22) be expressed as (22)3? \n\t\n(d)\t Is (22)3 = 22 + 3 or is (22)3 = 22 × 3? \n8.\t Simplify the following.\n\t\nExample: (102)2 = 102 × 102 = 102 + 2 = 104 = 10 000\n\t\n(a)\t (33)2 \n\t \t\n(b)\t (43)2 \n\t\n(c)\t (24)2 \n\t \t\n(d)\t (92)2 \n\t\n(e)\t (33)3 \n\t \t\n(f)\t (43)3 \n\t\n(g)\t (54)3 \n\t \t\n(h)\t(92)3 \n(am)n = am × n, where m and n are natural numbers and  \na is not equal to zero.\n9.\t Simplify.\n\t\n(a)\t (54)10 \t\n\t \t\n\t\n\t\n(b)\t (104)5 \t\n\t\n(c)\t (64)4 \t\n\t \t\n\t\n\t\n(d)\t (54)10 \t\n10.\tWrite 512 as a power of powers of 5 in two different ways. \nTo simplify (x2)5 we can write it out as a product of powers or we can use a shortcut.\n(x2)5\t= x2 × x2 × x2 × x2 × x2\n\t\n= x × x × x × x × x × x × x × x × x × x    = x10\n\t\n2 × 5 factors = 10 factors\n \n \n \n \n \nMaths_LB_gr8_bookA.indb   64\n2014/07/03   10:53:27 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n65\n11.\tComplete the table.\nExpression\nWrite as a product of the \npowers and then simplify\nUse the rule (am)n to \nsimplify\n(a)\n(a4)5\na4 × a4 × a4 × a4 × a4 \n= a4 + 4 + 4 + 4 + 4 = a20\n(a4)5 = a4 × 5 = a20\n(b)\n(b10)5\n(c)\n(x7)3\n(d)\ns6 × s6 × s6 × s6 \n= s6 + 6 + 6 + 6\n= s24\n(e)\ny3 × 7 = y21\npower of a product\n1.\t Complete the table. You may use your calculator when you are not sure of a value.\nx\n1\n2\n3\n4\n5\n(a)\n2x\n21 = 2\n(b)\n3x\n32 = 9\n(c)\n6x\n63 = 216\n2.\t Use the table in question 1 to answer the questions below. Are these statements true \nor false? If a statement is false rewrite it as a correct statement.\n\t\n(a)\t 62 = 22 × 32\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 63 = 23 × 33\n\t\n(c)\t 65 = 25 × 35 \t\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 68 = 24 × 34\nMaths_LB_gr8_bookA.indb   65\n2014/07/03   10:53:27 AM\n\n66\t\nMATHEMATICS Grade 8: Term 1\n3.\t Complete the table.\nExpression\nThe bases of the  \nexpression are factors  \nof …\nEquivalent  \nexpression\n(a)\n26 × 56\n10\n106\n(b)\n32 × 42\n(c)\n42 × 22\n(d)\n565\n(e)\n303\n(f)\n35 × x5\n3x\n(3x)5\n(g)\n72 × z2\n(h)\n43 × y3\n(i)\n(2m)6\n(j)\n(2m)3\n(k)\n210 × y10 \n(2y)10\n122 can be written in terms of its factors as (2 × 6)2 or as (3 × 4)2. \nWe already know that 122 = 144. \nWhat this tells us is that both (2 × 6)2 and (3 × 4)2 also equal 144.\nWe write\t\n122\t= (2 × 6)2\t\nor\t\n122\t= (3 × 4)2\n\t\n\t= 22 × 62\t\n\t\n\t= 32 × 42\n\t\n\t= 4 × 36\t\n\t\n\t= 9 × 16 \n\t\n\t= 144\t\n\t\n\t= 144\nA product raised to a power is the product of the \nfactors each raised to the same power. \nUsing symbols, we write (a × b)m = am × bm, where m \nis a natural number and a and b are not equal to zero.\nMaths_LB_gr8_bookA.indb   66\n2014/07/03   10:53:27 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n67\n4.\t Write each of the following expressions as an expression with one base:\n\t\nExample: 310 × 210 = (3 × 2)10 = 610\n\t\n(a)\t 32 × 52 \t\n(b)\t 53 × 23\t\n(c)\t 74 × 44 \n\t\n(d)\t 23 × 63\t\n(e)\t 44 × 24\t\n(f)\t 52 × 72 \n5.\t Write the following as a product of powers:\n\t\nExample: (3x)3 = 33 × x3 = 27x3\n\t\n(a)\t 63\t\n(b)\t 152\t\n(c)\t 214\n\t\n(d)\t 65\t\n(e)\t 182\t\n(f)\t (st)7\n\t\n(g)\t (ab)3\t\n(h)\t (2x)2\t\n(i)\t (3y)5\n\t\n(j)\t (3c)2\t\n(k)\t (gh)4\t\n(l)\t (4x)3\n6.\t Simplify the following expressions:\n\t\nExample: 32 × m2 = 9 × m2 = 9m2\n\t\n(a)\t 35 × b5\t\n(b)\t 26 × y6\t\n(c)\t x2 × y2\n\t\n(d)\t 104 × x4\t\n(e)\t 33 × x3\t\n(f)\t 52 × t2\n\t\n(g)\t 63 × m7\t\n(h)\t 122 × a2\t\n(i)\t n3 × p9\nMaths_LB_gr8_bookA.indb   67\n2014/07/03   10:53:27 AM\n\n68\t\nMATHEMATICS Grade 8: Term 1\na quotient of powers\nConsider the following table:\nx\n1\n2\n3\n4\n5\n6\n2x\n2\n4\n8\n16\n32\n64\n3x\n3\n9\n27\n81\n243\n729\n5x\n5\n25\n125\n625\n3 125\n15 625\nAnswer questions 1 to 4 by referring to the table when you need to.\n1.\t Give the value of each of the following:\n\t\n(a)\t 34\t\n(b)\t 25\t\n(c)\t 56\n2.\t (a)\t Calculate 36 ÷ 33 (Read the values of 36 and 33 from the table and then divide. \n\t\n\t\nYou may use a calculator where necessary.)\n\t\n(b)\t Calculate 36 − 3\n\t\n(c)\t Is 36 ÷ 33 equal to 33? Explain.\n3.\t (a)\t Calculate the value of 26 − 2\t\n\t\n\t\n(b)\t Calculate the value of 26 ÷ 22\n\t\n(c)\t Calculate the value of 26 ÷ 2\t\n\t\n\t\n(d)\t Read from the table the value of 23\n\t\n(e)\t Read from the table the value of 24\n\t\n(f)\t Which of the statements below is true? Give an explanation for your answer.\n\t\n\t\nA.\t 26 ÷ 22 = 26 − 2 = 24\t\n\t\n\t\n\t\n\t\nB.\t 26 ÷ 22 = 26 ÷ 2 = 23\nTo calculate 45 – 3 we first  \ndo the calculation in the \nexponent, that is, we  \nsubtract 3 from 5. Then  \nwe can calculate 42 as  \n4 × 4 = 16.\nMaths_LB_gr8_bookA.indb   68\n2014/07/03   10:53:27 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n69\n4.\t Say which of the statements below are true and which are false. If a statement is false \nrewrite it as a correct statement.\n\t\n(a)\t 56 ÷ 54 = 56 ÷ 4\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 34 − 1 = 34 ÷ 3\n\t\n(c)\t 56 ÷ 5 = 56 − 1\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 25 ÷ 23 = 22\nam ÷ an = am – n  \nwhere m and n are natural numbers and m is a \nnumber greater than n and a is not zero.\n5.\t Simplify the following. Do not use a calculator.\n\t\nExample: 317 ÷ 312 = 317 − 12 = 35 = 243\n\t\n(a)\t 212 ÷ 210\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 617 ÷ 614\t\n\t\n(c)\t 1020 ÷ 1014\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 511 ÷ 58\n6.\t Simplify:\n\t\n(a)\t x12 ÷ x10\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t y17 ÷ y14\n\t\n(c)\t t20 ÷ t14\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t n11 ÷ n8\nMaths_LB_gr8_bookA.indb   69\n2014/07/03   10:53:27 AM\n\n70\t\nMATHEMATICS Grade 8: Term 1\nthe power of zero\n1.\t Simplify the following:\n\t\n(a)\t 212 ÷ 212\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 617 ÷ 617\n\t\n(c)\t 614 ÷ 614\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 210 ÷ 210\nWe define a0 = 1.\nAny number raised to the power of zero is always \nequal to 1.\n2.\t Simplify the following:\n\t\n(a)\t 1000\t\n\t\n\t\n\t\n(b)\t x0\t \t\n\t\n\t\n   (c)  (100x)0\t\n\t\n\t\n  (d)  (5x3)0\n3.4\t Calculations\nmixed operations\nSimplify the following:\n1.\t 33 + −27\n3\n × 2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n2.\t 5 × (2 + 3)2 + (−1)0\n3.\t 32 × 23 + 5 × 100 \t \t\n\t\n\t\n\t\n\t\n\t\n4.\t\n1 000\n100\n3  \n + (4 − 1)2\nMaths_LB_gr8_bookA.indb   70\n2014/07/03   10:53:28 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n71\n5.\t\n16 × 16 + \n216\n3\n + 32 × 10\t\n\t\n\t\n\t\n6.\t 43 ÷ 23 + 144  \n3.5\t Squares, cubes and roots of rational numbers\nsquaring a fraction\nSquaring or cubing a fraction or a decimal fraction is \nno different from squaring or cubing an integer.\n1.\t Complete the table.\nFraction\nSquare the fraction\nValue of the square of  \nthe fraction\n(a)\n1\n2\n1\n2  × 1\n2\n1\n2  × 1\n2  = 1\n4\n(b)\n2\n3\n(c)\n3\n4\n(d)\n2\n5\n(e)\n3\n5\n(f)\n2\n6\n(g)\n3\n7\n(h)\n11\n12\nMaths_LB_gr8_bookA.indb   71\n2014/07/03   10:53:30 AM\n\n72\t\nMATHEMATICS Grade 8: Term 1\n2.\t Calculate the following:\n\t\n(a)\t\n3\n2\n2\n\n\n\n\t\n(b)\t\n4\n5\n2\n\n\n\n\t\n(c)\t\n7\n8\n2\n\n\n\n\n3.\t (a)\t Use the fact that 0,6 can be written as 6\n10  to calculate (0,6)2.\n\t\n(b)\t Use the fact that 0,I can be written as 8\n10  to calculate (0,8)2.\nfinding the square root of a fraction\n1.\t Complete the table.\nFraction\nWriting the fraction as a \nproduct of factors\nSquare root\n(a)\n81\n121\n(b)\n64\n81\n(c)\n49\n169\n(d)\n100\n225\n2.\t Determine the following:\n\t\n(a)\t\n25\n16 \t\n\t\n\t\n\t\n(b)\t\n81\n144 \t\n\t\n\t\n\t\n(c)\t\n400\n900 \t\n\t\n\t\n(d)\t\n36\n81\nMaths_LB_gr8_bookA.indb   72\n2014/07/03   10:53:33 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n73\n3.\t (a)\t Use the fact that 0,01 can be written as 1\n100  to calculate 0 01\n,\n.\n\t\n(b)\t Use the fact that 0,49 can be written as 49\n100  to calculate 0 49\n,\n. \n4.\t Calculate the following:\n\t\n(a)\t\n0 09\n,\n\t\n(b)\t\n0 64\n,\n\t\n(c)\t\n1 44\n,\ncubing a fraction\nOne half cubed is equal to one eighth.\nWe write this as 1\n2\n3\n\n\n\n = 1\n2  × 1\n2  × 1\n2  = 1\n8\n1.\t Calculate the following:\n\t\n(a)\t\n2\n3\n3\n\n\n\n\t\n\t\n\t\n\t\n(b)\t\n5\n10\n3\n\n\n\n\t\n\t\n\t\n\t\n(c)\t\n5\n6\n3\n\n\n\n\t\n\t\n\t\n\t\n(d)  4\n5\n3\n\n\n\n\n  \n2.\t (a)\t Use the fact that 0,6 can be written as 6\n10  to find (0,6)3.\n\t\n(b)\t Use the fact that 0,I can be written as 8\n10  to calculate (0,8)3.\n\t\n(c)\t Use the fact that 0,7 can be written as 7\n10  to calculate (0,7)3.\nMaths_LB_gr8_bookA.indb   73\n2014/07/03   10:53:36 AM\n\n74\t\nMATHEMATICS Grade 8: Term 1\n3.6\t Scientific notation\nvery large numbers\n1.\t Express each of the following as a single number. Do not use a calculator.\n\t\nExample: 7,56 × 100 can be written as 756.\n\t\n(a)\t 3,45 × 100\t\n(b)\t 3,45 × 10\t\n(c)\t 3,45 × 1 000\n\t\n(d)\t 2,34 × 102\t\n(e)\t 2,34 × 10\t\n(f)\t 2,34 × 103\n\t\n(g)\t 104 × 102\t\n(h)\t 100 × 106\t\n(i)\t 3,4 × 105\nWe can write 136 000 000 as 1,36 × 108.\n1,36 × 108 is called the scientific notation for  \n136 000 000. \nIn scientific notation, a number is expressed in two \nparts: a number between 1 and 10 multiplied by a \npower of 10. The exponent must always be an integer.\n2.\t Write the following numbers in scientific notation:\n\t\n(a)\t 367 000 000\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 21 900 000\n\t\n(c)\t 600 000 000 000\t \t\n\t\n\t\n\t\n\t\n(d)\t 178\n3.\t Write each of the following numbers in the ordinary way.\n\t\nFor example: 3,4 × 105 written in the ordinary way is 340 000.\n\t\n(a)\t 1,24 × 108\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 9,2074 × 104\n\t\n(c)\t 1,04 × 106\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 2,05 × 103\nMaths_LB_gr8_bookA.indb   74\n2014/07/03   10:53:36 AM\n\n\t\nCHAPTER 3: EXPONENTS\t\n75\n4.\t The age of the universe is 15 000 000 000 years. Express the age of the universe in \nscientific notation. \n5.\t The average distance from the Earth to the Sun is 149 600 000 km. Express this \ndistance in scientific notation. \nBecause it is easier to multiply powers of ten without \na calculator, scientific notation makes it possible \nto do calculations in your head.\n6.\t Explain why the number 24 × 103 is not in scientific notation.\n7.\t Calculate the following. Do not use a calculator.\n\t\nExample:    3 000 000 × 90 000 000 = 3 × 106 × 9 × 107 = 3 × 9 × 106 + 7 \n\t\n\t\n\t\n= 27 × 1013 = 270 000 000 000 000\n\t\n(a)\t 13 000 × 150 000 \t \t\n\t\n\t\n\t\n\t\n(b)\t 200 × 6 000 000\n\t\n(c)\t 120 000 × 120 000 000\t\n\t\n\t\n\t\n(d)\t 2,5 × 40 000 000 \t\n8.\t Use > or < to compare these numbers:\n\t\n(a)\t 1,3 × 109 \n  2,4 × 107\t\n\t\n\t\n\t\n(b)\t 6,9 × 102 \n  4,5 × 103 \n\t\n(c)\t 7,3 × 104 \n  7,3 × 102\t\n\t\n\t\n\t\n(d)\t 3,9 × 106 \n  3,7 × 107\nMaths_LB_gr8_bookA.indb   75\n2014/07/03   10:53:36 AM\n\nWorksheet\nEXPONENTS\n1.\t Calculate:\n\t\n(a)\t112\t\n\t\n(b)\t 32 × 42\t\n\t\n(c)\t63 \t\n\t\n(d)\t\n121\t\n\t\n(e)\t (−3)2\t\n\t\n(f)\t\n125\n3\n\t\n2.\t Simplify:\n\t\n(a)\t34 × m6\t\n\t\n(b)\t b2 × n6\t\n\t\n(c)\ty12 ÷ y5\t\n\t\n(d)\t (102)3\t\n\t\n(e)\t (2w2)3\t\n\t\n(f)\t (3d 5)(2d)3\t\n3.\t Calculate:\n\t\n(a)\t\n2\n5\n2\n\n\n\n\t\n\t\n(b)\t  \n9\n25 \t\n\t\n(c)\t(64y 2)0\t\n\t\n(d)\t (0,7)2\t\n4.\t Simplify:\n\t\n(a)\t(22 + 4)2 + 6\n3\n2\n2 \t\n(b)\t\n−125\n3\n − 5 × 32\n5.\t Write 3 × 109 in the ordinary way. \n6.\t The first birds appeared on Earth about 208 000 000 years ago. Write this \n\t\nnumber in scientific notation.\nMaths_LB_gr8_bookA.indb   76\n2014/07/03   10:53:37 AM\n\nChapter 4\nNumeric and geometric\npatterns\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n77\nIn this chapter you will learn to create, recognise, describe, extend and make  \ngeneralisations about numeric and geometric patterns. Patterns allow us to make  \npredictions. You will also work with different representations of patterns, such as flow  \ndiagrams and tables.\n4.1\t The term–term relationship in a sequence................................................................. 79\n4.2\t The position–term relationship in a sequence........................................................... 84\n4.3\t Investigate and extend geometric patterns............................................................... 86\n4.4\t Describe patterns in different ways........................................................................... 90\nMaths_LB_gr8_bookA.indb   77\n2014/07/03   10:53:37 AM\n\n78\t\nMATHEMATICS Grade 8: Term 1\nMaths_LB_gr8_bookA.indb   78\n2014/07/03   10:53:38 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n79",
      "chapter_id": "3"
    },
    {
      "title": "Numeric and geometric patterns",
      "content": "4\t Numeric and geometric patterns\n4.1\t The term–term relationship in a sequence\ngoing from one term to the next\nWrite down the next three numbers in each of  \nthe sequences below. Also explain in writing, in  \neach case, how you figured out what the numbers  \nshould be. \n1.\t Sequence A:\t\n2; 5; 8; 11; 14; 17; 20; 23; \n2.\t Sequence B:\t\n4; 5; 8; 13; 20; 29; 40; \n3.\t Sequence C:\t\n1; 2; 4; 8; 16; 32; 64; \n4.\t Sequence D:\t 3; 5; 7; 9; 11; 13; 15; 17; 19; \n5.\t Sequence E:\t\n4; 5; 7; 10; 14; 19; 25; 32; 40; \n6.\t Sequence F:\t\n2; 6; 18; 54; 162; 486; \n7.\t Sequence G:\t 1; 5; 9; 13; 17; 21; 25; 29; 33; \n8.\t Sequence H:\t 2; 4; 8; 16; 32; 64; \nNumbers that follow one another are said to be \nconsecutive.\nA list of numbers which form a \npattern is called a sequence. \nEach number in a sequence is \ncalled a term of the sequence. \nThe first number is the first term \nof the sequence.\nMaths_LB_gr8_bookA.indb   79\n2014/07/03   10:53:38 AM\n\n80\t\nMATHEMATICS Grade 8: Term 1\nadding or subtracting the same number\n1.\t Which sequences on the previous page are of the same kind as sequence A? Explain \nyour answer.\nAmanda explains how she figured out how to continue sequence A:\n\tI looked at the first two numbers in the sequence and saw that I needed 3 to go from  \n2 to 5. I looked further and saw that I also needed 3 to go from 5 to 8. I tested that and \nit worked for all the next numbers. \n\tThis gave me a rule I could use to extend the sequence: add 3 to each \nnumber to find the next number in the pattern. \nTamara says you can also find the pattern by working  \nbackwards and subtracting 3 each time: \n         14 − 3 = 11;  11 − 3 = 8;  8 − 3 = 5;  5 − 3 = 2\n2.\t Provide a rule to describe the relationship between the numbers in the sequence.  \nUse this rule to calculate the missing numbers in the sequence.\n\t\n(a)\t 1;  8;  15; \n ; \n ; \n ; \n ; \n ; ...\n\t\n(b)\t 10 020; \n ; \n ; \n ;  9 980;  9 970; \n ; \n ;  9 940;  9 930; ...\n\t\n(c)\t 1,5;  3,0;  4,5; \n ; \n ; \n ; \n ; \n ; ...\n\t\n(d)\t 2,2;  4,0;  5,8; \n ; \n ; \n ; \n ; \n ; ...\n\t\n(e)\t 45\n3\n4;  46\n1\n2;  47\n1\n4;  48;  \n ;  \n ;  \n ;  \n ;  \n ; …\n\t\n(f)\t\n ;  100,49;  100,38;  100,27;  \n ;  \n ;  99,94;  99,83;  99,72; …\n3.\t Complete the table below.\nInput number\n1\n2\n3\n4\n5\n12\nn\nInput number + 7\n8\n11\n15\n30\nWhen the differences between \nconsecutive terms of a sequence \nare the same, we say the  \ndifference is constant. \nMaths_LB_gr8_bookA.indb   80\n2014/07/03   10:53:39 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n81\nmultiplying or dividing with the same number \nTake another look at sequence F: 2; 6; 18; 54; 162; 486; ...\nPiet explains that he figured out how to continue the sequence F: \nI looked at the first two terms in the sequence and wrote 2 × ? = 6. \nWhen I multiplied the first number by 3, I got the second number: 2 × 3 = 6.\nI then checked to see if I could find the next number if I multiplied 6 by 3:  \n6 × 3 = 18.\nI continued checking in this way: 18 × 3 = 54;  54 × 3 = 162 and so on. \nThis gave me a rule I can use to extend the sequence and my rule was: \nmultiply each number by 3 to calculate the next number in the \nsequence. \nZinhle says you can also find the pattern by  \nworking backwards and dividing by 3 each  \ntime: \n\t\n54 ÷ 3 = 18;  18 ÷ 3 = 6;  6 ÷ 3 = 2\n1.\t Check whether Piet’s reasoning works for sequence H: 2; 4; 8; 16; 32; 64; ...\n2.\t Describe, in words, the rule for finding the next number in the sequence. Also write \ndown the next five terms of the sequence if the pattern is continued.\n\t\n(a)\t 1; 10; 100; 1 000; \n\t\n(b)\t 16; 8; 4; 2; \n\t\n(c)\t 7; −21; 63; −189; \n\t\n(d)\t 3; 12, 48; \n\t\n(e)\t 2 187; −729; 243; −81; \nThe number that we multiply with \nto get the next term in the sequence \nis called a ratio. If the number we \nmultiply with remains the same \nthroughout the sequence, we say it is \na constant ratio. \nMaths_LB_gr8_bookA.indb   81\n2014/07/03   10:53:39 AM\n\n82\t\nMATHEMATICS Grade 8: Term 1\n3.\t (a)\t Fill in the missing output and input numbers:\n1\n2\n3\n5\n7\n6\n24\n36\n× 6\n\t\n(b)\t Complete the table below:\nInput numbers\n1\n2\n3\n4\n5\n12\nx\nOutput numbers\n6\n24\n36\nneither adding nor multiplying by the same number\n1.\t Consider sequences A to H again and answer the questions that follow:\n\t\nSequence A:\t\n2; 5; 8; 11; 14; 17; 20; 23; ...\n\t\nSequence B:\t\n4; 5; 8; 13; 20; 29; 40; …\n\t\nSequence C:\t\n1; 2; 4; 8; 16; 32; 64; …\n\t\nSequence D:\t\n3; 5; 7; 9; 11; 13; 15; 17; 19; ...\n\t\nSequence E:\t\n4; 5; 7; 10; 14; 19; 25; 32; 40; …\n\t\nSequence F:\t\n2; 6; 18; 54; 162; 486; …\n\t\nSequence G:\t\n1; 5; 9; 13; 17; 21; 25; 29; 33; ...\n\t\nSequence H:\t\n2; 4; 8; 16; 32; 64; ….\n\t\n(a)\t Which other sequence(s) is/are of the same kind as sequence B? Explain.\n\t\n(b)\t In what way are sequences B and E different from the other sequences?\nThere are sequences where there is neither a constant \ndifference nor a constant ratio between consecutive \nterms and yet a pattern still exists, as in the case of \nsequences B and E.\nWhat is the term-to-term rule \nfor the output numbers here,  \n+ 6 or × 6?\nMaths_LB_gr8_bookA.indb   82\n2014/07/03   10:53:39 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n83\n2.\t Consider the sequence: 10; 17; 26; 37; 50; ...\n\t\n(a)\t Write down the next five numbers in the sequence. \n\t\n(b)\t Eric observed that he can calculate the next term in the sequence as follows:  \n    10 + 7 = 17;  17 + 9 = 26;  26 + 11 = 37. Use Eric’s method to check whether your \n    numbers in question (a) above are correct.\n3.\t Which of the statements below can Eric use to describe the relationship between the \nnumbers in the sequence in question 2? Test the rule for the first three terms of the \nsequence and then simply write “yes” or “no” next to each statement.\n\t\n(a)\t Increase the difference between consecutive terms by 2 each time \n \n\t\n(b)\t Increase the difference between consecutive terms by 1 each time \n\t\n(c)\t Add two more than you added to get the previous term \n4.\t Provide a rule to describe the relationship between the numbers in the sequences \nbelow. Use your rule to provide the next five numbers in the sequence.\n\t\n(a)\t 1; 4; 9; 16; 25; \n\t\n(b)\t 2; 13; 26; 41; 58; \n\t\n(c)\t 4; 14; 29; 49; 74; \n\t\n(d)\t 5; 6; 8; 11; 15; 20; \nMaths_LB_gr8_bookA.indb   83\n2014/07/03   10:53:39 AM\n\n84\t\nMATHEMATICS Grade 8: Term 1\n4.2\t The position–term relationship in a sequence\nusing position to make predictions\n1.\t Take another look at sequences A to H. Which sequence(s) are of the same kind as \nsequence A? Explain.\n\t\n\t\nSequence A:\t\n2; 5; 8; 11; 14; 17; 20; 23;...\n\t\nSequence B:\t\n4; 5; 8; 13; 20; 29; 40; …\n\t\nSequence C:\t\n1; 2; 4; 8; 16; 32; 64; …\n\t\nSequence D:\t\n3; 5; 7; 9; 11; 13; 15; 17; 19; ...\n\t\nSequence E:\t\n4; 5; 7; 10; 14; 19; 25; 32; 40; …\n\t\nSequence F:\t\n2; 6; 18; 54; 162; 486; …\n\t\nSequence G:\t\n1; 5; 9; 13; 17; 21; 25; 29; 33; ...\n\t\nSequence H:\t\n2; 4; 8; 16; 32; 64; …\nSizwe has been thinking about Amanda and Tamara’s explanations of how they worked \nout the rule for sequence A and has drawn up a table. He agrees with them but says that \nthere is another rule that will also work. He explains: \n\t\nMy table shows the terms in the sequence and the difference between consecutive terms:\n \n1st \nterm \n2nd \nterm\n3rd \nterm\n4th \nterm\nA:\n5\n8\n11\n14\ndifferences\n+3\n+3\n+3\n+3\n+3\n+3\n+3\n+3\n+3\nSizwe reasons that the following rule will also work:\n\t\nMultiply the position of the number by 3 and add 2 to the answer.\n\t\nI can write this rule as a number sentence: Position of the number × 3 + 2\n\t\nI use my number sentence to check: 1 × 3 + 2 = 5;  2 × 3 + 2 = 8;  3 × 3 + 2 = 11\n2.\t (a)\t What do the numbers in bold in Sizwe’s number sentence stand for?\n\t\n(b)\t What does the number 3 in Sizwe’s number sentence stand for?\nMaths_LB_gr8_bookA.indb   84\n2014/07/03   10:53:39 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n85\n3.\t Consider the sequence 5; 8; 11; 14; ...\n\t\nApply Sizwe’s rule to the sequence and determine: \n\t\n(a)\t term number 7 of the sequence\t\n\t\n(b)\t term number 10 of the sequence\n\t\n(c)\t the 100th term of the sequence\n4.\t Consider the sequence: 3; 5; 7; 9; 11; 13; 15; 17; 19;..\n\t\n(a)\t Use Sizwe’s explanation to find a rule for this sequence.\n\t\n(b)\t Determine the 28th term of the sequence.\nmore predictions\nComplete the tables below by calculating the missing terms.\n1.\nPosition in sequence\n1\n2\n3\n4\n10\n54\nTerm\n4\n7\n10\n13\n2.\nPosition in sequence\n1\n2\n3\n4\n8\n16\nTerm\n4\n9\n14\n19\nMaths_LB_gr8_bookA.indb   85\n2014/07/03   10:53:39 AM\n\n86\t\nMATHEMATICS Grade 8: Term 1\n3.\nPosition in sequence\n1\n2\n3\n4\n7\n30\nTerm\n3\n15\n27\n4.\t Use the rule Position in the sequence × (position in the sequence + 1) to \ncomplete the table below.\nPosition in sequence\n1\n2\n3\n4\n5\n6\nTerm\n2\n4.3\t Investigating and extending geometric patterns\nsquare numbers\nA factory makes window frames. Type 1 has one windowpane, type 2 has four \nwindowpanes, type 3 has nine windowpanes, and so on.\nType 1\t\nType 2\t\nType 3\t\nType 4\n1.\t How many windowpanes will there be in type 5? \n2.\t How many windowpanes will there be in type 6? \n3.\t How many windowpanes will there be in type 7? \n4.\t How many windowpanes will there be in type 12? Explain.\nMaths_LB_gr8_bookA.indb   86\n2014/07/03   10:53:40 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n87\n5.\t Complete the table. Show your calculations.\nFrame type\n1\n2\n3\n4\n15\n20\nNumber of windowpanes\n1\n4\n9\n16\nIn algebra we think of a square as a number that is \nobtained by multiplying a number by itself. So 1 is \nalso a square because 1 × 1 = 1.\n12\t\n 22\t\n 32\t\n 42\t\n 52\t\n...\t\nn2\ntriangular numbers\nTherese uses circles to form a pattern of triangular shapes:\nPicture 1\nPicture 2\nPicture 3\nPicture 4\n1.\t If the pattern is continued, how many circles must Therese have\n\t\n(a)\t in the bottom row of picture 5? \n\t\n(b)\t in the second row from the bottom of picture 5? \n\t\n(c)\t in the third row from the bottom of picture 5? \n\t\n(d)\t in the second row from the top of picture 5? \n\t\n(e)\t in the top row of picture 5? \n\t\n(f)\t in total in picture 5? Show your calculation.\nThe symbol n is used below to \nrepresent the position number \nin the expression that gives the \nrule (n2) when generalising. \nMaths_LB_gr8_bookA.indb   87\n2014/07/03   10:53:40 AM\n\n88\t\nMATHEMATICS Grade 8: Term 1\n2.\t How many circles does Therese need to form triangle picture 7?  \nShow the calculation.\n3.\t How many circles does Therese need to form triangle picture 8? \n4.\t Complete the table below. Show all your work.\nPicture  \nnumber\n1\n2\n3\n4\n5\n6\n12\n15\nNumber of \ncircles\n1\n3\n6\n10\nMore than 2 500 years ago, Greek mathematicians already knew that the numbers 3, 6, \n10, 15 and so on could form a triangular pattern. They represented these numbers with \ndots which they arranged in such a way that they formed equilateral triangles, hence \nthe name triangular numbers. Algebraically we think of them as sums of consecutive \nnatural numbers starting with 1.\nLet us revisit the activity on triangular numbers that we did on the previous page.\t\nPicture 1\nPicture 2\nPicture 3\nPicture 4\nSo far, we have determined the number of circles in the pattern by adding consecutive \nnatural numbers. If we were asked to determine the number of circles in picture 200, for \nexample, it would take us a very long time to do so. We need to find a quicker method  \nof finding any triangular number in the sequence.\nMaths_LB_gr8_bookA.indb   88\n2014/07/03   10:53:40 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n89\nConsider the arrangement below.\nPicture 1\nPicture 2\nPicture 3\nPicture 4\nWe have added the yellow circles to the original blue circles and then rearranged the \ncircles in such a way that they are in a rectangular form. \n5.\t Picture 2 is 3 circles long and 2 circles wide. Complete the following sentences:\n\t\n(a)\t Picture 3 is \n circles long and \n circles wide. \n\t\n(b)\t Picture 1 is \n circles long and \n circle wide.\n\t\n(c)\t Picture 4 is \n circles long and \n circles wide.\n\t\n(d)\t Picture 5 is \n circles long and \n circles wide.\n6.\t How many circles will there be in a picture that is: \n\t\n(a)\t 10 circles long and 9 circles wide? \n\t\n(b)\t 7 circles long and 6 circles wide? \n\t\n(c)\t 6 circles long and 5 circles wide? \n\t\n(d)\t 20 circles long and 19 circles wide? \nSuppose we want to have a quicker method to determine the number of circles in  \npicture 15. We know that picture 15 is 16 circles long and 15 circles wide. This gives a \ntotal of 15 × 16 = 240 circles. But we must compensate for the fact that the yellow circles \nwere originally not there by halving the total number of circles. In other words, the \noriginal figure has 240 ÷ 2 = 120 circles.\n7.\t Use the above reasoning to calculate the number of circles in: \n\t\n(a)\t picture 20\t \t\n\t\n(b)\t picture 35\nMaths_LB_gr8_bookA.indb   89\n2014/07/03   10:53:40 AM\n\n90\t\nMATHEMATICS Grade 8: Term 1\n4.4\t Describing patterns in different ways\nt-shaped numbers…\nThe pattern below is made from squares.\n1\t\n2\t\n3\t\n4\n1.\t (a)\t How many squares will there be in pattern 5? \n\t\n(b)\t How many squares will there be in pattern 15? \n\t\n(c)\t Complete the table.\nPattern number\n1\n2\n3\n4\n5\n6\n20\nNumber of squares\n1\n4\n7\n10\nBelow are three different methods or plans to calculate the number of squares for  \npattern 20. Study each one carefully. \nPlan A: \nTo get from 1 square to 4 squares, you have to add 3 squares. To get from 4 squares to  \n7 squares, you have to add 3 squares. To get from 7 squares to 10 squares, you have to  \nadd 3 squares. So continue to add 3 squares for each pattern until pattern 20.\nPlan B: \nMultiply the pattern number by 3, and subtract 2. So pattern 20 will have  \n20 × 3 − 2 squares.\nPlan C: \nThe number of squares in pattern 5 is 13. So pattern 20 will have 13 × 4 = 52 squares \nbecause 20 = 5 × 4.\n2.\t (a)\t Which method or plan (A, B or C) will give the right answer? Explain why.\n\t\n(b)\t Which of the above plans did you use? Explain why?\nMaths_LB_gr8_bookA.indb   90\n2014/07/03   10:53:40 AM\n\n\t CHAPTER 4: NUMERIC AND GEOMETRIC PATTERNS\t\n91\n\t\n(c)\t Can this flow diagram be used to calculate the number of squares? \nInput\n× 3\n– 2\nOutput\n… and some other shapes\n1.\t Three figures are given below. Draw the next figure in the tile pattern.\n1\n2\n3\n4\n2.\t (a)\t If the pattern is continued, how many tiles will there be in the 17th figure?  \n    Answer this question by analysing what happens.\n\t\n(b)\t Thato decides that it easier for him  \n\t\nto see the pattern when the tiles are  \n\t\nrearranged as shown here:  \n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\nUse Thato’s method to determine the number of tiles in the 23rd figure.\n\t\n(c)\t Complete the flow diagram below by writing the appropriate operators so that \n    it can be used to calculate the number of tiles in any figure of the pattern. \nInput\n1\n2\n3\n17\nOutput\n5\n8\n11\n53\n\t\n(d)\t How many tiles will there be in the 50th figure if the pattern is continued?\n3 × 1 + 2\n3 × 2 + 2\n3 × 3 + 2\nMaths_LB_gr8_bookA.indb   91\n2014/07/03   10:53:41 AM\n\nWorksheet\nNUMERIC AND GEOMETRIC PATTERNS\n1.\t Write down the next four terms in each sequence. Also explain, in each case, how \t\n\t\nyou figured out what the terms are.\n\t\n(a)\t2; 4; 8; 14; 22; 32; 44; \n\t\n(b)\t2; 6; 18; 54; 162; \n\t\n(c)\t1; 7; 13; 19; 25; \n2.\t (a)\tComplete the table below by calculating the missing terms.\nPosition in sequence\n1\n2\n3\n4\n5\n7\n10\nTerm\n3\n10\n17\n\t\n(b)\tWrite the rule to calculate the term from the position in the sequence \t \t\n\t\n\t\nin words.\n3.\t Consider the stacks below.\n\t\nStack 1\nStack 2\nStack 3\n\t\n(a)\tHow many cubes will there be in stack 5? \n\t\n(b)\tComplete the table.\nStack number\n1\n2\n3\n4\n5\n6\n10\nNumber of cubes\n1\n8\n27\n\t\n(c)\tWrite down the rule to calculate the number of cubes for any stack number.\nMaths_LB_gr8_bookA.indb   92\n2014/07/03   10:53:41 AM\n\nChapter 5\nFunctions and relationships\n\t\nCHAPTER 5: FUNCTIONS AND RELATIONSHIPS\t\n93\nIn this chapter you will learn about quantities that change, such as the height of a tree.  \nAs the tree grows, the height changes. A quantity that changes is called a variable quantity \nor just a variable. \nIt is often the case that when one quantity changes, another quantity also changes. For \nexample, as you make more and more calls on a phone, the total cost increases. In this \ncase, we say there is a relationship between the amount of money you have to pay and \nthe number of calls you make. You will learn how to describe a relationship between two \nquantities in different ways.\n5.1\t Constant and variable quantities............................................................................... 95\n5.2\t Different ways to describe relationships.................................................................. 100\n5.3\t Algebraic symbols for variables and relationships.................................................... 103\nMaths_LB_gr8_bookA.indb   93\n2014/07/03   10:53:41 AM\n\n94\t\nMATHEMATICS Grade 8: Term 1\nMaths_LB_gr8_bookA.indb   94\n2014/07/03   10:53:41 AM\n\n\t\nCHAPTER 5: FUNCTIONS AND RELATIONSHIPS\t\n95",
      "chapter_id": "4"
    },
    {
      "title": "Functions and relationships",
      "content": "5\t Functions and relationships\n5.1\t Constant and variable quantities\nLooking for connections between quantities\nConsider the following seven situations. There are two quantities in each situation. \nFor each quantity, state whether it is constant (always the same number) or whether it \nchanges. Also state, in each case, whether one quantity has an influence on the other. If \nit has, try to say how the one quantity will influence the other quantity.\n1.\t Your age and the number of fingers on your hands\n2.\t The number of calls you make and the airtime left on your cellphone\n3.\t The length of your arm and your ability to finish Mathematics tests quickly\n4.\t The number of identical houses to be built and the number of bricks required\n5.\t The number of learners at a school and the length of the school day\n6.\t The number of learners at a school and the number of classrooms needed\nMaths_LB_gr8_bookA.indb   95\n2014/07/03   10:53:41 AM\n\n96\t\nMATHEMATICS Grade 8: Term 1\n7.\t The number of matches in each arrangement here, and the number of triangles in the \narrangement\nIf one variable quantity is influenced by another,  \nwe say there is a relationship between the two \nvariables. It is sometimes possible to find out what \nvalue of the one quantity, in other words what \nnumber, is linked to a specific value of the other \nquantity.\n8.\t (a)\t Look at the match arrangements in question 7. If you know that there are 3  \n\t\ntriangles in an arrangement, can you say with certainty how many matches  \n\t\nthere are in that specific arrangement?\n\t\n(b)\t How many matches are there in the arrangement with 10 triangles? \n \n\t\n(c)\t Is there another possible answer for question (b)? \n9.\t Complete the flow diagram by filling in all the missing numbers. Do you see any \nconnections between the situation in question 7 and this flow diagram? If so, \ndescribe the connections.\n× 2\n+ 1\n2\n4\n8\n10\n3\n5\n9\n1\n2\n3\n4\n5\n6\n7\nInput numbers\nOutput numbers\nA quantity that changes is \ncalled a variable quantity \nor just a variable.\nMaths_LB_gr8_bookA.indb   96\n2014/07/03   10:53:42 AM\n\n\t\nCHAPTER 5: FUNCTIONS AND RELATIONSHIPS\t\n97\nCompleting some flow diagrams\nA relationship between two quantities can be shown with a flow diagram, such as those \nbelow. Unfortunately, only some of the numbers can be shown on a flow diagram.\n1.\t Calculate the output numbers for the flow diagram below. Some input numbers are \nmissing. Choose and insert your own input numbers.\n\t\n(a)\t\n+ 5\n8\n7\n4\n3\n1\n0\n–1\n\t\n(b)\t What type of numbers are the given input numbers?\n\t\n(c)\t In the above flow diagram, the output number 8 corresponds to the input  \n\t\nnumber 3. Complete the following sentences:\n\t\nIn the relationship shown in the above flow diagram, the output number \n \ncorresponds to the input number −1.\n\t\nThe input number \n corresponds to the output number 7.\n\t\nIf more places are added to the flow diagram, the input number \n will correspond to \nthe output number 31. \n2.\t (a)\t Complete this flow diagram.\n– 5\n7\n6\n5\n4\n3\n\t\n(b)\t Compare this flow diagram to the flow diagram in question 1. What link do you \n\t\nfind between the two?\nEach input number in a flow \ndiagram has a corresponding \noutput number. The first \n(top) input number  \ncorresponds to the first \noutput number. The second \ninput number corresponds to \nthe second output number \nand so on.\nWe call +5 the operator.\nMaths_LB_gr8_bookA.indb   97\n2014/07/03   10:53:42 AM\n\n98\t\nMATHEMATICS Grade 8: Term 1\n3.\t Complete the flow diagrams below. You have to find out what the operator for (b) is, \nand fill it in yourself.\n\t\n(a)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\n– 5\n–105\n–100\n–15\n0\n15\n100\n105\n–23\n5\n–80\n–4\n–15\n–12\n53\n45\n\t\n(c)\t What number can you add in (a), instead of subtracting 5,  \n\t\nthat will produce the same output numbers?\t\t\n\t\n\t\n\t\n\t\n(d)\t What number can you subtract in (b), instead of adding a  \n\t\nnumber, that will produce the same output numbers?\t\n\t\n4.\t Complete the flow diagram:\n× 6\n+ 40\n–1\n–2\n–3\n–4\n–5\nA completed flow diagram shows two kinds of information:\n• It shows what calculations are done to produce the \noutput numbers.\n• It shows which output number is connected to \nwhich input number.\nThe flow diagram that you have completed in question 4 shows the following \ninformation:\n• Each input number is multiplied by 6, then 40 is added to produce the output \nnumbers.\n• The input and output numbers are connected as shown in the table below.\nInput numbers\n−1\n−2\n−3\n−4\n−5\nOutput numbers\n34\n28\n22\n16\n10\nMaths_LB_gr8_bookA.indb   98\n2014/07/03   10:53:42 AM\n\n\t\nCHAPTER 5: FUNCTIONS AND RELATIONSHIPS\t\n99\n5.\t (a)\t Describe in words how the output numbers below can be calculated.\n10\n20\n30\n40\n50\n× 4\n– 15\n\t\n(b)\t Use the table below to show which output numbers are connected to which  \n\t\ninput numbers in the above flow diagram.\n6.\t The following information is available about the floor space and cost of new houses \nin a new development. The cost of an empty stand is R180 000.\nFloor space in  \nsquare metres\n90\n120\n150\n180\n210\nCost of house and stand\n540 000\n660 000\n780 000\n900 000\n1 020 000\n\t\n(a)\t Represent the above information on the flow diagram below.\n\t\n(b)\t Show what the houses only will cost, if you get the stand for free.\n\t\n(c)\t Try to figure out what the cost of a house and stand will be, if there are exactly  \n\t\none hundred 1 m by 1 m sections of floor space in the house. \nMaths_LB_gr8_bookA.indb   99\n2014/07/03   10:53:42 AM\n\n100\t MATHEMATICS Grade 8: Term 1\n5.2\t Different ways to describe relationships\nA relationship between red dots and blue dots\nHere is an example of a relationship between two quantities:\n \nIn each arrangement there are some red dots and some blue dots.\n1.\t How many blue dots are there if there is one red dot? \n2.\t How many blue dots are there if there are two red dots? \n\t\n3.\t How many blue dots are there if there are three red dots? \n4.\t How many blue dots are there if there are four red dots? \n5.\t How many blue dots are there if there are five red dots? \n6.\t How many blue dots are there if there are six red dots? \n7.\t How many blue dots are there if there are seven red dots? \n8.\t How many blue dots are there if there are ten red dots? \n9.\t How many blue dots are there if there are twenty red dots? \n10. How many blue dots are there if there are one hundred red dots? \n11.\tWhich of the descriptions on the next page correctly \ndescribe the relationship between the number of blue  \ndots and the number of red dots in the above \narrangements? Test each description thoroughly for all  \nthe above arrangements. List them on the dotted line \nbelow. Write only the letters, for example (d).\nSomething to think about\nAre there different possibilities \nfor the number of blue dots \nif there are 3 red dots in the \narrangement?\nAre there different possibili-\nties for the number of blue \ndots if there are 2 red dots in \nthe arrangement?\nAre there different possibili-\nties for the number of blue \ndots if there are 20 red dots \nin the arrangement?\nMaths_LB_gr8_bookA.indb   100\n2014/07/03   10:53:42 AM\n\n\t\nCHAPTER 5: FUNCTIONS AND RELATIONSHIPS\t\n101\n\t\n(a)\t the number of red dots  \n× 4\n+ 2\n the number of blue dots\n\t\n(b)\t to calculate the number of blue dots you multiply the number of red dots  \n\t\nby 2, add 1 and multiply the answer by 2\n\t\n(c)\t number of blue dots = 2 × the number of red dots + 4\n(d)\nNumber of red dots\n1\n2\n3\n4\n5\n6\nNumber of blue dots\n6\n10\n14\n18\n22\n26\n\t\n(e)\n× 4\n+ 2\n1\n2\n3\n4\n5\n\t\n(f)\n× 2\n+ 1\n1\n2\n3\n4\n5\n× 2\n\t\n(g)\t number of blue dots = 4 × the number of red dots + 2\n\t\n(h)\t number of blue dots = 2 × (2 × the number of red dots + 1) \n\t\n(Remember that the calculations inside the brackets are done first.)\nThe descriptions in (c), (g) and (h) above are called \nword formulae.\nMaths_LB_gr8_bookA.indb   101\n2014/07/03   10:53:42 AM\n\n102\t MATHEMATICS Grade 8: Term 1\nTranslating between different languages of description\nA relationship between two quantities can be \ndescribed in different ways, including the following:\n• a table of values of the two quantities\n• a flow diagram\n• a word formula\n• a symbol formula (or symbolic formula)\n1.\t The relationship between two quantities is described as follows:\n\t\nThe second quantity is always 3 times the first quantity plus 8.\n\t\nThe first quantity varies between 1 and 5, and it is always a whole number.\n\t\n(a)\t Describe this relationship with the flow diagram.\n\t\n(b)\t Describe the relationship between the two quantities with this table.\n\t\n(c)\t Describe the relationship between the two quantities with a word formula.\n2.\t The relationship between two quantities is described as follows:\n\t\nThe input numbers are the first five odd numbers.\nvalue of the one \nquantity\n \n+ 5\n× 3\nthe corresponding value of \nthe other quantity\n\t\n(a)\t Describe this relationship with a table.\n\t\n(b)\t Describe the relationship with a word formula.\nYou will learn about symbolic \nformulae in section 5.3.\nMaths_LB_gr8_bookA.indb   102\n2014/07/03   10:53:43 AM\n\n\t\nCHAPTER 5: FUNCTIONS AND RELATIONSHIPS\t\n103\n5.3\t Algebraic symbols for variables and relationships\nDescribing procedures in different ways\n1.\t In each case do four things: \n• Complete the table. \n• Describe the relationship with a word formula.\n• Describe the input numbers in words.\n• Describe the output numbers in words.\n\t\n(a)\t input number  \n× 10\n+ 15\n output number\nInput number\n5\n10\n15\n20\n25\n30\nOutput number\n\t\n\t\nthe output number = \n\t\n(b)\t input number  \n+ 15\n× 10\n output number\nInput number\n5\n10\n15\n20\n25\n30\nOutput number\n\t\n(c)\t input number \n× 2\n+ 3\n× 5\n output number\nInput number\n5\n10\n15\n20\n25\n30\nOutput number\nMaths_LB_gr8_bookA.indb   103\n2014/07/03   10:53:43 AM\n\n104\t MATHEMATICS Grade 8: Term 1\nFormulae with symbols\nInstead of writing “input number” and “output number” in formulae, one may just write \na single letter symbol as an abbreviation. \nMathematicians have long ago adopted the  \nconvention of using the letter symbol x as an  \nabbreviation for “input number”, and the letter  \nsymbol y as an abbreviation for “output number”.\nThe word formula you wrote for question 1(a) can be written more shortly as\n\t\n\t\ny = 10 × x + 15\nMathematicians have also long ago agreed that one  \nmay leave the ×-sign out when writing symbolic \nformulae.\nSo, instead of y = 10 × x + 15 we may write y = 10x + 15.\n2.\t Rewrite your word formulae in questions 1(b) and 1(c) as symbolic formulae.\n3.\t Write a word formula for each of the following relationships:\n\t\n(a)\t y = 7x + 10  \n\t\n\t\n(b)\t y = 7(x + 10)  \n\t\n(c)\t y = 7(2x + 10)  \nWriting symbolic FORMULAE\nDescribe each of the following relationships with a symbolic formula:\n1.\t To calculate the output number, the input number is multiplied by 4 and 7 is \nsubtracted from the answer.\n2.\t To calculate the output number, 7 is subtracted from the input number and the \nanswer is multiplied by 5.\n3.\t To calculate the output number, 7 is subtracted from the input number, the answer is \nmultiplied by 5 and 3 is added to this answer.\nOther letter symbols than x \nand y are also used to indicate \nvariable quantities.\nIt is not at all wrong to use \nthe multiplication sign in \nsymbolic formulae. \nMaths_LB_gr8_bookA.indb   104\n2014/07/03   10:53:43 AM\n\nChapter 6\nAlgebraic expressions 1\n\t\nCHAPTER 6: ALGEBRAIC EXPRESSIONS 1\t\n105\nAn algebraic expression is a description of certain calculations that have to be done in a  \ncertain order. In this chapter, you will be introduced to the language of algebra. You will \nalso learn about expressions that appear to be different but that produce the same results \nwhen evaluated. When we evaluate an expression, we choose or are given a value of the \nvariable in the expression. Because now we have an actual value, we can carry out the \noperations (+, −, ×, ÷) in the expression using this value.\n6.1\t Algebraic language................................................................................................. 107\n6.2\t Add and subtract like terms.................................................................................... 112\nMaths_LB_gr8_bookA.indb   105\n2014/07/03   10:53:43 AM\n\n106\t MATHEMATICS Grade 8: Term 1\nMaths_LB_gr8_bookA.indb   106\n2014/07/03   10:53:43 AM\n\n\t\nCHAPTER 6: ALGEBRAIC EXPRESSIONS 1\t\n107",
      "chapter_id": "5"
    },
    {
      "title": "Algebraic expressions 1",
      "content": "6\t Algebraic expressions 1\n6.1\t Algebraic language\nwords, diagrams and symbols\n1.\t Complete this table.\nWords\nFlow diagram\nExpression\nMultiply a number by two  \nand add six to the answer.\n× 2\n+ 6\n2 × x + 6\n(a)\nAdd three to a number  \nand then multiply the answer \nby two.\n(b)\n× 5\n– 1\n(c)\n7 + 4 × x \n(d)\n10 − 5 × x\nAn algebraic expression indicates a sequence of \ncalculations that can also be described in words or \nwith a flow diagram.\nThe flow diagram illustrates the order in which the \ncalculations must be done.\nIn algebraic language the multiplication sign \nis usually omitted. So we write 2x instead of 2 × x. \nWe also write x × 2 as 2x.\n2.\t Write the following expressions in ‘normal’ algebraic language:\n\t\n(a)\t −2 × a + b\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t a2\nMaths_LB_gr8_bookA.indb   107\n2014/07/03   10:53:44 AM\n\n108\t MATHEMATICS Grade 8: Term 1\nLooking different but yet the same\n1.\t Complete the table by calculating the numerical values of the expressions for the \nvalues of x. Some answers for x = 1 have been done for you as an example.\nx\n1\n3\n7\n10\n(a)\n2x + 3x \n2 × 1 + 3 × 1 \n2 + 3 = 5\n(b)\n5x\n(c)\n2x + 3\n(d)\n5x2\n5 × (1)2\n5 × 1 = 5\n2.\t Do the expressions 2x + 3x and 5x, in question 1 above, produce different answers or \nthe same answer for:\n\t\n(a)\t x = 3?\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t x = 10?\n3.\t Do the expressions 2x + 3 and 5x produce different answers or the same answer for:\n\t\n(a)\t x = 3?\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t x = 10?\n4.\t Write down all the algebraic expressions in question 1 that have the same numerical \nvalue for the same value(s) of x, although they may look different. Justify your answer.\nOne of the things we do in algebra is to evaluate \nexpressions. When we evaluate an expression we \nchoose or are given a value of the variable in the\nMaths_LB_gr8_bookA.indb   108\n2014/07/03   10:53:44 AM\n\n\t\nCHAPTER 6: ALGEBRAIC EXPRESSIONS 1\t\n109\nexpression. Because now we have an actual value we \ncan carry out the operations in the expression using \nthis value, as in the examples given in the table.\nAlgebraic expressions that have the same  \nnumerical value for the same value of x but look  \ndifferent are called equivalent expressions.\n5.\t Say whether the following statements are true or false. Explain your answer in  \neach case.\n\t\n(a)\t The expressions 2x + 3x and 5x are equivalent.\n\t\n(b)\t The expressions 2x + 3 and 5x are equivalent.\n6.\t Consider the expressions 3x + 2z + y and 6xyz.\n\t\n(a)\t What is the value of 3x + 2z + y for x = 4, y = 7 and  \n\t\nz = 10?\n\t\n(b)\t What is the value of 6xyz for x = 4, y = 7 and z = 10?\n\t\n(c)\t Are the expressions 3x + 2z + y and 6xyz equivalent? Explain.\nTo show that the two expressions in question 5(a) are \nequivalent we write 2x + 3x = 5x. \nWe can explain why this is so:\n2x + 3x = (x + x) + (x + x + x) = 5x \nWe say the expression 2x + 3x simplifies to 5x. \t\nRemember that 6xyz is  \nthe same as 6 × x × y × z.\nThe term 3x is a product.  \nThe number 3 is called the \ncoefficient of x.\nMaths_LB_gr8_bookA.indb   109\n2014/07/03   10:53:44 AM\n\n110\t MATHEMATICS Grade 8: Term 1\n7.\t In each case below, write down an expression equivalent to the one given.\n\t\n(a)\t 3x + 3x\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 3x + 8x + 2x\n\t\n(c)\t 8b + 2b + 2b\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 7m + 2m + 10m\n\t\n(e)\t 3x2 + 3x2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 3x2 + 8x2 + 2x2\n8.\t What is the coefficient of x2 for the expression equivalent to 3x2 + 8x2 + 2x2?\nIn an expression that can be written as a sum, the \ndifferent parts of the expression are called the terms \nof the expression. For example, 3x, 2z and y are the \nterms of the expression 3x + 2z + y.\nAn expression can have like terms or unlike \nterms or both.\nLike terms are terms that have the same \nvariable(s) raised to the same power. The terms \n2x and 3x are examples of like terms.\n9.\t (a)\t Calculate the numerical value of 10x + 2y for x = 3 and y = 2 by completing the  \n\t\nempty spaces in the diagram.\n\t\n\t\nInput value: 3\n\t\n +  \n = \n    (Output value)\n\t\n\t\nInput value: 2\n\t\n(b)\t What is the output value for the expression 12xy for x = 3 and y = 2?\n\t\n(c)\t Are the expressions 10x + 2y and 12xy equivalent? Explain.\n\t\n(d)\t Are the terms 10x and 2y like or unlike terms? Explain.\n× 10\n× 2\nMaths_LB_gr8_bookA.indb   110\n2014/07/03   10:53:44 AM\n\n\t\nCHAPTER 6: ALGEBRAIC EXPRESSIONS 1\t\n111\n10.\t(a)\t Which of the following algebraic expressions do you think will give the same  \n\t\nresults?\n\t\n\t\nA.  6x + 4x\t\nB.  10x\t\nC.  10x2\t\nD.  9x + x\n\t\n(b)\t Test the algebraic expressions you have identified for the following values of x:\n\t\n\t\nx = 10\t\nx = 17\t\nx = 54\n\t\n(c)\t Are the terms 6x and 4x like or unlike terms? Explain.\n\t\n(d)\t Are the terms 10x and 10x2 like or unlike terms? Explain.\n11.\tAshraf and Hendrik have a disagreement about whether the terms 7x2y3 and 301y3x2 \nare like terms or not. Hendrik thinks they are not, because in the first term the x2 \ncomes before the y3 whereas in the second term the y3 comes before the x2.\n\t\nExplain to Hendrik why his argument is not correct.\n12.\tExplain why the terms 5abc, 10acb and 15cba are like terms.\nMaths_LB_gr8_bookA.indb   111\n2014/07/03   10:53:44 AM\n\n112\t MATHEMATICS Grade 8: Term 1\n6.2\t Add and subtract like terms\nrearrange terms and then combine like terms\n1.\t Complete the table by evaluating the expressions for the given values of x.\nx\n1\n2\n10\n30x + 80\n30 × 1 + 80 \n= 30 + 80 = 110\n5x + 20\n30x + 80 + 5x + 20\n35x + 100\n135x\n2.\t Write down all the expressions in the table that are equivalent.\n3.\t Tim thinks that the expressions 135x and 35x + 100 are equivalent because for x = 1 \nthey both have the same numerical value 135. \n\t\nExplain to Tim why the two expressions are not equivalent.\nWe have already come across the commutative and associative properties of operations. \nWe will now use these properties to help us form equivalent algebraic expressions.\nCommutative property\nThe order in which we add or multiply numbers does \nnot change the answer: a + b = b + a and ab = ba\nMaths_LB_gr8_bookA.indb   112\n2014/07/03   10:53:44 AM\n\n\t\nCHAPTER 6: ALGEBRAIC EXPRESSIONS 1\t\n113\nAssociative property\nThe way in which we group three or more numbers \nwhen adding or multiplying does not change the \nanswer: (a + b) + c = a + (b + c) and (ab)c = a(bc)\nWe can find an equivalent expression by rearranging and combining like terms, as \nshown below: \n30x + 80 + 5x + 20\nHence 30x + (80 + 5x) + 20 \nHence 30x + (5x + 80) + 20 \n= (30x + 5x) + (80 + 20)\n= 35x + 100\nLike terms are combined to form a  \nsingle term.\nThe terms 30x and 5x are combined to get the new term 35x, and the terms 80 and 20 are \ncombined to form the new term 100. We say that the expression 30x + 80 + 5x + 20 is \nsimplified to a new expression 35x + 100.\n4.\t Simplify the following expressions:\n\t\n(a)\t 13x + 7 + 6x − 2\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 21x − 8 + 7x + 15\n\t\n(c)\t 18c − 12d + 5c − 7c\t \t\n\t\n\t\n\t\n\t\n(d)\t 3abc + 4 + 7abc − 6\n\t\n(e)\t 12x2 + 2x − 2x2 + 8x\t\n\t\n\t\n\t\n\t\n(f)\t 7m3 + 7m2 + 9m3 + 1\nWhen you are not sure about whether you correctly \nsimplified an expression, it is always advisable \nto check your work by evaluating the original \nexpression and the simplified expression for some \nvalues. This is a very useful habit to have.\nThe terms 80 and 20 are \ncalled constants. The \nnumbers 30 and 5 are called \ncoefficients.\nBrackets are used in the \nexpression on the left to \nshow how the like terms \nhave been rearranged.\nWhen we use a value of the \nvariable in the expression we \ncall that substitution.\nMaths_LB_gr8_bookA.indb   113\n2014/07/03   10:53:44 AM\n\n114\t MATHEMATICS Grade 8: Term 1\n5.\t Make a simpler expression that is equivalent to the given expression. Test your answer \nfor three different values of x, and redo your work until you get it right.\n\t\n(a)\t Simplify (15x + 7y) + (25x + 3 + 2y)\t\n(b)\t Simplify 12mn + 8mn \nIn questions 6 to 8 below, write down the letter representing the correct answer. Also \nexplain why you think your answer is correct.\n6.\t The sum of 5x2 + x + 7 and x − 9 is:\n\t\nA.  x2 − 2\t\nB.  5x2 + 2x +16\t\nC.  5x2 + 16\t\nD.  5x2 + 2x − 2\n7.\t The sum of 6x2 − x + 4 and x2 − 5 is equivalent to:\n\t\nA.  7x2 − x + 9\t\nB.  7x2 − x − 1\t\nC.  6x4 − x − 9\t\nD.  7x4 − x − 1\n8.\t The sum of 5x2 + 2x + 4 and 3x2 − 5x − 1 can be expressed as:\n\t\nA.  8x2 + 3x + 3\t\nB.  8x2 + 3x − 3\t\nC.  8x2 − 3x + 3\t\nD.  8x2 − 3x − 3\nCombining like terms is a useful algebraic habit. It allows us to replace an expression \nwith another expression that may be convenient to work with. \nDo the following questions to get a sense of what we are talking about.\nconvenient replacements\n1.\t Consider the expression x + x + x + x + x + x + x + x + x + x. What is the value of the \nexpression in each of the following cases?\n\t\n(a)\t x = 2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t x = 50\nMaths_LB_gr8_bookA.indb   114\n2014/07/03   10:53:44 AM\n\n\t\nCHAPTER 6: ALGEBRAIC EXPRESSIONS 1\t\n115\n2.\t Consider the expression x + x + x + z + z + y. What is the value of the expression in \neach of the following cases?\n\t\n(a)\t x = 4, y = 7, z = 10\t \t\n\t\n\t\n\t\n\t\n(b)\t x = 0, y = 8, z = 22\n3.\t Suppose you have to evaluate 3x + 7x for x = 20. Will calculating 10 × 20 give the \ncorrect answer? Explain.\nSuppose we evaluate the expression 3x + 7x for x = 20 without first combining the like \nterms. We will have to do three calculations, namely 3 × 20, then 7 × 20 and then find \nthe sum of the two: 3 × 20 + 7 × 20 = 60 + 140 = 200.\nBut if we first combine the like terms 3x and 7x into one term 10x, we only have to do \none calculation: 10 × 20 = 200. This is one way of thinking about the convenience or  \nusefulness of simplifying an algebraic expression.\n4.\t The expression 5x + 3x is given and you are required to evaluate it for x = 8.  \nWill calculating 8 × 8 give the correct answer? Explain.\n5.\t Suppose you have to evaluate 7x + 5 for x = 10. Will calculating 12 × 10 give the correct \nanswer? Explain.\nMaths_LB_gr8_bookA.indb   115\n2014/07/03   10:53:44 AM\n\n116\t MATHEMATICS Grade 8: Term 1\n6.\t The expression 5x + 3 is given and you have to evaluate it for x = 8.  \nWill calculating 8 × 8 give the correct answer? Explain.\nSamantha was asked to evaluate the expression 12x2 + 2x − 2x2 + 8x for x = 12. She \nthought to herself that just substituting the value of x directly into the terms  \nwould require a lot of work. She first combined the like terms as shown below:\n12x2 − 2\n2 + 2  + 8x \n= 10x2 + 10x\nThen for x = 10, Samantha found the value of  \n10x2 + 10x by calculating \n10 × 102 + 10 × 10 \n= 1 000 + 100\n= 1 100\nUse Samatha’s way of thinking for questions 7 to 9.\n7.\t What is the value of 12x + 25x + 75x + 8x when x = 6?\n8.\t Evaluate 3x2 + 7 + 2x2 + 3 for x = 5.\n9.\t When Zama was asked to evaluate the expression 2n − 1 + 6n for n = 4, she wrote \ndown the following:\n\t\n  2n − 1 + 6n = n + 6n = 6n2\n\t\n  Hence for n = 4: 6 × (4)2 = 6 × 8 = 48\n\t\nExplain where Zama went wrong and why.\nThe terms +2x and –2x2 change \npositions by the commutative \nproperty of operations.\nMaths_LB_gr8_bookA.indb   116\n2014/07/03   10:53:44 AM\n\nWorksheet\nALGEBRAIC EXPRESSIONS 1\n1.\t Complete the table.\nWords\nFlow diagram\nExpression\n(a)\nMultiply a number by three \nand add two to the answer.\n× 3\n+ 2\n(b)\n9x − 6\n(c)\n× 7\n– 3\n7x – 3\n2.\t Which of the following pairs consist of like terms? Explain.\n\t\nA.  3y; −7y\t\nB.  14e2; 5e\t\nC.  3y2z; 17y2z\t\nD.  −bcd; 5bd\n3.\t Write the following in the ‘normal’ algebraic way:\n\t\n(a)\tc2 + d3\t\n(b)\t 7 × d × e × f\n4.\t Consider the expression 12x2 − 5x + 3.\n\t\n(a)\tWhat is the number 12 called? \n\t\n(b)\tWrite down the coefficient of x. \n\t\n(c)\tWhat name is given to the number 3? \n5.\t Explain why the terms 5pqr and −10prq and 15qrp are like terms.\n6.\t If y = 7, what is the value of each of the following?\n\t\n(a)\ty + 8\t\n(b)\t9y\t\n(c)\t 7 − y\nMaths_LB_gr8_bookA.indb   117\n2014/07/03   10:53:45 AM\n\nWorksheet\nALGEBRAIC EXPRESSIONS 1\n7.\t Simplify the following expressions:\n\t\n(a)\t18c + 12d + 5c − 7c\t\n(b)\t 3def + 4 + 7def − 6\n8.\t Evaluate the following expressions for y = 3, z = −1:\n\t\n(a)\t2y2 + 3z\t\n(b)\t (2y)2 + 3z\n9.\t Write each algebraic expression in the simplest form.\n\t\n(a)\t5y + 15y\t\n(b)\t 5c + 6c − 3c + 2c\n\t\n(c)\t4b + 3 + 16b − 5\t\n(d)\t 7m + 3n + 2 − 6m\n\t\n(e)\t 5h2 + 17 − 2h2 + 3\t\n(f)\t 7e2f + 3ef + 2 + 4ef\n10. Evaluate the following expressions:\n\t\n(a)\t3y + 3y + 3y + 3y + 3y + 3y for y = 18\n\t\n(b)\t13y + 14 − 3y + 6 for y = 200\n\t\n(c)\t20 − y2 + 101y2 + 80 for y = 1\n\t\n(d)\t12y2 + 3yz + 18y2 + 2yz for y = 3 and z = 2\nMaths_LB_gr8_bookA.indb   118\n2014/07/03   10:53:45 AM\n\nChapter 7\nAlgebraic equations 1\n\t\nCHAPTER 7: algebraic EQUATIONS 1\t\n119\nIn this chapter you will learn to find numbers that make certain statements true.  \nA statement about an unknown number is called an equation. When we work to find out \nwhich number will make the equation true, we say we solve the equation. The number that \nmakes the equation true is called the solution of the equation.\n7.1\t Setting up equations............................................................................................... 121 \n7.2\t Solving equations by inspection.............................................................................. 123\n7.3\t More examples....................................................................................................... 124\nMaths_LB_gr8_bookA.indb   119\n2014/07/03   10:53:45 AM\n\n120\t MATHEMATICS Grade 8: Term 1\nMaths_LB_gr8_bookA.indb   120\n2014/07/03   10:53:45 AM\n\n\t\nCHAPTER 7: algebraic EQUATIONS 1\t\n121",
      "chapter_id": "6"
    },
    {
      "title": "Algebraic equations 1",
      "content": "7\t Algebraic equations 1\n7.1\t\nSetting up equations\nAn equation is a mathematical sentence that is true \nfor some numbers but false for other numbers. The \nfollowing are examples of equations:\n\t \t\n\t\nx + 3 = 11  and  2x = 8 \nx + 3 = 11 is true if x = 8, but false if x = 3.\nWhen we look for a number or numbers that make \nan equation true we say that we are solving the \nequation. For example, x = 4 is the solution of  \n2x = 8 because it makes 2x = 8 true. (Check: 2 × 4 = 8)\nlooking for numbers to make statements true\n1.\t Are the following statements true or false? Justify your answer.\n\t\n(a)\t x − 3 = 0, if x = −3\n\t\n(b)\t x3 = 8, if x = −2\n\t\n(c)\t 3x = −6, if x = −3\n\t\n(d)\t 3x = 1, if x = 1\n\t\n(e)\t 6x + 5 = 47, if x = 7\n2.\t Find the original number. Show your reasoning.\n\t\n(a)\t A number multiplied by 10 is 80.\n\t\n(b)\t Add 83 to a number and the answer is 100.\n\t\n(c)\t Divide a number by 5 and the answer is 4.\nMaths_LB_gr8_bookA.indb   121\n2014/07/03   10:53:45 AM\n\n122\t MATHEMATICS Grade 8: Term 1\n\t\n(d)\t Multiply a number by 4 and the answer is 20.\n\t\n(e)\t Twice a number is 100.\n\t\n(f)\t A number raised to the power 5 is 32.\n\t\n(g)\t A number raised to the power 4 is −81. \n\t\n(h)\t Fifteen times a number is 90.\n\t\n(i)\t 93 added to a number is −3.\n\t\n(j)\t Half a number is 15.\n3.\t Write the equations below in words using “a number” in place of the letter symbol x. \nThen write what you think “the number” is in each case. \n\t\nExample: 4 + x = 23. Four plus a number equals twenty-three. The number is 19.\n\t\n\t\n(a)\t 8x = 72\n\t\n(b)\t 2\n5\nx = 2 \n\t\n(c)\t 2x + 5 = 21\n\t\n(d)\t 12 + 9x = 30\n\t\n(e)\t 30 − 2x = 40\n\t\n(f)\t 5x + 4 = 3x + 10\nMaths_LB_gr8_bookA.indb   122\n2014/07/03   10:53:46 AM\n\n\t\nCHAPTER 7: algebraic EQUATIONS 1\t\n123\n7.2\t\nSolving equations by inspection\nTHE ANSWER IS IN PLAIN SIGHT\n1.\t Seven equations are given below the table. Use \nthe table to find out for which of the given values \nof x it will be true that the left-hand side of the \nequation is equal to the right-hand side.\nx\n−3\n−2\n−1\n0\n1\n2\n3\n4\n2x + 3\n−3\n−1\n1\n3\n5\n7\n9\n11\nx + 4\n1\n2\n3\n4\n5\n6\n7\n8\n9 − x\n12\n11\n10\n9\n8\n7\n6\n5\n3x − 2\n−11\n−8\n−5\n−2\n1\n4\n7\n10\n10x − 7\n−37\n−27\n−17\n−7\n3\n13\n23\n33\n5x + 3\n−12\n−7\n−2\n3\n8\n13\n18\n23\n10 − 3x\n19\n16\n13\n10\n7\n4\n1\n−2\n\t\n(a)\t 2x + 3 = 5x + 3\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t 5x + 3 = 9 − x\n\t\n(c)\t 2x + 3 = x + 4\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 10x − 7 = 5x + 3\n\t\n(e)\t 3x − 2 = x + 4\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 9 − x = 2x + 3\n\t\n(g)\t 10 − 3x = 3x − 2\nTwo or more equations can have the same solution.  \nFor example, 5x = 10 and x + 2 = 4 have the same  \nsolution; x = 2 is the solution for both equations.\n2.\t Which of the equations in question 1 have the same solutions? Explain.\nYou can read the solutions of \nan equation from a table.\nTwo equations are called \nequivalent if they have the \nsame solution.\nMaths_LB_gr8_bookA.indb   123\n2014/07/03   10:53:46 AM\n\n124\t MATHEMATICS Grade 8: Term 1\n3.\t Complete the table below.  \n\t\nThen answer the questions that follow.\nYou can also do a search by \nnarrowing down the possible \nsolution to an equation.\nx\n0\n5\n10\n15\n20\n25\n30\n35\n40\n2x + 3\n3x − 10\n\t\n(a)\t Can you find a solution for 2x + 3 = 3x − 10 in the table? \n\t\n(b)\t What happens to the values of 2x + 3 and 3x − 10 as x increases? Do they become \n\t\nbigger or smaller?  \n\t\n\t\n \n\t\n(c)\t Is there a point where the value of 3x − 10  \n\t\nbecomes bigger or smaller than the value of  \n\t\n2x + 3 as the value of x increases? If so,  \n\t\nbetween which x-values does this happen?\n\t\n(d)\t Now that you narrowed down where  \n\t\nthe possible solution can be, try other  \n\t\npossible values for x until you find out \n\t\nfor what value of x the statement  \n\t\n2x + 3 = 3x − 10 is true. \n7.3\t\nMore examples\nlooking for and checking solutions\n1.\t What is the solution for the equations below?\n\t\n(a)\t x − 3 = 4\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t x + 2 = 9\n\t\n(c)\t 3x = 21\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 3x + 1 = 22\nThis point where the two \nexpressions are equal is called \nthe break-even point.\n“Searching” for the  \nsolution of an equation by \nusing tables or by narrowing \ndown to the possible  \nsolution is called solution \nby inspection.\nMaths_LB_gr8_bookA.indb   124\n2014/07/03   10:53:46 AM\n\n\t\nCHAPTER 7: algebraic EQUATIONS 1\t\n125\nWhen a certain number is the solution of an equation \nwe say that the number satisfies the equation. For \nexample, x = 4 satisfies the equation 3x = 12 because  \n3 × 4 = 12.\n2.\t Choose the number in brackets that satisfies the equation. Explain your choice. \n\t\n(a)\t 12x = 84\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n{5; 7; 10; 12}\n\t\n(b)\t 84\nx  = 12\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n{−7; 0; 7; 10}\n\t\n(c)\t 48 = 8k + 8\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n{−5; 0; 5; 10}\n\t\n(d)\t 19 − 8m = 3\t\n\t\n\t\n\t\n\t\n\t\n\t\n{−2; −1; 0; 1; 2}\n\t\n(e)\t 20 = 6y − 4\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n{3; 4; 5; 6}\n\t\n(f)\t x3 = −64\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n{−8; −4; 4; 8}\n\t\n(g)\t 5x =125\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n{−3; −1; 1; 3}\n\t\n(h)\t 2x = 8\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n{1; 2; 3; 4}\n\t\n(i)\t x2 = 9\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n{1; 2; 3; 4}\n3.\t What makes the following equations true? Check your answers.\n\t\n(a)\t m + 8 = 100\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 80 = x + 60\n\t\n(c)\t 26 − k = 0\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 105 × y = 0\nMaths_LB_gr8_bookA.indb   125\n2014/07/03   10:53:46 AM\n\n126\t MATHEMATICS Grade 8: Term 1\n\t\n(e)\t k ×10 = 10\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 5x = 100\n\t\n(g)\t 15\nt  = 5\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(h)\t3 = t\n5 \t \t\n4.\t Solve the equations below by inspection. Check your answers.\n\t\n(a)\t 12x + 14 = 50\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 100 = 15m + 25\n\t\n(c)\t 100\nx\n = 20\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 7m + 5 = 40\n\t\n(e)\t 2x + 8 = 10\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 3x + 10 = 31\n\t\n(g)\t −1 + 2x = −11\t\n\t\n\t\n\t\n\t\n\t\n\t\n(h)\t2 + x\n7  = 5\n\t\n(i)\t 100 = 64 + 9x\t \t\n\t\n\t\n\t\n\t\n\t\n(j)\t 2\n6\nx  = 4\nMaths_LB_gr8_bookA.indb   126\n2014/07/03   10:53:47 AM\n\nTerm 1\nRevision and assessment",
      "chapter_id": "7"
    },
    {
      "title": "Term 1: Revision and assessment",
      "content": "TERm 1: REVISION AND ASSESSMENT\t\n127\nRevision  ......................................................................................................................... 128\n• Whole numbers  ........................................................................................................ 128\n• Integers  .................................................................................................................... 130\n• Exponents  ................................................................................................................ 132\n• Numeric and geometric patterns  .............................................................................. 133\n• Functions and relationships  ...................................................................................... 136\n• Algebraic expressions 1  ............................................................................................ 137\n• Algebraic equations 1  ............................................................................................... 138\nAssessment  .................................................................................................................... 140\nMaths_LB_gr8_bookA.indb   127\n2014/07/03   10:53:47 AM\n\n128\t MATHEMATICS Grade 8: Term 1\nRevision\nShow all your steps of working.\nwhole numbers\n1.\t (a)\t Write both 300 and 160 as products of prime factors.\n\t\n(b)\t Determine the HCF and LCM of 300 and 160.\n2.\t Tommy, Thami and Timmy are given birthday money by their grandmother in the \nratio of their ages. They are turning 11, 13 and 16 years old, respectively. If the total \namount of money given to all three boys is R1 000, how much money does Thami  \nget on his birthday?\n3.\t Tshepo and his family are driving to the coast on holiday. The distance is 1 200 km \nand they must reach their destination in 12 hours. After 5 hours, they have travelled \n575 km. Then one of their tyres bursts. It takes 45 minutes to get the spare wheel on, \nbefore they can drive again. At what average speed must they drive the remainder of \nthe journey to reach their destination on time?\n4.\t The number of teachers at a school has increased in the ratio 5 : 6. If there used to be \n25 teachers at the school, how many teachers are there now?\nMaths_LB_gr8_bookA.indb   128\n2014/07/03   10:53:48 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n129\n5.\t ABC for Life needs to have their annual statements audited. They are quoted  \nR8 500 + 14% VAT by Audits Inc. How much will ABC for Life have to pay Audits Inc. \nin total?\n6.\t Reshmi invests R35 000 for three years at an interest rate of 8,2% per annum. \nDetermine how much money will be in her account at the end of the investment \nperiod.\n7.\t Lesebo wants to buy a lounge suite that costs R7 999 cash. He does not have enough \nmoney and so decides to buy it on hire purchase. The store requires a 15% deposit  \nup front, and 18 monthly instalments of R445.\n\t\n(a)\t Calculate the deposit that Lesebo must pay.\n\t\n(b)\t How much extra does Lesebo pay because he buys the lounge suite on hire  \n\t\npurchase, rather than in cash?\nMaths_LB_gr8_bookA.indb   129\n2014/07/03   10:53:48 AM\n\n130\t MATHEMATICS Grade 8: Term 1\n8.\t Consider the following exchange rates table:\nSouth African Rand\n1.00 ZAR\n inv. 1.00 ZAR\nEuro\n0.075370\n13.267807\nUS Dollar\n0.098243\n10.178807\nBritish Pound\n0.064602\n15.479409\nIndian Rupee\n5.558584\n  0.179902\nAustralian Dollar\n0.102281\n  9.776984\nCanadian Dollar\n0.101583\n  9.844200\nEmirati Dirham\n0.360838\n  2.771327\nSwiss Franc\n0.093651\n10.677960\nChinese Yuan Renminbi\n0.603065\n  1.658195\nMalaysian Ringgit\n0.303523\n  3.294646\n\t\n(a)\t Write down the amount in rand that needs to be exchanged to get 1 Swiss franc.  \n\t\nGive your answer to the nearest cent.\n\t\n(b)\t Write down the only currency for which an exchange of R100 will give you  \n\t\nmore than 100 units of that currency.\n\t\n(c)\t Ntsako is travelling to Dubai and converts R10 000 into Emirati dirhams. How  \n\t\nmany dirhams does Ntsako receive (assume no commission)?\nintegers\nDon’t use a calculator for any of the questions in this section.\n1.\t Write a number in each box to make the calculations correct.\n\t\n(a)\t\n + \n = −11\t \t\n\t\n\t\n\t\n\t\n(b)\t\n − \n = −11\n2.\t Fill <, > or = into each block to show the relationships.\n\t\n(a)\t −23 \n 20\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −345 \n −350\n\t\n(c)\t 4 − 3 \n 3 − 4 \t\n\t\n\t\n\t\n\t\n\t\n(d)\t 5 − 7 \n −(7 − 5)\nMaths_LB_gr8_bookA.indb   130\n2014/07/03   10:53:48 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n131\n\t\n(e)\t −9 × 2 \n −9 × −2\t\n\t\n\t\n\t\n\t\n(f)\t −4 × 5 \n 4 × −5\n\t\n(g)\t −10 ÷ 5 \n −10 ÷ −2\t \t\n\t\n\t\n\t\n(h)\t−15 × −15 \n 224\n3.\t Follow the pattern to complete the number sequences.\n\t\n(a)\t 8; 5; 2; \n\t\n(b)\t 2; −4; 8; \n\t\n(c)\t −289; −293; −297; \n4.\t Look at the number lines. In each case, the missing number is halfway between the \nother two numbers. Fill in the correct values in the boxes.\n\t\n(a)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n–456\n–448\n–11\n5\n5.\t Calculate the following:\n\t\n(a)\t −5 − 7 \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 7 − 10\n\t\n(c)\t 8 − (−9)\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t (−5)(−2)(−4)\n\t\n(e)\t 5 + 4 × −2\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t (\n)(\n)\n4\n2\n4\n2\n−\n−\n\t\n(g)\t −−\n−\n(\n)\n(\n)( )\n3\n125\n9 3\n3 3\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n(h)\t\n−\n−−\n64\n3\n1\n3\n6.\t (a)\t Write down two numbers that multiply to give −15. (One of the numbers must  \n\t\nbe positive and the other negative.)\n\t\n(b)\t Write down two numbers that add to 15. One of the numbers must be positive \n\t\nand the other negative.\nMaths_LB_gr8_bookA.indb   131\n2014/07/03   10:53:48 AM\n\n132\t MATHEMATICS Grade 8: Term 1\n7.\t At 5 a.m., the temperature in Kimberley was −3 °C. At 1 p.m., it was 17 °C. By how \nmany degrees had the temperature risen?\n8.\t A submarine is 220 m below the surface of the sea. It travels 75 m upwards. How far \nbelow the surface is it now?\nexponents\nYou should not use a calculator for any of the questions in this section.\n1.\t Write down the value of the following:\n\t\n(a)\t (−3)3\t\n\t\n\t\n(b)\t −52\t\t\n\t\n\t\n(c)\t (−1)200\t \t\n\t\n(d)\t (102)2\t\n\t\n2.\t Write the following numbers in scientific notation:\n\t\n(a)\t 200 000\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 12,345\n3.\t Write the following numbers in ordinary notation:\n\t\n(a)\t 1,3 × 102\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 7,01 × 107\n4.\t Which of the following numbers is bigger: 5,23 × 1010 or 2,9 × 1011?\nMaths_LB_gr8_bookA.indb   132\n2014/07/03   10:53:48 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n133\n5.\t Simplify the following:\n\t\n(a)\t 27 × 23 \n\t\n(b)\t 2x3 × 4x4 \n\t\n(c)\t (−8y6) ÷ (4y3) \n\t\n(d)\t (3x8)3 \n\t\n(e)\t (2x5)(0,5x−5) \n\t\n(f)\t (−3a2b3c)(−4abc2)2 \n\t\n(g)\t (\n)(\n)\n2\n5\n20\n2\n3\n2\n2\n8\n4\nxy z\ny z\nxy z\n−\n \n6.\t Write down the values of each of the following:\n\t\n(a)\t (0,6)2\t \t\n\t\n(b)\t (0,2)3\t \t\n\t\n(c)\t\n1\n2\n5\n\n\n\n\t\n\t\n\t\n(d)\t\n1\n4 \t\n\t\n  \n\t\n(e)\t 4\n9\n64 \t \t\n\t\n(f)\t\n0 001\n3\n,\n\t\t\nnumeric and geometric patterns\n1.\t For each of the following sequences, write the rule for the relationship between each \nterm and the following term in words. Then use the rule to write the next three terms \nin the sequence.\n\t\n(a)\t 12; 7; 2; \n ; \n ; \n\t\n(b)\t −2; −6; −18; \n ; \n ; \n\t\n(c)\t 100; −50; 25; \n ; \n ; \n\t\n(d)\t 3; 4; 7; 11; \n ; \n ; \nMaths_LB_gr8_bookA.indb   133\n2014/07/03   10:53:50 AM\n\n134\t MATHEMATICS Grade 8: Term 1\n2.\t In this question, you are given the rule by which each term of the sequence can be \nfound. In all cases, n is the position of the term. \n\t\nDetermine the first three terms of each of the sequences. (Hint: Substitute n = 1 to  \nfind the value of the first term.)\n\t\n(a)\t n × 4\n\t\n(b)\t n × 5 − 12\n\t\n(c)\t 2 × n2\n\t\n(d)\t 3n ÷ 3 × −2\n3.\t Write down the rule by which each term of the sequence can be found (in a similar \nformat to those given in question 2, where n is the position of the term).\n\t\n(a)\t 2; 4; 6; …\n\t\n(b)\t –7; –3; 1; ...\n\t\n(c)\t 2; 4; 8; …\n\t\n(d)\t 9; 16; 23; …\n4.\t Use the rules you have found in question 3 to find the value of the 20th term of the \nsequences in questions 3(a) and 3(b).\n\t\n(a)\t\n\t\n(b)\t\nMaths_LB_gr8_bookA.indb   134\n2014/07/03   10:53:50 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n135\n5.\t Find the relationship between the position of the term in the sequence and the value \nof the term, and use it to fill in the missing values in the tables.\n(a)\nPosition in \nsequence\n1\n2\n3\n4\n25\nValue of the \nterm\n−8\n−11\n−14\n(b)\nPosition in \nsequence\n1\n2\n3\nValue of the \nterm\n1\n3\n9\n243\n19 683\n6.\t The image below shows a series of patterns created by matches. \n\t\n(a)\t Write in words the rule that describes the number of matches needed for  \n\t\neach new pattern.\n\t\n(b)\t Use the rule to determine the missing values in the table below, and fill them in.\nNumber of the pattern\n1\n2\n3\n4\n20\nNumber of matches needed\n4\n7\n151\nMaths_LB_gr8_bookA.indb   135\n2014/07/03   10:53:50 AM\n\n136\t MATHEMATICS Grade 8: Term 1\nfunctions and relationships\n1.\t Fill in the missing input values, output values or rule in these flow diagrams. Note \nthat p and t are integers.\n\t\n(a)\n1\n2\n4\n10\np\nt\nt = –2(p – 3)\n\t\n(b)\n–1\n–3\n–7\np\nt\nt = p2 – 1\n\t\n(c)\n2\n4\n8\n30\np\nt = (p ÷   \n   ) +   \n50\n100\n0\n–1\n–3\n–14\nt\n2.\t Consider the values in the following table. The rule for finding y is: divide x by −2  \nand subtract 4. Use the rule to determine the missing values in the table, and write \nthem in.\nx\n−2\n0\n2\n5\ny\n−1\n−3\n−4\n48\nMaths_LB_gr8_bookA.indb   136\n2014/07/03   10:53:50 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n137\n3.\t Consider the values in the following table:\nx\n−2\n−1\n0\n1\n2\n4\n15\ny\n1\n3\n5\n7\n9\n61\n\t\n(a)\t Write in words the rule for finding the y-values in the table.\n\t\n(b)\t Use the rule to determine the missing values in the table, and write them in.\nalgebraic expressions 1\n1.\t Look at this algebraic expression: 5x3 − 9 + 4x − 3x2.\n\t\n(a)\t How many terms does this expression have? \n\t\n(b)\t What is the variable in this expression? \n\t\n(c)\t What is the coefficient of the x2 term? \n\t\n(d)\t What is the constant in this expression? \n\t\n(e)\t Rewrite the expression so that the terms are in order of decreasing powers of x. \n2.\t In this question, x = 6 and y = 17. Complete the rules to show different ways to \ndetermine y if x is known. The first way is done for you:\n\t\n\t\nWay 1:\t Multiply x by 2 and add 5. This can be written as y = 2x + 5\n\t\n(a)\t Way 2: Multiply x by \n and then subtract \n. This can be written as \n\t\n(b)\t Way 3: Divide x by \n and then add \n. This can be written as \n\t\n(c)\t Way 4: Add \n to x, and then multiply by \n. This can be written as \nMaths_LB_gr8_bookA.indb   137\n2014/07/03   10:53:50 AM\n\n138\t MATHEMATICS Grade 8: Term 1\n3.\t Simplify:\n\t\n(a)\t 2x2 + 3x2 \n\t\n(b)\t 9xy − 12yx \n\t\n(c)\t 3y2 − 4y + 3y − 2y2 \n\t\n(d)\t 9m3 + 9m2 + 9m3 − 3 \n4.\t Calculate the value of the following expressions if a = −2; b = 3; c = −1 and d = 0:\n\t\n(a)\t abc\n\t\n(b)\t −a2\n\t\n(c)\t (abc)d\n\t\n(d)\t a + b − 2c\n\t\n(e)\t (a + b)10\nalgebraic equations 1\n1.\t Write equations that represent the given information:\n\t\n(a)\t Nandi is x years old. Shaba, who is y years old, is three years older than Nandi.\n\t\n(b)\t The temperature at Colesberg during the day was x °C. But at night, the  \n\t\ntemperature dropped by 15 degrees to reach −2 °C.\n2.\t Solve the following equations for x:\n\t\n(a)\t x + 5 = 2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t  7 − x = 9\nMaths_LB_gr8_bookA.indb   138\n2014/07/03   10:53:50 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n139\n\t\n(c)\t 3x − 1 = −10\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 2x3 = −16\n\t\n(e)\t 2x = 16\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 2(3)x = 6\n3.\t If 3n − 1 = 11, what is the value of 4n?\n4.\t If c = a + b and a + b + c = 16, determine the value of c. \n5.\t (a)\t If 2a + 3 = b, write down values for a and b that will make the equation true.\n\t\n(b)\t Write down a different pair of values to make the equation true.\nMaths_LB_gr8_bookA.indb   139\n2014/07/03   10:53:50 AM\n\n140\t MATHEMATICS Grade 8: Term 1\nAssessment\nIn this section, the numbers in brackets at the end of a question indicate the number \nof marks the question is worth. Use this information to help you determine how much \nworking is needed. The total number of marks allocated to the assessment is 60.\n1.\t The profits of GetRich Inc. have decreased in the ratio 5 : 3 due to the recession  \nin the country. If their profits used to be R1 250 000, how much are their profits  \nnow?\t\n\t\n\t\n\t\n\t\n    (2)\n2.\t Which car has the better rate of petrol consumption: Ashley’s car, which drove  \n520 km on 32 ℓ of petrol, or Zaza’s car, which drove 880 km on 55 ℓ of  \npetrol? Show all your working.\t\n\t\n\t\n(3)\n3.\t Hanyani took out a R25 000 loan from a lender that charges him 22% interest each \nyear. How much will he owe in one year’s time?\t\n\t\n\t\n(3)\n4.\t Consider the following exchange rates table:\nSouth African Rand\n1.00 ZAR\ninv. 1.00 ZAR\nIndian Rupee\n5.558584\n   0.179902\nAustralian Dollar\n0.102281\n   9.776984\nCanadian Dollar\n0.101583\n   9.844200\nEmirati Dirham\n0.360838\n   2.771327\nChinese Yuan Renminbi\n0.603065\n   1.658195\nMalaysian Ringgit\n0.303523\n   3.294646\nMaths_LB_gr8_bookA.indb   140\n2014/07/03   10:53:50 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n141\n\t\nChen returns from a business trip to Malaysia with 2 500 ringgit in his wallet.  \nIf he changes this money into rand in South Africa, how much will he receive?\t\n(2)\n5.\t Fill <, > or = into the block to show the relationship between the number  \n\t\nexpressions:\n\t\n(a)\t 6 − 4 \n 4 − 6\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n(1)\n\t\n(b)\t 2 × −3 \n −32\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   \t\n\t\n\t\n (1)\n6.\t Look at the number sequence below. Fill in the next term into the block.\n\t\n−5; 10; −20; \n\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   \t  (1)\n7.\t Calculate the following:\n\t\n(a)\t (−4)2 − 20\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n  \t   (2)\n\t\n(b)\t\n−8\n3\n + 14 ÷ 2\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   \t  (2)\n8.\t Julius Caesar was a Roman emperor who lived from 100 BC to 44 BC. How old was  \nhe when he died?\t\n\t\n\t\n\t\n\t\n    (2)\n9.\t (a)\t Write down two numbers that divide to give an answer of −8. One of the \n\t\nnumbers must be positive, and the other negative.\t\n\t\n\t\n\t\n      \t \t\n\t\n\t\n(1)\n\t\n(b)\t Write down two numbers that subtract to give an answer of 8. One of the  \n\t\nnumbers must be positive and the other negative.\t\n\t\n\t\n\t\n \t\n\t\n\t\n\t     (1)\n10.\tWrite the following number in scientific notation: 17 million.\t\n    (2)\nMaths_LB_gr8_bookA.indb   141\n2014/07/03   10:53:51 AM\n\n142\t MATHEMATICS Grade 8: Term 1\n11.\tWhich of the following numbers is bigger: 3,47 × 1021 or 7,99 × 1020?\t\n    (1)\n12.\tSimplify the following, leaving all answers with positive exponents:\n\t\n(a)\t 37 × 3–2 \n    \t(1)\n\t\n(b)\t (−12y8) ÷ (−3y2) \n  \t (2)\n\t\n(c)\t\n(\n)(\n)\n3\n15\n2\n3\n2\n5\n4\n7\nxy z\nyz\nx y z\n−\n \n\t    (4)\n13.\tWrite down the values of each of the following:\n\t\n(a)\t (0,3)3 \n    \t(1)\n\t\n(b)\t 8\n25\n16  \n\t  (2)\n14.\tConsider the following number sequence: 2; −8; 32; … \n\t\n(a)\t Write in words the rule by which each term of the sequence can be found.\t \t\n(1)\n\t\n(b)\t Write the next three terms in this sequence.\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(2)\n15.\tThe picture below shows a series of patterns created by matches. \n\t\n(a)\t Write a formula for the rule that describes the relationship between the number  \n\t\nof matches and the position of the term in the sequence (pattern number). Let n \n\t\nbe the position of the term.\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(2)\nMaths_LB_gr8_bookA.indb   142\n2014/07/03   10:53:51 AM\n\n\t\nTERM 1: REVISION AND ASSESSMENT\t\n143\n\t\n(b)\t Use the rule to determine the values of a to c in the following table:\t \t\n\t\n   \t  (3)\nNumber of the pattern\n1\n2\n3\n4\n15\nc\nNumber of matches needed\n8\n15\n22\na\nb\n148\n16.\tConsider the values in the following table:\nx\n−2\n−1\n0\n1\n2\n5\n12\ny\n−7\n−4\n−1\n2\n5\n98\n\t\n(a)\t Write in words the rule for finding the y-values in the table.\t \t\n\t\n    \t \t\n\t\n(2)\n\t\n(b)\t Use the rule to determine the missing values in the table, and fill them in.\t \t\n(3)\n17.\tSimplify:\n\t\n(a)\t 2z2 − 3z2 \n\t\n(1)\n\t\n(b)\t 8y2 − 6y + 4y − 7y2 \n    \t(2)\n18.\tDetermine the value of 2a2 − 10 if a = −2.\t\n\t\n\t\n    (2)\nMaths_LB_gr8_bookA.indb   143\n2014/07/03   10:53:51 AM\n\n144\t MATHEMATICS Grade 8: Term 1\n19.\tIf c + 2d = 27, give the value of the following:\n\t\n(a)\t 2c + 4d \n  \t  (1)\n\t\n(b)\t c\nd\n+\n−\n2\n9\n \n    (1)\n\t\n(c)\t\nc\nd\n+ 2\n3\n \n\t (1)\n20.\tSolve the following for x:\t \t\n\t\n\t\n\t\n    (5)\n\t\n(a)\t −x = −11\t\n(b)\t 2x − 5 = −11\t\n(c)\t 4x3 = 32\nMaths_LB_gr8_bookA.indb   144\n2014/07/03   10:53:52 AM\n\nChapter 8\nAlgebraic expressions 2\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n145\nIn this chapter, you will learn about simplifying algebraic expressions by expanding them. \nExpanding an algebraic expression allows you to change the form of an expression without \nchanging the output values it gives.\nRewriting an expression in a different form can be useful for simplifying calculations and \ncomparing expressions. We use two main tools to simplify expressions: we combine like \nterms and/or use the distributive property.\n8.1\t Expanding algebraic expressions............................................................................. 147\n8.2\t Simplifying algebraic expressions............................................................................ 152\n8.3\t Simplifying quotient expressions............................................................................. 155\n8.4\t Squares, cubes and roots of expressions.................................................................. 160\nMaths_LB_gr8_bookA.indb   145\n2014/07/03   10:53:52 AM\n\n146\t MATHEMATICS Grade 8: Term 2\nMaths_LB_gr8_bookA.indb   146\n2014/07/03   10:53:53 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n147",
      "chapter_id": "8"
    },
    {
      "title": "Algebraic expressions 2",
      "content": "8\t Algebraic expressions 2\n8.1\t Expanding algebraic expressions\nmultiply often or multiply once: it is your choice \n1.\t (a)\t Calculate 5 × 13 and 5 × 87 and add the two answers.\n\t\n(b)\t Add 13 and 87, and then multiply the answer by 5.\n\t\n(c)\t If you do not get the same answer for questions 1(a) and 1(b), you have made a \n\t\nmistake. Redo your work until you get it right.\nThe fact that, if you work correctly, you get the same \nanswer for questions 1(a) and 1(b) is an example of a \ncertain property of addition and multiplication called  \nthe distributive property. You use this property each \ntime you multiply a number in parts. For example, you \nmay calculate 3 × 24 by calculating \n3 × 20 and 3 × 4, and then add the two answers:\n3 × 24 = 3 × 20 + 3 × 4\nWhat you saw in question 1 was that 5 × 100 = 5 × 13 + 5 × 87.\nThis can also be expressed by writing 5(13 + 87).\n2.\t (a)\t Calculate 10 × 56.\n\t\n(b)\t Calculate 10 × 16 + 10 × 40.\n3.\t Write down any two numbers smaller than 100. Let us call them x and y.  \n(a)\t Add your two numbers, and multiply the answer by 6.\n\t\n(b)\t Calculate 6 × x and 6 × y and add the two answers.\n\t\n(c)\t If you do not get the same answers for (a) and (b) you have made a mistake  \n\t\nsomewhere. Correct your work. \t\nThe word distribute means “to \nspread out”. The distributive \nproperties may be described \nas follows:\na(b + c) = ab + ac and \na(b − c) = ab – ac,\nwhere a, b and c can be any \nnumbers.\nMaths_LB_gr8_bookA.indb   147\n2014/07/03   10:53:53 AM\n\n148\t MATHEMATICS Grade 8: Term 2\n4.\t Complete the table.\n(a)\nx\n1\n2\n3\n4\n5\n3(x + 2)\n3x + 6\n3x + 2\n3(x − 2)\n3x − 6\n3x − 2\n\t\n(b)\t If you do not get the same answers for the expressions 3(x + 2) and 3x + 6, and  \n\t\nfor 3(x − 2) and 3x − 6, you have made a mistake somewhere. Correct your work.\nIn algebra we normally write 3(x + 2) instead of 3 × (x + 2). The expression 3 × (x + 2) \ndoes not mean that you should first multiply by 3 when you evaluate the expression for \na certain value of x. The brackets tell you that the first thing you should do is add the \nvalue(s) of x to 2 and then multiply the answer by 3.\nHowever, instead of first adding the values within the brackets and then multiplying \nthe answer by 3 we may just do the calculation 3 × x + 3 × 2 = 3x + 6 as shown in the table.\n\t\n(c)\t Which expressions amongst those given in the table are equivalent? Explain.\n\t\n(d)\t For what value(s) of x is 3(x + 2) = 3x + 2?\n\t\n(e)\t Try to find a value of x such that 3(x + 2) ≠ 3x + 6.\nIf multiplication is the last step in evaluating an \nalgebraic expression, then the expression is called \na product expression or, briefly, a product. The \nway you evaluated the expression 3(x + 2) in the table \nis an example of a product expression.\nMaths_LB_gr8_bookA.indb   148\n2014/07/03   10:53:53 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n149\n5.\t (a)\t Determine the value of 5x + 15 if x = 6. \n\t\n(b)\t Determine the value of 5(x + 3) if x = 6. \n\t\n(c)\t Can we use the expression 5x + 15 to calculate the value of 5(x + 3) for any values \n\t\nof x? Explain.\n6.\t Complete the flow diagrams.\n\t\n(a)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\n× 2\n2\n4\n8\n10\n12\n12\n+ 8\n× 8\n2\n4\n8\n10\n12\n+ 2\n\t\n(c)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\n+ 8\n2\n4\n8\n10\n12\n× 2\n+ 4\n2\n4\n8\n10\n12\n× 2\n\t\n(e)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\n+ 1\n2\n4\n8\n10\n12\n× 8\n× 2\n2\n4\n8\n10\n12\n+ 16\nMaths_LB_gr8_bookA.indb   149\n2014/07/03   10:53:53 AM\n\n150\t MATHEMATICS Grade 8: Term 2\n7.\t (a)\t Which of the above flow diagrams produce the same output numbers?\n\t\n(b)\t Write an algebraic expression for each of the flow diagrams in question 6.\nproduct expressions AND sum expressions\n1.\t Complete the following:\n\t\n(a)\t (3 + 6) + (3 + 6) + (3 + 6) + (3 + 6) + (3 + 6)\n\t\n\t\n= \n × (\n) \n\t\n(b)\t (3 + 6) + (3 + 6) + (3 + 6) + (3 + 6) + (3 + 6)\n\t\n\t\n= (3 + 3 +\n) + (\n)\n\t\n\t\n= (\n×\n) + (\n×\n) \n2.\t Complete the following:\n\t\n(a)\t (3x + 6) + (3x + 6) + (3x + 6) + (3x + 6) + (3x + 6) \n\t\n\t\n= \n(\n)\n\t\n(b)\t (3x + 6) + (3x + 6) + (3x + 6) + (3x + 6) + (3x + 6)\n\t\n\t\n= (3x + 3x +\n) + (\n)\n\t\n\t\n= (\n×\n) \n (\n×\n) \n3.\t In each case, write an expression without brackets that will give the same results as \nthe given expression.\n\t\n(a)\t 3(x + 7)\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 10(2x + 1)\n\t\n(c)\t x(4x + 6)\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 3(2p + q) \n\t\n(e)\t t(t + 9)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t x(y + z)\n\t\n(g)\t 2b(b + a − 4) \t\n\t\n\t\n\t\n\t\n\t\n\t\n(h)\tk2(k − m)\nMaths_LB_gr8_bookA.indb   150\n2014/07/03   10:53:53 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n151\nThe process of writing product expressions as sum \nexpressions is called expansion. It is sometimes \nalso referred to as multiplication of algebraic \nexpressions.\n4.\t (a)\t Complete the table for the given values of x, y and z.\n3(x + 2y + 4z)\n3x + 6y + 12z\n3x + 2y + 4z\nx = 1\ny = 2\nz = 3\nx = 10\ny = 20\nz = 30\nx = 23\ny = 60\nz = 100\nx = 14\ny = 0\nz = 1\nx = 5\ny = 9\nz = 32\n\t\n(b)\t Which sum expression and product expression are equivalent?\n5.\t For each expression, write an equivalent expression without brackets.\n\t\n(a)\t 2(x2 + x + 1)\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t p(q + r + s)\n\t\n(c)\t −3(x + 2y + 3z)\t \t\n\t\n\t\n\t\n\t\n\t\n(d)\t x(2x2 + x + 7)\n\t\n(e)\t 6x(8 − 2x)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t 12x(4 − x)\n\t\n(g)\t 3x(8x − 5) − 4x(6x − 5)\t \t\n\t\n\t\n\t\n(h)\t10x(3x(8x − 5) − 4x(6x − 5))\nMaths_LB_gr8_bookA.indb   151\n2014/07/03   10:53:53 AM\n\n152\t MATHEMATICS Grade 8: Term 2\n8.2\t Simplifying algebraic expressions\nexpand, rearrange and then combine like terms\n1.\t Write the shortest possible equivalent expression without brackets. \n\t\n(a)\t x + 2(x + 3)\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 5(4x + 3) + 5x\n\t\n(c)\t 5(x + 5) + 3(2x + 1)\t\t\n\t\n\t\n\t\n\t\n(d)\t (5 + x)2\n\t\n(e)\t −3(x2 + 2x − 3) + 3(x2 + 4x)\t \t\n\t\n\t\n(f)\t x(x − 1) + x + 2\nWhen you are not sure whether you simplified an \nexpression correctly, you should always check  \nyour work by evaluating the original expression  \nand the simplified expression for some values of  \nthe variables.\n2.\t (a)\t Evaluate x(x + 2) + 5x2 − 2x for x = 10.\n\t\n(b)\t Evaluate 6x2 for x = 10.\n\t\n(c)\t Can we use the expression 6x2 to calculate the values of the expression  \n\t\nx(x + 2) + 5x2 – 2x for any given value of x? Explain.\nMaths_LB_gr8_bookA.indb   152\n2014/07/03   10:53:53 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n153\nThis is how a sum expression for x(x + 2) + 5x2 − 2x can be made:\nx(x + 2) + 5x2 − 2x\t= x × x + x × 2 + 5x2 − 2x\n\t\n= x2 + 2x + 5x2 − 2x\n\t\n= x2 + 5x2 + 2x − 2x        [Rearrange and combine like terms]      \n\t\n= 6x2 + 0\n\t\n= 6x2\n3.\t Evaluate the following expressions for x = −5: \n\t\n(a)\t x + 2(x + 3)\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 5(4x + 3) + 5x\n\t\n(c)\t 5(x + 5) + 3(2x + 1)\t\t\n\t\n\t\n\t\n\t\n(d)\t (5 + x)2\n\t\n(e)\t −3(x2 + 2x − 3) + 3(x2 + 4x)\t \t\n\t\n\t\n(f)\t x(x − 1) + x + 2\n4.\t Complete the table for the given values of x, y and z.\nx\n100\n80\n10\n20\n30\ny\n50\n40\n5\n5\n20\nz\n20\n30\n2\n15\n10\nx + (y − z)\nx − (y − z)\nx − y − z\nx − (y + z)\nx + y − z\nx − y + z\nMaths_LB_gr8_bookA.indb   153\n2014/07/03   10:53:54 AM\n\n154\t MATHEMATICS Grade 8: Term 2\n5.\t Say whether the following statements are true or false. Refer to the table in question 4. \nFor any values of x, y and z:\n\t\n(a)\t x + (y − z) = x + y – z\t\n\t\n\t\n\t\n\t\n(b)\t x − (y − z) = x − y − z\n6.\t Write the expressions without brackets. Do not simplify.\n\t\n(a)\t 3x − (2y + z)\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −x + 3(y − 2z)\nWe can simplify algebraic expressions by using properties of operations as shown:\n\t\n    (5x + 3) − 2(x + 1) \t\nHence 5x + 3 − 2x − 2 \nHence 5x − 2x + 3 − 2\nHence 3x + 1\n7.\t Write an equivalent expression without brackets for each of the following expressions \nand then simplify:  \n\t\n(a)\t 22x + (13x − 5)\t\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 22x − (13x − 5)\n\t\n(c)\t 22x − (13x + 5)\t\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 4x − (15 − 6x)\n8.\t Simplify.\n\t\n(a)\t 2(x2 + 1) − x − 2\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −3(x2 + 2x − 3) + 3x2\nHere are some of the techniques we have used so far \nto form equivalent expressions:\n• Remove brackets\n• Rearrange terms\n• Combine like terms\nx – (y + z) = x – y – z\nAddition is both associative \nand commutative.\nMaths_LB_gr8_bookA.indb   154\n2014/07/03   10:53:54 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n155\n8.3\t Simplifying quotient expressions\nFROM QUOTIENT EXPRESSIONS TO SUM EXPRESSIONS\n1.\t Complete the table for the given values of x.\nx\n1\n7\n−3\n−10\n7x2 + 5x\n7\n5\n2\nx\nx\nx\n+\n7x + 5\n7x + 5x\n7x2 + 5\n2.\t (a)\t What is the value of 7x + 5 for x = 0?\n\t\n(b)\t What is the value of 7\n5\n2\nx\nx\nx\n+\n for x = 0?\n\t\n(c)\t Which of the two expressions, 7x + 5 or 7\n5\n2\nx\nx\nx\n+\n, requires fewer calculations?  \n\t\nExplain.\nMaths_LB_gr8_bookA.indb   155\n2014/07/03   10:53:54 AM\n\n156\t MATHEMATICS Grade 8: Term 2\n\t\n(d)\t Are the expressions 7x + 5 and 7\n5\n2\nx\nx\nx\n+\n equivalent, x = 0 excluded? Explain.\n\t\n(e)\t Are there any other expressions that are equivalent to 7\n5\n2\nx\nx\nx\n+\n from those given  \n\t\nin the table? Explain.\nIf division is the last step in evaluating an algebraic \nexpression, then the expression is called a quotient \nexpression or an algebraic fraction.\nThe expression 7\n5\n2\nx\nx\nx\n+\n is an example of a quotient  \nexpression or algebraic fraction.\n3.\t Complete the table for the given values of x.\nx\n5\n10\n–5\n–10\n10x − 5x2\n5x\n \n10\n5\n5\n2\nx\nx\nx\n−\n2 − x \n\t\n(a)\t What is the value of 2 − x for x = 0?\n\t\n(b)\t What is the value of 10\n5\n5\n2\nx\nx\nx\n−\n for x = 0?\nMaths_LB_gr8_bookA.indb   156\n2014/07/03   10:53:55 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n157\n\t\n(c)\t Are the expressions 2 − x and 10\n5\n5\n2\nx\nx\nx\n−\n equivalent, x = 0 excluded? Explain.\n\t\n(d)\t Which of the two expressions 2 − x or 10\n5\n5\n2\nx\nx\nx\n−\n requires fewer calculations?  \n\t\nExplain.\nWe have found that quotient expressions such as \n10\n5\n5\n2\nx\nx\nx\n−\n can sometimes be manipulated to give \nequivalent expressions such as 2 − x.\nThe value of this is that these equivalent  \nexpressions require fewer calculations.\nThe expressions 10\n5\n5\n2\nx\nx\nx\n−\n and 2 − x are not quite \nequivalent because for x = 0, the value of 2 − x can be \ncalculated, while the first expression has no value. \nHowever, we can say that the two expressions are \nequivalent if they have the same values for all values \nof x admissible for both expressions.\nHow is it possible that 7\n5\n2\nx\nx\nx\n+\n = 7x + 5 and 10\n5\n5\n2\nx\nx\nx\n−\n = 2 − x for all admissible values\nof x? We say x = 0 is not an admissible value of x because division by 0 is not allowed. \nOne of the methods for finding equivalent expressions for algebraic fractions is by \nmeans of division:\n\t\n7\n5\n2\nx\nx\nx\n+\n\t = 1\nx (7x2 + 5x) \t\n[just as 3\n5  = 3 × 1\n5 ]\n\t\n\t = ( 1\nx  × 7x2) + ( 1\nx  × 5x)\t\n[distributive property]\n\t\n\t = 7\n2\nx\nx  + 5x\nx \t\n\t\n\t\n\t = 7x + 5\t\n[provided x ≠ 0]\nMaths_LB_gr8_bookA.indb   157\n2014/07/03   10:53:56 AM\n\n158\t MATHEMATICS Grade 8: Term 2\n4.\t Use the method shown on the previous page to simplify each fraction below.\n\t\n(a)\t 8\n10\n6\n2\nx\nz\n+\n+\n\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t 20\n16\n4\n2\nx\nx\n+\n\t\n(c)\t 9\n2\nx y\nxy\nxy\n+\n\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t 21\n14\n7\n2\nab\na\na\n−\nSimplifying a quotient expression can sometimes \nlead to a result which still contains quotients, as you \ncan see in the example below.\n5\n3\n2\n2\nx\nx\nx\n+\n= 5\n2\n2\nx\nx\n + 3\n2\nx\nx\n= 5 + 3\nx\n5.\t (a)\t Evaluate 5\n3\n2\n2\nx\nx\nx\n+\n for x = −1.\n\t\n       \n\t\n(b)\t For the expression 5\n3\n2\n2\nx\nx\nx\n+\n to be equivalent to 5 + 3\nx  which value of x must be  \n\t\nexcluded? Why?\nMaths_LB_gr8_bookA.indb   158\n2014/07/03   10:53:58 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n159\n6.\t Simplify the following expressions:\n\t\n(a)\t 8\n2\n4\n2\n2\nx\nx\nx\n+\n+\n\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t 4\n1\nn\nn\n+\n7.\t Evaluate:\n\t\n(a)\t 8\n2\n4\n2\n2\nx\nx\nx\n+\n+\n for x = 2\t \t\n\t\n\t\n\t\n(b)\t 4\n1\nn\nn\n+\n for n = 4\n8.\t Simplify. \n\t\n(a)   6\n12\n2\n2\n4\n3\nx\nx\nx\n−\n+\n\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t −\n−\n6\n4\n6\n4\nn\nn\nn\n9.\t When Natasha and Lebogang were asked to evaluate the expression x\nx\nx\n2\n2\n1\n+\n+\n \nfor x = 10, they did it in different ways.\n\t\n\t\nNatasha’s calculation:\t \t\n\t\n\t\n\t\n\t\nLebogang’s calculation:\n\t\n10 + 2 + 1\n10\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n100\n20\n1\n10\n+\n+\n\t\n= 12\t 1\n10 \t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n= 121\n10\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n= 12 1\n10   \n\t\nExplain how each of them thought about evaluating the given expression.\nMaths_LB_gr8_bookA.indb   159\n2014/07/03   10:54:00 AM\n\n160\t MATHEMATICS Grade 8: Term 2\n8.4\t Squares, cubes and roots of expressions\nsimplifying squares and cubes\nStudy the following example:\n(3x)2\t = 3x × 3x\t\nMeaning of squaring\n\t\n= 3 × x × 3 × x\t\n\t\n= 3 × 3 × x × x\t\nMultiplication is commutative: a × b = b × a\n\t\n= 9x2\t\nWe say that (3x)2 simplifies to 9x2\n1.\t Simplify the expressions.\n\t\n(a)\t (2x)2\t\n(b)\t (2x2)2\t\n(c)\t (−3y)2\n2.\t Simplify the expressions.\n\t\n(a)\t 25x − 16x\t\n(b)\t 4y + y + 3y\t\n(c)\t a + 17a − 3a\n3.\t Simplify.\n\t\n(a)\t (25x − 16x)2\t\n(b)\t (4y + y + 3y)2\t\n(c)\t (a + 17a − 3a)2\nStudy the following example:\n(3x)3\t = 3x × 3x × 3x\t\nMeaning of cubing\n\t\n= 3 × x × 3 × x × 3 × x\n\t\n= 3 × 3 × 3 × x × x × x\t\nMultiplication is commutative: a × b = b × a\n\t\n= 27x3 \t\nWe say that (3x)3 simplifies to 27x3\n\t\n\t\nMaths_LB_gr8_bookA.indb   160\n2014/07/03   10:54:00 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n161\n4.\t Simplify the following:\n\t\n(a)\t (2x)3\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t (−x)3\t\n\t\n(c)\t (5a)3\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t (7y2)3\n\t\n(e)\t (−3m)3\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t (2x3)3\n5.\t Simplify.\n\t\n(a)\t 5a − 2a\t\n(b)\t 7x + 3x\t\n(c)\t 4b + b\n6.\t Simplify.\n\t\n(a)\t (5a − 2a)3\t\n(b)\t (7x + 3x)3\t\n(c)\t (4b + b)3\n\t\n(d)\t (13x − 6x)3\t\n(e)\t (17x + 3x)3\t\n(f)\t (20y −14y)3 \nAlways remember to test whether the simplified \nexpression is equivalent to the given expression for at \nleast three different values of the given variable.\nMaths_LB_gr8_bookA.indb   161\n2014/07/03   10:54:00 AM\n\n162\t MATHEMATICS Grade 8: Term 2\nsquare and cube roots of expressions\n1.\t Thabang and his friend Vuyiswa were asked to simplify \n2\n2\n2\n2\na\na\n×\n. \n\t\nThabang reasoned as follows:\nTo find the square root of a number is the same as asking yourself the question: \n“Which number was multiplied by itself?” The number that is multiplied by  \nitself is 2a2 and therefore \n2\n2\n2\n2\na\na\n×\n = 2a2\n\t\nVuyiswa reasoned as follows: \nI should first simplify 2a2 × 2a2 to get 4a4 and then calculate \n4\n4\na  = 2a2\n\t\nWhich of the two methods do you prefer? Explain why.\n2.\t Say whether each of the following is true or false. Give a reason for your answer.\n\t\n(a)\t\n6\n6\nx\nx\n×\n = 6x\t \t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n5\n5\n2\n2\nx\nx\n×\n = 5x2\n3.\t Simplify.\n\t\n(a)\t y6 × y6\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 125x2 + 44x2\n4.\t Simplify.\n\t\n(a)\t\ny12 \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n125\n44\n2\n2\nx\nx\n+\n\t\n(c)\t\n25\n16\n2\n2\na\na\n−\n\t \t\n\t\n\t\n\t\n\t\n\t\n(d)\t\n121\n2\ny\n\t\n(e)\t\n16\n9\n2\n2\na\na\n+\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(f)\t\n25\n9\n2\n2\na\na\n−\nMaths_LB_gr8_bookA.indb   162\n2014/07/03   10:54:02 AM\n\n\t\nCHAPTER 8: ALGEBRAIC EXPRESSIONS 2\t\n163\n5.\t What does it mean to find the cube root of 8x3 written as \n8\n3\n3\nx ?\n6.\t Simplify the following:\n\t\n(a)\t 2a × 2a ×2a\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t 10b3 × 10b3 × 10b3\n\t\n(c)\t 3x3 × 3x3 × 3x3\t \t\n\t\n\t\n\t\n\t\n\t\n(d)\t −3x3 × −3x3 × −3x3\n7.\t Determine the following:\n\t\n(a)\t\n1000\n9\n3\nb \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n2\n2\n2\n3\na\na\na\n×\n×\n\t\n(c)\t\n27\n3\n3\nx \t\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t\n−27\n3\n3\nx\n8.\t Simplify the following expressions:\n\t\n(a)\t 6x3 + 2x3\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t −m3 − 3m3 − 4m3\n9.\t Determine the following:\n\t\n(a)\t\n6\n2\n3\n3\n3\nx\nx\n+\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\t\n−8\n3\n3\nm\n\t\n(c)\t\n125\n3\n3\ny \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(d)\t\n93\n123\n3\n3\n3\na\na\n+\n\t\nMaths_LB_gr8_bookA.indb   163\n2014/07/03   10:54:04 AM\n\nWorksheet\nALGEBRAIC EXPRESSIONS 2\n1.\t Simplify the following:\n\t\n(a)\t2(3b + 1) + 4\t\n(b)\t 6 − (2 + 5e)\n\t\n(c)\t18mn + 22mn + 70mn\t\n(d)\t 4pqr + 3 + 9pqr\n2.\t Evaluate each of the following expressions for m = 10:\n\t\n(a)\t3m2 + m + 10\t\n(b)\t 5(m2 − 5) + m2 + 25\n3.\t (a)\tSimplify: 4\n6\n2\nb +\n \n\t\n(b)\tEvaluate the expression 4\n6\n2\nb +\n for b = 100.\n4.\t Simplify.\n\t\n(a)\t(4g)2\t\n(b)\t(6y)3\t\n(c)\t(7s + 3s)2\n5.\t Determine the following:\n\t\n(a)\t\n121 2\nb \t\n(b)\t\n64\n3\n3\ny\n\t\n(c)\t\n63\n18\n2\n2\nd\nd\n+\nMaths_LB_gr8_bookA.indb   164\n2014/07/03   10:54:06 AM\n\nChapter 9\nAlgebraic equations 2\n\t\nCHAPTER 9: ALGEBRAIC EQUATIONS 2\t\n165\nIn this chapter you will solve equations by applying inverse operations. You will also solve \nequations that contain exponents.\n9.1\t Thinking forwards and backwards........................................................................... 167\n9.2\t Solving equations using the additive and multiplicative inverses............................. 170\n9.3\t Solving equations involving powers........................................................................ 172\nMaths_LB_gr8_bookA.indb   165\n2014/07/03   10:54:06 AM\n\n166\t MATHEMATICS Grade 8: Term 2\nMaths_LB_gr8_bookA.indb   166\n2014/07/03   10:54:06 AM\n\n\t\nCHAPTER 9: ALGEBRAIC EQUATIONS 2\t\n167",
      "chapter_id": "8"
    },
    {
      "title": "Algebraic equations 2",
      "content": "9\t Algebraic equations 2\n9.1\t Thinking forwards and backwards\ndoing and undoing what has been done\n1.\t Complete the flow diagram by finding the output values.\n–3\n–2\n0\n5\n17\n–6\n× 2\n2.\t Complete the table.\nx\n−3\n−2\n0\n5\n17\n2x\n3.\t Evaluate 4x if:\n\t\n(a)\t x = −7\t\n(b)\t x = 10\t\n(c)\t x = 0\n4.\t (a)\t Complete the flow diagram by finding the input values.\n4\n12\n28\n32\n40\n× 4\n\t\n(b)\t Puleng put another integer into the flow diagram and got −68 as an answer. \t\n\t\n\t\nWhich integer did she put in? Show your calculation.\n\t\n(c)\t Explain how you worked to find the input numbers when you did question (a).\nMaths_LB_gr8_bookA.indb   167\n2014/07/03   10:54:07 AM\n\n168\t MATHEMATICS Grade 8: Term 2\n5.\t (a)\t Complete the table.\nx\n5x\n5\n15\n25\n40\n90\n\t\n(b)\t Complete the flow diagrams.\n5\n15\n25\n40\n90\n÷ 5\n1\n3\n5\n8\n18\n× 5\n\t\n(c)\t Explain how you completed the table.\nOne of the things we do in algebra is to evaluate \nexpressions. When we evaluate expressions we \nreplace a variable in the expression with an input \nnumber to obtain the value of the expression called \nthe output number. We will think of this process \nas a doing process.\nHowever, in other cases we may need to undo what \nwas done. When we know what output number was \nobtained but do not know what input number was \nused, we have to undo what was done in evaluating \nthe expression. In such a case we say we are solving \nan equation.\n6.\t Look again at questions 1 to 5. For each question, say whether the question required \na doing or an undoing process. Give an explanation for your answer (for example: \ninput to output).\nMaths_LB_gr8_bookA.indb   168\n2014/07/03   10:54:07 AM\n\n\t\nCHAPTER 9: ALGEBRAIC EQUATIONS 2\t\n169\n7.\t (a)\t Complete the flow diagrams below.\n1\n+ 6\n7\n– 6\n\t\n(b)\t What do you observe?\n8.\t (a)\t Complete the flow diagrams below.\n2\n× 5\n10\n÷ 5\n\t\n(b)\t What do you observe?\n9.\t (a)\t Complete the flow diagrams below.\n20\n× 5\n+ 5\n105\n– 5\n÷ 5\n\t\n(b)\t What do you observe?\n10.\t(a)\t Complete the flow diagram below.\n64\n÷ 8\n+ 12\n\t\n(b)\t What calculations will you do to determine what the input number was when \t \t\n\t\nthe output number is 20?\nSolve the following problems by undoing what was done to get the answer: \n11.\tWhen a certain number is multiplied by 10 the answer is 150. What is the number?\n12.\tWhen a certain number is divided by 5 the answer is 1. What is the number?\n13.\tWhen 23 is added to a certain number the answer is 107. What is the original number?\n14.\tWhen a certain number is multiplied by 5 and 2 is subtracted from the answer, the \nfinal answer is 13. What is the original number?\nMoving from the output value to the input value is \ncalled solving the equation for the unknown.\nMaths_LB_gr8_bookA.indb   169\n2014/07/03   10:54:08 AM\n\n170\t MATHEMATICS Grade 8: Term 2\n9.2\t Solving equations using the additive and\n\t\nmultiplicative inverses\nfinding the unknown\nConsider the equation 3x + 2 = 23.\nWe can represent the equation 3x + 2 = 23 in a flow diagram, where x represents an \nunknown number:\nx\n23\n× 3\n+ 2\nWhen you reverse the process in the flow diagram, you start with the output number 23, \nthen subtract 2 and then divide the answer by 3:\n23\n– 2\n÷ 3\n \nWe can write all of the above reverse process as follows: \nSubtract 2 from both sides of the equation:\n\t\n3x + 2 – 2\t= 23 − 2\n\t\n3x\t= 21\nDivide both sides by 3:\n\t\n3\n3\nx\t= 21\n3\n\t\nx\t= 7\nWe say x = 7 is the solution of 3x + 2 = 23 because 3 × 7 + 2 = 23. We say that x = 7 makes \nthe equation 3x + 2 = 23 true.\nThe numbers +2 and −2 are additive inverses of \neach other. When we add a number and its additive \ninverse we always get 0.\nThe numbers 3 and 1\n3 are multiplicative  \ninverses of each other. When we multiply a number \nand its multiplicative inverse we always get 1, so  \n3 × 1\n3 = 1\nThe additive and  \nmultiplicative inverses help  \nus to isolate the unknown \nvalue  or the input value.\nAlso remember:\n• The multiplicative \nproperty of 1: the  \nproduct of any number \nand 1 is that number.\n• The additive property \nof 0: the sum of any number \nand 0 is that number. \nMaths_LB_gr8_bookA.indb   170\n2014/07/03   10:54:08 AM\n\n\t\nCHAPTER 9: ALGEBRAIC EQUATIONS 2\t\n171\nSolve the equations below by using the additive and multiplicative inverses. Check  \nyour answers. \n1.\t x + 10 = 0\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n2.\t 49x + 2 = 100\n3.\t 2x = 1\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n4.\t 20 = 11 − 9x\nIn some cases you need to collect like terms before \nyou can solve the equations using additive and \nmultiplicative inverses, as in the example below:\nExample: \t\nSolve for x: 7x + 3x = 10\n\t\n10x\t= 10\n\t\n10\n10\nx\t= 10\n10\n\t\nx\t= 1\n5.\t 4x + 6x = 20\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n6.\t 5x = 40 + 3x\n7x and 3x are like terms and \ncan be replaced with one \nequivalent expression  \n(7 + 3)x = 10x.\nMaths_LB_gr8_bookA.indb   171\n2014/07/03   10:54:09 AM\n\n172\t MATHEMATICS Grade 8: Term 2\n7.\t 3x + 1 − x = 0\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n8.\t x + 20 + 4x = −55\n\t\n \n9.3\t Solving equations involving powers\nSolving an exponential equation is the same as asking \nthe question: To what exponent must the base \nbe raised in order to make the equation true?\n1.\t Complete the table.\nx\n1\n3\n5\n7\n2x\n2.\t Complete the table.\nx\n2\n5\n3x\n1\n27\nKarina solved the equation 3x = 27 as follows:\n\t\n3x\t= 27\nHence\t3x\t= 33\nHence\t x\t= 3\n3.\t Now use Karina’s method and solve for x in each of the following:\n\t\n(a)\t 2x = 32\t \t\n\t\n\t\n(b)\t 4x = 16\t \t\n\t\n(c)\t 6x = 216\t \t\n\t\n(d)\t 5x + 1 = 125\nThe number 27 can be  \nexpressed as 33 because  \n33 = 27.\nMaths_LB_gr8_bookA.indb   172\n2014/07/03   10:54:09 AM\n\nChapter 10\nConstruction of geometric\nfigures\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n173\nIn this chapter, you will learn how to construct, or draw, different lines, angles and shapes. \nYou will use drawing instruments, such as a ruler, to draw straight lines, a protractor to \nmeasure and draw angles, and a compass to draw arcs that are a certain distance from  \na point. Through the various constructions, you will investigate some of the properties of \ntriangles and quadrilaterals; in other words, you will find out more about what is always \ntrue about all or certain types of triangles and quadrilaterals.\n10.1\t Bisecting lines......................................................................................................... 175\n10.2\t Constructing perpendicular lines............................................................................ 177\n10.3\t Bisecting angles...................................................................................................... 179\n10.4\t Constructing special angles without a protractor.................................................... 181\n10.5\t Constructing triangles............................................................................................. 182\n10.6\t Properties of triangles............................................................................................. 185\n10.7\t Properties of quadrilaterals...................................................................................... 187\n10.8\t Constructing quadrilaterals..................................................................................... 189\nMaths_LB_gr8_bookA.indb   173\n2014/07/03   10:54:09 AM\n\n174\t MATHEMATICS Grade 8: Term 2\nCan two circles be drawn so that the red lines do not cross at right angles?\nMaths_LB_gr8_bookA.indb   174\n2014/07/03   10:54:09 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n175",
      "chapter_id": "9"
    },
    {
      "title": "Construction of geometric figures",
      "content": "10\tConstruction of geometric figures\n10.1\t Bisecting lines\nWhen we construct, or draw, geometric figures, we often need to bisect lines or angles. \nBisect means to cut something into two equal parts. There are different ways to bisect a \nline segment.\nBisecting a line segment with a ruler\n1.\t Read through the following steps.\nStep 1: Draw line segment AB and determine its midpoint.\nA\nB\n0\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n1\nStep 2: Draw any line segment through the midpoint.\nA\nB\nC\nD\nF\nCD is called a bisector because it  \nbisects AB. AF = FB.\n2.\t Use a ruler to draw and bisect the following line segments:  \nAB = 6 cm and XY = 7 cm.\nThe small marks on AF  \nand FB show that AF and  \nFB are equal.\nMaths_LB_gr8_bookA.indb   175\n2014/07/03   10:54:09 AM\n\n176\t MATHEMATICS Grade 8: Term 2\nIn Grade 6, you learnt how to use a compass to draw circles, and parts of circles called \narcs. We can use arcs to bisect a line segment.\nBisecting a line segment with a compass and ruler\n1.\t Read through the following steps.\nStep 1\nPlace the compass on one endpoint of the line  \nsegment (point A). Draw an arc above and below  \nthe line. (Notice that all the points on the arc above \nand below the line are the same distance from point A.)\nStep 2\nWithout changing the compass width, place the \ncompass on point B. Draw an arc above and below \nthe line so that the arcs cross the first two. (The two \npoints where the arcs cross are the same distance \naway from point A and from point B.)\nStep 3\nUse a ruler to join the points where the arcs intersect. \nThis line segment (CD) is the bisector of AB.\nNotice that CD is also perpendicular to AB. So it  \nis also called a perpendicular bisector.\n2.\t Work in your exercise book. Use a compass and a  \nruler to practise drawing perpendicular bisectors  \non line segments.\nA\nB\n0°\n240\nA\nB\nA\nB\nC\n90°\nD\nIntersect means to cross  \nor meet.\nA perpendicular is a line \nthat meets another line at  \nan angle of 90°.\nTry this!\nWork in your exercise book. \nUse only a protractor and \nruler to draw a perpendicular \nbisector on a line segment. \n(Remember that we use a  \nprotractor to measure angles.)\nMaths_LB_gr8_bookA.indb   176\n2014/07/03   10:54:09 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n177\n10.2\tConstructing perpendicular lines\nA perpendicular line from a given point\n1.\t Read through the following steps.\nStep 1\nPlace your compass on the given point \n(point P). Draw an arc across the line \non each side of the given point. Do not \nadjust the compass width when  \ndrawing the second arc.\nA\nB\nP\nStep 2\nFrom each arc on the line, draw  \nanother arc on the opposite side of the \nline from the given point (P). The two \nnew arcs will intersect.\nA\nB\nP\nQ\nStep 3\nUse your ruler to join the given point (P) \nto the point where the arcs intersect (Q).\nP\nB\nA\nQ\nPQ is perpendicular to AB.  \nWe also write it like this: PQ ⊥ AB. \n2.\t Use your compass and ruler to draw a perpendicular  \nline from each given point to the line segment:\nMaths_LB_gr8_bookA.indb   177\n2014/07/03   10:54:10 AM\n\n178\t MATHEMATICS Grade 8: Term 2\nA perpendicular line at a given point on a line\n1.\t Read through the following steps.\nStep 1\nPlace your compass on the given  \npoint (P). Draw an arc across  \nthe line on each side of the given \npoint. Do not adjust the compass \nwidth when drawing the second arc.\nA\nB\nP\nStep 2\nOpen your compass so that it is wider \nthan the distance from one of the arcs to  \nthe point P. Place the compass on each \narc and draw an arc above or below the \npoint P. The two new arcs will intersect.\nA\nB\nP\nQ\nStep 3\nUse your ruler to join the  \ngiven point (P) and the point  \nwhere the arcs intersect (Q).\nPQ ⊥ AB\n2.\t Use your compass and ruler to draw a perpendicular at the given point on each line:\nP\nB\nA\nQ\nMaths_LB_gr8_bookA.indb   178\n2014/07/03   10:54:10 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n179\n10.3\tBisecting angles\nAngles are formed when any two lines meet. We use degrees (°) to measure angles.\nMeasuring and classifying angles\nIn the figures below, each angle has a number from 1 to 9. \n1.\t Use a protractor to measure the sizes of all the angles in each figure. Write your \nanswers on each figure.\n\t\n(a)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(b)\n1\n2\n3\n4\n6\n7 8\n9\n5\n2.\t Use your answers to fill in the angle sizes below.\n\t\n1 = \n° \n\t \t\n\t\n\t\n\t\n6 = \n° \n\t\n1 + 2 = \n° \n\t \t\n\t\n\t\n7 + 8 = \n° \n \n\t\n1 + 4 = \n° \n\t \t\n\t\n\t\n6 + 7 + 8 = \n° \n\t\n2 + 3 = \n° \n\t\t\n\t\n\t\n5 + 6 + 7 = \n° \n\t\n3 + 4 = \n° \n\t\t\n\t\n\t\n6 + 5 = \n° \n\t\n1 + 2 + 4 = \n° \n\t\t\n\t\n5 + 6 + 7 + 8 = \n° \n\t\n1 + 2 + 3 + 4 = \n° \n\t\t\n5 + 6 + 7 + 8 + 9 = \n° \n\t\n\t\n3.\t Next to each answer above, write down what type of angle it is, namely acute, obtuse, \nright, straight, reflex or a revolution.\nMaths_LB_gr8_bookA.indb   179\n2014/07/03   10:54:12 AM\n\n180\t MATHEMATICS Grade 8: Term 2\nBisecting angles without a protractor\n1.\t Read through the following steps.\nStep 1\nPlace the compass on the vertex of  \nthe angle (point B). Draw an arc  \nacross each arm of the angle.\nStep 2\nPlace the compass on the point where \none arc crosses an arm and draw an arc \ninside the angle. Without changing the \ncompass width, repeat for the other \narm so that the two arcs cross.\n80°\nA\nB\nC\nStep 3\nUse a ruler to join the vertex  \nto the point where the arcs  \nintersect (D). \nDB is the bisector of ABC.\n2.\t Use your compass and ruler to bisect the angles below. \nK\nL\nM\nP\nQ\nR\n50°\n80°\nA\nB\nC\n80°\nA\nB\nC\nD\nYou could measure each of \nthe angles with a protractor \nto check if you have bisected \nthe given angle correctly.\nMaths_LB_gr8_bookA.indb   180\n2014/07/03   10:54:12 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n181\n10.4\tConstructing special angles without a protractor\nconstructing angles of 60°, 30° and 120° \n1.\t Read through the following steps.\nStep 1\nDraw a line segment (JK). With the \ncompass on point J, draw an arc across \nJK and up over above point J.\nJ\nK\nStep 2\nWithout changing the compass width, \nmove the compass to the point where \nthe arc crosses JK, and draw an arc that \ncrosses the first one. \nJ\nK\nStep 3\nJoin point J to the point where the two  \narcs meet (point P). \nPJK = 60°\nJ\nK\n60°\nP\n2.\t (a)\t Construct an angle of 60° at point B on the  \n\t\nnext page.\n\t\n(b)\t Bisect the angle you constructed.\n\t\n(c)\t Do you notice that the bisected angle consists  \n\t\nof two 30° angles?\n\t\n(d)\t Extend line segment BC to A.  \n\t\nThen measure the angle adjacent to the  \n\t\n60° angle. \n\t\n\t\nWhat is its size? \n\t\n(e)\t The 60° angle and its adjacent angle add up to \n\t\nWhen you learn more about \nthe properties of triangles \nlater, you will understand why \nthe method above creates \na 60° angle. Or can you \nalready work this out now? \n(Hint: What do you know \nabout equilateral triangles?)\nAdjacent means “next to”.\nMaths_LB_gr8_bookA.indb   181\n2014/07/03   10:54:12 AM\n\n182\t MATHEMATICS Grade 8: Term 2\nA\nB\nC\nConstructing angles of 90° and 45°\n1.\t Construct an angle of 90° at point A. Go back to section 10.2 if you need help. \n2.\t Bisect the 90° angle, to create an angle of 45°. Go back to section 10.3 if you  \nneed help.\nChallenge\nWork in your exercise  \nbook. Try to construct the \nfollowing angles without \nusing a protractor: 150°, \n210° and 135°.\n10.5\tConstructing triangles\nIn this section, you will learn how to construct triangles. You will need a pencil, a \nprotractor, a ruler and a compass.\nA triangle has three sides and three angles. We can construct a triangle when we know \nsome of its measurements, that is, its sides, its angles, or some of its sides and angles. \t\nA\nMaths_LB_gr8_bookA.indb   182\n2014/07/03   10:54:12 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n183\nConstructing triangles\nConstructing triangles when three sides are given\n1.\t Read through the following steps. They describe how to construct ∆ABC with side \nlengths of 3 cm, 5 cm and 7 cm.\nStep 1\nDraw one side of the triangle using a \nruler. It is often easier to start with the \nlongest side.\nA\nB\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n1\nStep 2\nSet the compass width to 5 cm.  \nDraw an arc 5 cm away from point A. \nThe third vertex of the triangle will  \nbe somewhere along this arc.\nA\nB\n7 cm\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n1\nStep 3\nSet the compass width to 3 cm. Draw \nan arc from point B. Note where this \narc crosses the first arc. This will be the \nthird vertex of the triangle.\nA\nB\n7 cm\n0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n1\nStep 4\nUse your ruler to join points A and B to \nthe point where the arcs intersect (C).\n7 cm\nA\nB\nC\n5 cm\n2.\t Work in your exercise book. Follow the steps above to construct the following \ntriangles:\n\t\n(a)\t ∆ABC with sides 6 cm, 7 cm and 4 cm\n\t\n(b)\t ∆KLM with sides 10 cm, 5 cm and 8 cm\n\t\n(c)\t ∆PQR with sides 5 cm, 9 cm and 11 cm\t\n\t\n \nMaths_LB_gr8_bookA.indb   183\n2014/07/03   10:54:13 AM\n\n184\t MATHEMATICS Grade 8: Term 2\nConstructing triangles when certain angles and sides are given\n3.\t Use the rough sketches in (a) to (c) below to construct accurate triangles, using a  \nruler, compass and protractor. Do the construction next to each rough sketch.\n• The dotted lines show where you have to use a compass to measure the length  \nof a side.\n• Use a protractor to measure the size of the given angles.\n\t\n(a)\t Construct ∆ABC, with two angles and \n\t\none side given.\n7 cm\nA\nB\nC\n50°\n45°\n\t\n(b)\t Construct a ∆KLM, with two sides and \n\t\nan angle given. \n9 cm\n8 cm\nK\nL\nM\n35°\n\t\n(c)\t Construct right-angled ∆PQR, with the \n\t\nhypotenuse and one other side given.\n6 cm\n8,1 cm\nQ\nR\nP\nMaths_LB_gr8_bookA.indb   184\n2014/07/03   10:54:13 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n185\n4.\t Measure the missing angles and sides of each  \ntriangle in 3(a) to (c) on the previous page.  \nWrite the measurements at your completed \nconstructions.\n5.\t Compare each of your constructed triangles in  \n3(a) to (c) with a classmate’s triangles. Are the  \ntriangles exactly the same?\nIf triangles are exactly the same, we say they  \nare congruent.\n10.6\tProperties of triangles\nThe angles of a triangle can be the same size or different sizes. The sides of a triangle can \nbe the same length or different lengths.\nproperties of equilateral triangles\n1.\t (a)\t Construct ∆ABC next to its rough sketch below.\n\t\n(b)\t Measure and label the sizes of all its  \n\t\nsides and angles.\t\n5 cm\n5 cm\n5 cm\nA\nB\nC\n2.\t Measure and write down the sizes of the sides  \nand angles of ∆DEF on the right. \n3.\t Both triangles in questions 1 and 2 are called  \nequilateral triangles. Discuss with a classmate  \nif the following is true for an equilateral triangle:\n• All the sides are equal.\n• All the angles are equal to 60°.\t\nD \nF \nE \nChallenge\n1.\t Construct these triangles:\n(a)\t∆STU, with three angles  \ngiven: S= 45°, T= 70° and \nU= 65°.\n(b)\t∆XYZ, with two sides and \nthe angle opposite one of \nthe sides given: X= 50°,  \nXY = 8 cm and XZ = 7 cm.\n2.\t Can you find more than  \n\t\none solution for each triangle \n\t\nabove? Explain your findings to \n\t\na classmate.\nMaths_LB_gr8_bookA.indb   185\n2014/07/03   10:54:14 AM\n\n186\t MATHEMATICS Grade 8: Term 2\nProperties of isosceles triangles\n1.\t (a)\t Construct ∆DEF with EF = 7 cm, E= 50° and F= 50°. \n\t\n\t\nAlso construct ∆JKL with JK = 6 cm, KL = 6 cm and J= 70°.\n\t\n(b)\t Measure and label all the sides and angles of each triangle.\n2.\t Both triangles above are called isosceles triangles. Discuss with a classmate \nwhether the following is true for an isosceles triangle:\n• Only two sides are equal.\n• Only two angles are equal.\n• The two equal angles are opposite the two equal sides.\nThe sum of the angles in a triangle\n1.\t Look at your constructed triangles ∆ABC, ∆DEF and ∆JKL above and on the previous \npage. What is the sum of the three angles each time? \n2.\t Did you find that the sum of the interior angles of each triangle is 180°? Do the \nfollowing to check if this is true for other triangles.\n\t\n(a)\t On a clean sheet of paper, construct any  \n\t\ntriangle. Label the angles A, B and C  \n\t\nand cut out the triangle.\n\t\n(b)\t Neatly tear the angles off the triangle  \n\t\nand fit them next to one another.\n\t\n(c)\t Notice that A, B and C form a straight  \n\t\nangle. Complete: A + B + C = \n°\nWe can conclude that the interior angles of a triangle \nalways add up to 180°.\n\t\n\t\n\t\nA\nB\nC\nC\nB\nA\nMaths_LB_gr8_bookA.indb   186\n2014/07/03   10:54:15 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n187\n10.7\tProperties of quadrilaterals\nA quadrilateral is any closed shape with four straight sides. We classify quadrilaterals \naccording to their sides and angles. We note which sides are parallel, perpendicular or \nequal. We also note which angles are equal.\nProperties of quadrilaterals\n1.\t Measure and write down the sizes of all the angles and the lengths of all the sides of \n\t\neach quadrilateral below.\nSquare\nA\nD\nB\nC\nRectangle\nF\nG\nE\nD\nParallelogram\nN\nM\nL\nK\nRhombus\nM\nL\nK\nJ\nTrapezium\nKite\nA\nD\nC\nB\nR\nT\nS\nU\nMaths_LB_gr8_bookA.indb   187\n2014/07/03   10:54:15 AM\n\n188\t MATHEMATICS Grade 8: Term 2\n2.\t Use your answers in question 1. Place a ✓ in the correct box below to show which \nproperty is correct for each shape.\nProperties\nParallelogram Rectangle Rhombus Square\nKite\nTrapezium\nOnly one pair of \nsides are parallel\nOpposite sides are \nparallel\nOpposite sides are \nequal\nAll sides are equal\nTwo pairs of adjacent \nsides are equal\nOpposite angles are \nequal\nAll angles are equal\nSum of the angles in a quadrilateral\n1.\t Add up the four angles of each quadrilateral on the previous page. What do you \nnotice about the sum of the angles of each quadrilateral?\n2.\t Did you find that the sum of the interior angles of each quadrilateral equals 360°?  \nDo the following to check if this is true for other quadrilaterals.\n\t\n(a)\t On a clean sheet of paper, use a ruler to construct any quadrilateral. \n\t\n(b)\t Label the angles A, B, C and D. Cut out the quadrilateral.\n\t\n(c)\t Neatly tear the angles off the quadrilateral and fit them next to one another. \n\t\n(d)\t What do you notice?\nWe can conclude that the interior angles of a \nquadrilateral always add up to 360°.\n\t \t\n\t\nMaths_LB_gr8_bookA.indb   188\n2014/07/03   10:54:15 AM\n\n\t CHAPTER 10: CONSTRUCTION OF GEOMETRIC FIGURES\t\n189\n10.8\tConstructing quadrilaterals\nYou learnt how to construct perpendicular lines in section 10.2. If you know how to \nconstruct parallel lines, you should be able to construct any quadrilateral accurately.\nConstructing parallel lines to draw quadrilaterals\n1.\t Read through the following steps.\nStep 1\nFrom line segment AB, mark a point D.  \nThis point D will be on the line that will  \nbe parallel to AB. Draw a line from A \nthrough D.\nA\nB\nD\nStep 2\nDraw an arc from A that crosses AD and \nAB. Keep the same compass width and \ndraw an arc from point D as shown.\nA\nB\nD\nStep 3\nSet the compass width to the distance  \nbetween the two points where the first \narc crosses AD and AB. From the point \nwhere the second arc crosses AD, draw a \nthird arc to cross the second arc.\nStep 4\nDraw a line from D through the point \nwhere the two arcs meet. DC is parallel \nto AB.\n2.\t Practise drawing a parallelogram, square and rhombus in your exercise book.\n3.\t Use a protractor to try to draw quadrilaterals with at least one set of parallel lines.\nA\nB\nD\nA\nB\nD\nA\nD\nB\nC\nMaths_LB_gr8_bookA.indb   189\n2014/07/03   10:54:16 AM\n\nWorksheet\nCONSTRUCTION OF GEOMETRIC FIGURES\n1.\t Do the following construction in your exercise book.\n\t\n(a)\tUse a compass and ruler to construct equilateral ∆ABC with sides 9 cm.\n\t\n(b)\tWithout using a protractor, bisect B. Let the bisector intersect AC at point D.\n\t\n(c)\tUse a protractor to measure ADB. Write the measurement on the drawing.\n2.\t Name the following types of triangles and quadrilaterals.\n\t\nA\t \t\nB\t \t\nC\n\t\nD\t \t\nE\t \t\nF\n3.\t Which of the following quadrilaterals matches each description below? (There may \n\t\nbe more than one answer for each.)\n\t\n\t\n\t\nparallelogram; rectangle; rhombus; square; kite; trapezium\n\t\n(a)\tAll sides are equal and all angles are equal. \n\t\n(b)\tTwo pairs of adjacent sides are equal. \n\t\n(c)\tOne pair of sides is parallel. \n\t\n(d)\tOpposite sides are parallel. \n\t\n(e)\t Opposite sides are parallel and all angles are equal. \n\t\n(f)\t All sides are equal. \nMaths_LB_gr8_bookA.indb   190\n2014/07/03   10:54:16 AM\n\nChapter 11\nGeometry of 2D shapes\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n191\nIn this chapter, you will learn more about different kinds of triangles and quadrilaterals,  \nand their properties. You will explore shapes that are congruent and shapes that are  \nsimilar. You will also use your knowledge of the properties of 2D shapes in order to solve \ngeometric problems.\n11.1\t Types of triangles.................................................................................................... 193\n11.2\t Unknown angles and sides of triangles.................................................................... 195\n11.3\t Types of quadrilaterals and their properties............................................................. 200\n11.4\t Unknown angles and sides of quadrilaterals............................................................ 204\n11.5\t Congruency............................................................................................................ 205\n11.6\t Similarity................................................................................................................ 207\nMaths_LB_gr8_bookA.indb   191\n2014/07/03   10:54:16 AM\n\n192\t MATHEMATICS Grade 8: Term 2\nMaths_LB_gr8_bookA.indb   192\n2014/07/03   10:54:17 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n193",
      "chapter_id": "10"
    },
    {
      "title": "Geometry of 2D shapes",
      "content": "11\tGeometry of 2D shapes\n11.1\t Types of triangles\nBy now, you know that a triangle is a closed 2D shape with three straight sides. We can \nclassify or name different types of triangles according to the lengths of their sides and \naccording to the sizes of their angles.\nNaming triangles according to their sides\n1.\t Match the name of each type of triangle with its correct description.\nName of triangle\nDescription of triangle\nIsosceles triangle\nAll the sides of a triangle are equal.\nScalene triangle\nNone of the sides of a triangle are equal.\nEquilateral triangle\nTwo sides of a triangle are equal.\n2.\t Name each type of triangle by looking at its sides.\nI\n \nQ\nR\nP\nX\nZ\nY\nG\nH\nNaming triangles according to their angles\nRemember the following types of angles: \nA\nC\nB\nE\nD\nF\nO\nP\nQ\n\t\nAcute angle\t\nRight angle\t\nObtuse angle\n\t\n(< 90°)\t\n(= 90°)\t\n(between 90° and 180°)\nMaths_LB_gr8_bookA.indb   193\n2014/07/03   10:54:17 AM\n\n194\t MATHEMATICS Grade 8: Term 2\nStudy the following triangles; then answer the questions: \n\t\nAcute triangle\t\nRight-angled triangle\t\nObtuse triangle\n1.\t Are all the angles of a triangle always equal? \n2.\t When a triangle has an obtuse angle, it is called an \n triangle.\n3.\t When a triangle has only acute angles, it is called an \n triangle.\n4.\t When a triangle has an angle equal to \n, it is called a right-angled triangle.\ninvestigating the angles and sides of triangles\n1.\t (a)\t What is the sum of the  \n\t\ninterior angles of a triangle? \n\t\n(b)\t Can a triangle have two right angles?  \n\t\nExplain your answer.\n\t\n(c)\t Can a triangle have more than one obtuse angle? Explain your answer.\n2.\t Look at the triangles below. The arcs show which angles are equal.\nA\nB\nC\nE\nM\nF\nL\nK\nJ\n\t\nEquilateral triangle\t\nIsosceles triangle\t\nRight-angled triangle\n\t\n(a)\t ∆ABC is an equilateral triangle. What do you notice about its angles?\n\t\n(b)\t ∆FEM is an isosceles triangle. What do you notice about its angles?\nIf you cannot work out the \nanswers in 1(b) and (c), try \nto construct the triangles to \nfind the answers.\nMaths_LB_gr8_bookA.indb   194\n2014/07/03   10:54:18 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n195\n\t\n(c)\t ∆JKL is a right-angled triangle. Is its longest side opposite the 90° angle? \n\t\n(d)\t Construct any three right-angled triangles on a sheet of paper. Is the longest \n\t\nside always opposite the 90° angle?\nProperties of triangles:\n• The sum of the interior angles of a triangle  \nis 180°.\n• An equilateral triangle has all sides equal and \neach interior angle is equal to 60°.\n• An isosceles triangle has two equal sides and the \nangles opposite the equal sides are equal.\n• A scalene triangle has no sides equal.\n• A right-angled triangle has a right angle (90°).\n• An obtuse triangle has one obtuse angle \n(between 90° and 180°).\n• An acute triangle has three acute angles (< 90°).\n11.2\t Unknown angles and sides of triangles\nYou can use what you know about triangles to obtain other information. When you work \nout new information, you must always give reasons for the statements you make.\nLook at the examples below of working out unknown angles and sides when certain \ninformation is given. The reason for each statement is written in square brackets.\nA\nB\nC\nE\nD\nF\nL\nK\nJ\n40°\n85°\nA = B = C = 60°\t\n[Angles in an equilateral ∆ = 60°]\nDE = DF\t\n\t\n[Given]\nE = F\t \t\n\t\n[Angles opposite the equal sides of an isosceles ∆ are equal]\nJ = 55°\t\n\t\n[The sum of the interior angles of a ∆ = 180°; so J = 180° – 40° – 85°]\nInterior angles are the \nangles inside a closed shape, \nnot the angles outside of it.\nMaths_LB_gr8_bookA.indb   195\n2014/07/03   10:54:19 AM\n\n196\t MATHEMATICS Grade 8: Term 2\nYou can shorten the following reasons in the ways shown:\n• Sum of interior angles (∠s) of a triangle (∆) = 180°: Interior ∠s of ∆\n• Isosceles triangle has 2 sides and 2 angles equal: Isosceles ∆\n• Equilateral triangle has 3 sides and 3 angles equal: Equilateral ∆\n• Angles forming a straight line = 180°: Straight line\nWorking out unknown angles and sides\nFind the sizes of unknown angles and sides in the following triangles. Always give \nreasons for every statement.\n1.\t What is the size of C?\nC\nB\nA\n50°\n95°\n\t\nA + B + C\t= \n\t\n[Interior ∠s of a ∆]\n\t50° + \n + C\t= \n\t\n145° + C\t= \n\t\nC\t= \n  – 145°\n\t\nC\t= \n2.\t Determine the size of P.\nP\nR\nQ\n60°\n45°\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n3.\t (a)\t What is the length of KM?\n\t\n(b)\t Find the size of K.\nL\nK\nM\n50 mm\n38°\n4.\t What is the size of S?\nT\nS\nR\n32°\n\t\nMaths_LB_gr8_bookA.indb   196\n2014/07/03   10:54:19 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n197\n5.\t (a)\t Find CB.\n\t\n(b)\t Find C if A = 50°.\nC\nB\nA\n6 cm\n8 cm\n\n\n6.\t (a)\t Find DF.\n\t\n(b)\t Find E if D = 100°.\n7 mm\nE\nD\nF\n4 mm\nworking out more unknown angles and sides\n1.\t Calculate the size of X and Z.\n24°\nY\nX\nZ\nMaths_LB_gr8_bookA.indb   197\n2014/07/03   10:54:20 AM\n\n198\t MATHEMATICS Grade 8: Term 2\n2.\t Calculate the size of x.\n80°\nx\nG\nE\nF\n3.\t KLM is a straight line. Calculate the size of x and y. \ny\n100°\nx\n50°\nM\nL\nJ\nK\n4.\t Angle b and an angle with size 130° form a straight angle. Calculate the size of a and b.\nb\n130°\na\n30°\n5.\t m and n form a straight angle. Calculate the size of m and n.\nn\nm\nMaths_LB_gr8_bookA.indb   198\n2014/07/03   10:54:20 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n199\n6.\t BCD is a straight line segment. Calculate the size of x.\n68°\nA\nD\nC\nB\nx\n7.\t Calculate the size of x and then the size of H.\nx + 20°\nG\nI\nH\nx\n2x + 40°\n8.\t Calculate the size of N.\nx + 10°\nN\nP\nM\n2x – 30°\n2x – 50°\n9.\t DNP is a straight line. Calculate the size of x and of y.\n56°\nM\nP\nN\nD\nx\ny\nMaths_LB_gr8_bookA.indb   199\n2014/07/03   10:54:21 AM\n\n200\t MATHEMATICS Grade 8: Term 2\n11.3\t Types of quadrilaterals and their properties\nA quadrilateral is a figure with four straight sides which  \nmeet at four vertices. In many quadrilaterals all the sides  \nare of different lengths and all the angles are of different  \nsizes.\nYou have previously worked with these types of  \nquadrilaterals, in which some sides have the same lengths,  \nand some angles may be of the same size.\nparallelograms\nrectangles\nkites\nrhombuses\nsquares \ntrapeziums\nthe properties of different types of quadrilaterals\n1.\t In each question below, different examples of a certain type of quadrilateral are  \ngiven. In each case identify which kind of quadrilateral it is. Describe the properties \nof each type by making statements about the lengths and directions of the sides and \nthe sizes of the angles of each type. You may have to take some measurements to be \nable to do this.\nQuestion 1(a)\nMaths_LB_gr8_bookA.indb   200\n2014/07/03   10:54:21 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n201\nQuestion 1(b)\nQuestion 1(c)\nQuestion 1(d)\nMaths_LB_gr8_bookA.indb   201\n2014/07/03   10:54:21 AM\n\n202\t MATHEMATICS Grade 8: Term 2\nQuestion 1(e)\nQuestion 1(f)\nMaths_LB_gr8_bookA.indb   202\n2014/07/03   10:54:21 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n203\n2.\t Use your completed lists and the drawings in question 1 to determine if the following \nstatements are true (T) or false (F).\n\t\n(a)\t A rectangle is a parallelogram. \n\t\n(b)\t A square is a parallelogram. \n\t\n(c)\t A rhombus is a parallelogram. \n\t\n(d)\t A kite is a parallelogram. \n\t\n(e)\t A trapezium is a parallelogram. \n\t (f)\t A square is a rhombus. \n\t\n(g)\t A square is a rectangle. \n\t\n\t\n\t\n(h)\tA square is a kite. \n\t\n(i)\t A rhombus is a kite. \n\t \t\n\t\n\t\n(j)\t A rectangle is a rhombus. \n\t\n(k)\t A rectangle is a square. \nIf a quadrilateral has all the properties of another \nquadrilateral, you can define it in terms of the other \nquadrilateral, as you have found above.\n3.\t Here are some conventional definitions of quadrilaterals:\n• A parallelogram is a quadrilateral with two opposite sides parallel.\n• A rectangle is a parallelogram that has all four angles equal to 90°.\n• A rhombus is a parallelogram with all four sides equal.\n• A square is a rectangle with all four sides equal.\n• A trapezium is a quadrilateral with one pair of opposite sides parallel.\n• A kite is a quadrilateral with two pairs of adjacent sides equal.\n\t\nWrite down other definitions that work for these quadrilaterals.\n\t\n(a)\t Rectangle: \n\t\n(b)\t Square: \n\t\n(c)\t Rhombus: \n\t\n(d)\t Kite: \n\t\n(e)\t Trapezium: \n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\nA convention is something \n(such as a definition or \nmethod) that most people \nagree on, accept and follow.\nMaths_LB_gr8_bookA.indb   203\n2014/07/03   10:54:21 AM\n\n204\t MATHEMATICS Grade 8: Term 2\n11.4\t Unknown angles and sides of quadrilaterals\nfinding unknown angles and sides\nFind the length of all the unknown sides and angles in the following quadrilaterals. \nGive reasons to justify your statements. (Also recall that the sum of the angles of a \nquadrilateral is 360°.)\n1.\nB\nC\nA\n100°\n80°\nD\n4 cm\n7 cm\n2.  \nL\nM\nK\n102°\n78°\nJ\n6 cm\n3.\t ABCD is a kite. \nC\nB\nA\n105°\n55°\nD\n4 cm\n7 cm\n4.\t The perimeter of RSTU is 23 cm. \nT\nU\nS\n105°\n120°\n75°\nR\n4 cm\n3 cm\n7 cm\nMaths_LB_gr8_bookA.indb   204\n2014/07/03   10:54:22 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n205\n5.\t PQRS is a rectangle and has a perimeter of 40 cm.\nP\nQ\nS\nR\n(3x) cm\n(x) cm\n11.5\t Congruency\nwhat is congruency?\n1.\t ∆ABC is reflected in the vertical line  \n\t\n(mirror) to give ∆KLM.\n\t\nAre the sizes and shapes of the two  \n\t\ntriangles exactly the same?\n2.\t ∆MON is rotated 90° around point F  \n\t\nto give you ∆TUE. \n\t\nAre the sizes and shapes of ∆MON  \n\t\nand ∆TUE exactly the same?\n3.\t Quadrilateral ABCD is translated \n\t\n6 units to the right and 1 unit down to  \n\t\ngive quadrilateral XRZY.\n\t\nAre ABCD and XRZY exactly the same?\nK\nM\nA\nB\nN\nC\nA\nB\nZ\n1 unit down\n6 units right\nC\nL\nMirror\nT\nE\nU\nF\nM\nO\nR\nY\nX\nD\nMaths_LB_gr8_bookA.indb   205\n2014/07/03   10:54:22 AM\n\n206\t MATHEMATICS Grade 8: Term 2\nIn the previous activity, each of the figures was transformed (reflected, rotated or \ntranslated) to produce a second figure. The second figure in each pair has the same \nangles, side lengths, size and area as the first figure. The second figure is thus an \naccurate copy of the first figure.\nWhen one figure is an image of another figure,  \nwe say that the two figures are congruent. \nThe symbol for congruent is: ≡\nNotation of congruent figures\nWhen we name shapes that are congruent, we name them so that the matching, or \ncorresponding, angles are in the same order. For example, in ∆ABC and ∆KLM on the \nprevious page:\n\t\nAis congruent to (matches and is equal to) K. \n\t\nB is congruent to M. \n\t\nCis congruent to L.\nWe therefore use this notation: ∆ABC ≡ ∆KML.\nSimilarly for the other pairs of figures on the previous page: \n\t\n∆MON ≡ ∆ETU and ABCD ≡ XRZY.\nThe notation of congruent figures also shows which sides of the two figures correspond \nand are equal. For example, ∆ABC ≡ ∆KML shows that:\n\t\nAB = KM, BC = ML and AC = KL\nThe incorrect notation ∆ABC ≡ ∆KLM will show the following incorrect information: \n\t\nB = L, C = M, AB = KL, and AC = KM.\nidentifying congruent angles and sides\nWrite down which angles and sides are equal between each pair of congruent figures.\n1.\t ∆PQR ≡ ∆UCT\n2.\t ∆KLM ≡ ∆UWC\n3.\t ∆GHI ≡ ∆QRT\n4.\t ∆KJL ≡ ∆POQ\nThe word congruent comes \nfrom the Latin word congruere,  \nwhich means “to agree”. \nFigures are congruent if they \nmatch up perfectly when laid \non top of each other.\nWe cannot assume that, \nwhen the angles of polygons \nare equal, the polygons are \ncongruent. You will learn \nabout the conditions of  \ncongruence in Grade 9.\nMaths_LB_gr8_bookA.indb   206\n2014/07/03   10:54:22 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n207\n11.6\t Similarity\nIn Grade 7, you learnt that two figures are similar when they have the same shape \n(their angles are equal) but they may be different sizes. The sides of one figure are \nproportionally longer or shorter than the sides of the other figure; that is, the length \nof each side is multiplied or divided by the same number. We say that one figure is an \nenlargement or a reduction of the other figure. \nchecking for similarity\n1.\t Look at the rectangles below and answer the questions that follow.\nA\nB\nC\nF\nG\nI\nK\nJ\nH\nD\nL\nM\nRectangle 2\nRectangle 1\n\t\n(a)\t Look at rectangle 1 and ABCD:\n\t\n\t\nHow many times is FH longer than BC?\t\n\t\n\t\n\t\nHow many times is GF longer than AB?\t\n\t\n\t\n(b)\t Look at rectangle 2 and ABCD:\n\t\n\t\nHow many times is IL longer than BC?\t\n\t\n\t\n\t\nHow many times is LM longer than CD?\t \t\n\t\n(c)\t Is rectangle 1 or rectangle 2 an enlargement of rectangle ABCD? Explain your  \n\t\nanswer.\n2.\t Look at the triangles below and answer the questions that follow.\nA\nB\nC\nF\nG\nI\nK\nJ\nH\nMaths_LB_gr8_bookA.indb   207\n2014/07/03   10:54:22 AM\n\n208\t MATHEMATICS Grade 8: Term 2\n\t\n(a)\t How many times is:\n• FG longer than BC?\t\n\t\n\t HF longer than AB?\t\n• HG longer than AC?\t\n\t\n\t IK shorter than BC?\t\n• JI shorter than AB?\t\n\t\n\t JK shorter than AC?\t\n\t\n(b)\t Is ∆HFG an enlargement of ∆ABC? Explain your answer.\n\t\n(c)\t Is ∆JIK a reduction of ∆ABC? Explain your answer.\nIn the previous activity, rectangle KILM is an enlargement \nof rectangle ABCD. Therefore, ABCD is similar to KILM. The \nsymbol for ‘is similar to’ is: ///. So we write: ABCD /// KILM.\nThe triangles on the previous page are also similar. ∆HFG is  \nan enlargement of ∆ABC and ∆JIK is a reduction of ∆ABC.\nIn ∆ABC and ∆HFG, A = H, B = F and C = G. We therefore  \nwrite it like this: ∆ABC /// ∆HFG.\nIn the same way, ∆ABC /// ∆JIK.\nSimilar figures are figures that have the same \nangles (same shape) but are not necessarily the  \nsame size.\nusing properties of similar and congruent figures\n1.\t Are the triangles in each pair similar or congruent? Give a reason for each answer.\nB\nC\nA\nD\nE\nF\n4\n4\n4\n8\n8\n8\n60°\n60°\n60°\n60°\n60°\n60°\nK\nL\nJ\nT\nR\nS\n4\n5\n3\n6\n8\n10\nWhen you enlarge or \nreduce a polygon, you \nneed to enlarge or \nreduce all its sides  \nproportionally, or by  \nthe same ratio. This \nmeans that you multiply \nor divide each length by \nthe same number.\nMaths_LB_gr8_bookA.indb   208\n2014/07/03   10:54:23 AM\n\n\t\nCHAPTER 11: GEOMETRY OF 2D SHAPES\t\n209\n2.\t Is ∆RTU /// ∆EFG? Give a reason for your answer.\n3.\t ∆PQR /// ∆XYZ. Determine the length of XZ and XY.\n(ZY is twice as long as QR.)\n4.\t Are the following statements true or false? Explain your answers.\n\t\n(a)\t Figures that are congruent are similar.\n\t\n(b)\t Figures that are similar are congruent.\n\t\n(c)\t All rectangles are similar.\n\t\n(d)\t All squares are similar.\nR\nU\nT\n3\n4\n5\nF\nG\nE\n13\n12\n5\nP\nR\nQ\nX\nY\nZ\n8 cm\n8 cm\n5 cm\n4 cm\nMaths_LB_gr8_bookA.indb   209\n2014/07/03   10:54:23 AM\n\nWorksheet\nGEOMETRY OF 2D SHAPES\n1.\t Study the triangles below and answer the following questions:\n\t\n(a)\tTick the correct answer. ∆ABC is:\n\t\n\t\n \n\t\nacute and equilateral\n\t\n\t\n \n\t\nobtuse and scalene\n\t\n\t\n \n\t\nacute and isosceles\n\t\n\t\n \n\t\nright-angled and isosceles.\n\t\n(b)\tIf AB = 40 mm, what is the length of AC? \n\t\n(c)\tIf B = 80°, what is the size of C and of A? \n\t\n(d)\t∆ABC ≡ ∆FDE. Name all the sides in the two triangles that are equal to AB.\n\t\n(e)\t Name the side that is equal to DE. \n\t\n(f)\t If F is 40°, what is the size of B? \n2.\t Look at figures JKLM and PQRS. (Give reasons for your answers below.)\nJ\n6 cm\nK\nL\nM\n3 cm\n65°\n115°\nP\n2 cm\nQ\nR\nS\n1 cm\n\t\n(a)\tWhat type of quadrilateral is JKLM? \n\t\n(b)\tIs JKLM /// PQRS? \n\t\n(c)\tWhat is the size of L? \n\t\n(d)\tWhat is the size of S? \n\t\n(e)\t What is the length of KL? \nA\nC\nB\nF\nE\nD\nMaths_LB_gr8_bookA.indb   210\n2014/07/03   10:54:24 AM\n\nChapter 12\nGeometry of straight lines\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n211\nIn this chapter, you will explore the relationships between pairs of angles that are created \nwhen straight lines intersect (meet or cross). You will examine the pairs of angles that are \nformed by perpendicular lines, by any two intersecting lines, and by a third line that cuts \ntwo parallel lines. You will come to understand what is meant by vertically opposite angles, \ncorresponding angles, alternate angles and co-interior angles. You will be able to identify \ndifferent angle pairs, and then use your knowledge to help you work out unknown angles \nin geometric figures.\n12.1\t Angles on a straight line......................................................................................... 213\n12.2\t Vertically opposite angles........................................................................................ 216\n12.3\t Lines intersected by a transversal............................................................................ 219\n12.4\t Parallel lines intersected by a transversal................................................................. 222\n12.5\t Finding unknown angles on parallel lines................................................................ 224\n12.6\t Solving more geometric problems.......................................................................... 227\nMaths_LB_gr8_bookA.indb   211\n2014/07/03   10:54:24 AM\n\n212\t MATHEMATICS Grade 8: Term 2\n[to come]\nMaths_LB_gr8_bookA.indb   212\n2014/07/03   10:54:25 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n213",
      "chapter_id": "11"
    },
    {
      "title": "Geometry of straight lines",
      "content": "12\tGeometry of straight lines\n12.1\t Angles on a straight line\nSum of angles on a straight line\nIn the figures below, each angle is given a label from 1 to 5. \n1.\t Use a protractor to measure the sizes of all the angles in each figure. Write your \nanswers on each figure.\n\t\nA\t\n\t\n\t\n\t\n\t\nB\n2.\t Use your answers to fill in the angle sizes below.\n\t\n(a)\t 1 + 2 = \n°\t\t\n\t\n\t\n(b)\t 3 + 4 + 5 = \n°\nThe sum of angles that are formed on a straight line is \nequal to 180°. (We can shorten this property as: ∠s  \non a straight line.) \nTwo angles whose sizes add up to 180° are also \ncalled supplementary angles, for example 1 + 2.\nAngles that share a vertex and a common side are \nsaid to be adjacent. So  1 + 2 are therefore also  \ncalled supplementary adjacent angles.\n \t\n1\n2\n5\n3\n4\nB\nA\nC\n1 2\nD\nsame vertex\ncommon side\nMaths_LB_gr8_bookA.indb   213\n2014/07/03   10:54:25 AM\n\n214\t MATHEMATICS Grade 8: Term 2\nWhen two lines are perpendicular, their adjacent \nsupplementary angles are each equal to 90°.\nIn the drawing alongside, DCA and DCB are  \nadjacent supplementary angles because they are  \nnext to each other (adjacent) and they add up to  \n180° (supplementary).\nFinding unknown angles on straight lines\nWork out the sizes of the unknown angles below. Build an equation each time as you \nsolve these geometric problems. Always give a reason for every statement you make. \n1.\t Calculate the size of a.\na\n63°\n\ta + 63°\t = \n\t\n[∠s on a straight line]\n \t\na\t = \n − 63°\n\t\n\t = \n2.\t Calculate the size of x.\nx\n29°\n3.\t Calculate the size of y.\n52°\n2y\n48°\nB\nC\nD\nA\nMaths_LB_gr8_bookA.indb   214\n2014/07/03   10:54:26 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n215\nFinding more unknown angles on straight lines\n1.\t Calculate the size of:\n\t\n(a)\t x\t\n\t\n\t\n\t\n(b)\t ECB\nA\nC\nB\nD\nE\nx\n2x\n3x\n2.\t Calculate the size of:\n\t\n(a)\t m\t \t\n\t\n\t\n(b)\t SQR\nP\nT\nS\nQ\nR\n70°\n80°\nm + 10°\n3.\t Calculate the size of:\n\t\n(a)\t x\t\n\t\n\t\n\t\n(b)\t HEF\nD\nG\nH\nE\nF\n2x + 10°\nx + 40°\nx + 30°\n4.\t Calculate the size of:\n\t\n(a)\t k\t\n\t\n\t\n\t\n(b)\t TYP\nX\nT\nY\nZ\nk + 65°\n2k\nP\nMaths_LB_gr8_bookA.indb   215\n2014/07/03   10:54:26 AM\n\n216\t MATHEMATICS Grade 8: Term 2\n5.\t Calculate the size of:\n\t\n(a)\t p\t\n\t\n\t\n\t\n\t\n(b)\t JKR\nJ\nR\nK\nL\n2p – 55°\n3p\nU\n70°\n12.2\t Vertically opposite angles\nWhat are vertically opposite angles?\n1.\t Use a protractor to measure the sizes  \nof all the angles in the figure.  \nWrite your answers on the figure.\n2.\t Notice which angles are equal and how these equal angles are formed.\nVertically opposite angles (vert. opp. ∠s) are the \nangles opposite each other when two lines intersect. \nVertically opposite angles are always equal.\t\nb\nc\nd\na\nMaths_LB_gr8_bookA.indb   216\n2014/07/03   10:54:27 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n217\nFinding unknown angles\nCalculate the sizes of the unknown angles in the following figures. Always give a reason \nfor every statement you make.\n1.\t Calculate x, y and z.\n105°\ny\nz\nx\n\t\nx\t = \n°\t\n[vert. opp. ∠s]\n\ty + 105°\t = \n°\t\n[∠s on a straight line]\n\t\ny\t = \n − 105°\n\t\n\t = \n\t\nz\t = \n\t\n[vert. opp. ∠s]\n2.\t Calculate j, k and l.\n64°\nj\nk\nl\n3.\t Calculate a, b, c and d.\n88°\n62°\na\nb\nc\nd\nMaths_LB_gr8_bookA.indb   217\n2014/07/03   10:54:27 AM\n\n218\t MATHEMATICS Grade 8: Term 2\nequations using vertically opposite angles\nVertically opposite angles are always equal. We can use this property to build an \nequation. Then we solve the equation to find the value of the unknown variable.\n1.\t Calculate the value of m.\nm + 20°\n100°\n\tm + 20°\t =\t 100°\t\n[vert. opp. ∠s]\n\t\nm\t =\t 100° − 20°\n\t\n\t =\t\n2.\t Calculate the value of t.\n3t + 12°\n66°\n3.\t Calculate the value of p.\n2p + 30°\n108°\n4.\t Calculate the value of z.\n2z – 10°\n58°\nMaths_LB_gr8_bookA.indb   218\n2014/07/03   10:54:27 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n219\n5.\t Calculate the value of y.\n102° – 2y\n78°\n6.\t Calculate the value of r.\n180° – 3r\n126°\n12.3\t Lines intersected by a transversal\npairs of angles formed by a transversal\nA transversal is a line that crosses at least two other lines.\nThe blue line is the transversal\nWhen a transversal intersects two lines, we can compare the sets of angles on the two \nlines by looking at their positions.\nMaths_LB_gr8_bookA.indb   219\n2014/07/03   10:54:27 AM\n\n220\t MATHEMATICS Grade 8: Term 2\nThe angles that lie on the same side of the transversal and are in matching positions are \ncalled corresponding angles (corr. ∠s). In the figure,  \nthese are corresponding angles:\n• a and e\n• b and f\n• d and h\n• c and g.\n1.\t In the figure, a and e are both left of the  \ntransversal and above a line.\n\t\nWrite down the location of the following corresponding angles. The first one is done \n\t\nfor you.\n\t\nb and f:  Right of the transversal and above the lines.\n\t\nd and h: \n\t\nc and g: \nAlternate angles (alt. ∠s) lie on opposite sides of the transversal, but are not adjacent \nor vertically opposite. When the alternate angles lie between the two lines, they are \ncalled alternate interior angles. In the figure, these are alternate interior angles:\n• d and f\n• c and e.\nWhen the alternate angles lie outside of  \nthe two lines, they are called alternate  \nexterior angles. In the figure, these are  \nalternate exterior angles:\n• a and g\n• b and h.\n2.\t Write down the location of the following alternate angles:\n\t\nd and f:  \n\t\nc and e:  \n\t\na and g:  \n\t\nb and h:  \ne\nf\ng\nh\nb\nc\nd\na\ne\nf\ng\nh\nb\nc\nd\na\nMaths_LB_gr8_bookA.indb   220\n2014/07/03   10:54:28 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n221\nCo-interior angles (co-int. ∠s) lie on  \nthe same side of the transversal and  \nbetween the two lines. In the figure, these  \nare co-interior angles:\n• c and f\n• d and e.\n3.\t Write down the location of the following  \nco-interior angles:\n\t\nd and e: \n\t\nc and f: \nidentifying types of angles\nTwo lines are intersected by a transversal as shown below.\n \na\nb c\nd\ne\nf\ng\nh\nWrite down the following pairs of angles:\n1.\t two pairs of corresponding angles:  \n2.\t two pairs of alternate interior angles:  \n3.\t two pairs of alternate exterior angles:  \n4.\t two pairs of co-interior angles:  \n5.\t two pairs of vertically opposite angles:  \ne\nf\ng\nh\nb\nc\nd\na\nMaths_LB_gr8_bookA.indb   221\n2014/07/03   10:54:28 AM\n\n222\t MATHEMATICS Grade 8: Term 2\n12.4\t Parallel lines intersected by a transversal\ninvestigating angle sizes\nIn the figure below left, EF is a transversal to AB and CD. In the figure below right, PQ is a \ntransversal to parallel lines JK and LM. \n5\n8 7\n6\nC\n1\n4\n3\n2\nD\nE\nF\nB\nA\n9\n11\n10\n12\n13\n15\n14\n16\nQ\nJ\nK\nL\nP\nM\n1.\t Use a protractor to measure the sizes of all the angles in each figure. Write the \nmeasurements on the figures.\n2.\t Use your measurements to complete the following table.\n\t\nAngles\nWhen two lines are not parallel\nWhen two lines are parallel\nCorr. ∠s\n1 = \n ; 5 = \n \n4 = \n ; 8 = \n \n2 = \n ; 6 = \n \n3 = \n ; 7 = \n \n9 = \n ; 13\n = \n \n12\n = \n ; 16\n = \n \n10\n = \n ; 14\n = \n \n11\n = \n ; 15\n = \n \nAlt. int. ∠s\n4 = \n ; 6 = \n \n3 = \n ; 5 = \n \n12\n = \n ; 14\n = \n \n11\n = \n ; 13\n = \n \nAlt. ext. ∠s\n1 = \n ; 7 = \n \n2 = \n ; 8 = \n \n9 = \n ; 15\n = \n \n10\n = \n ; 16\n = \n \nCo-int. ∠s\n4 + 5 = \n \n3 + 6 = \n \n12\n + 13\n = \n \n11\n + 14\n = \n \nMaths_LB_gr8_bookA.indb   222\n2014/07/03   10:54:30 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n223\n3.\t Look at your completed table in question 2. What do you notice about the angles \nformed when a transversal intersects parallel lines?\nWhen lines are parallel:\n• corresponding angles are equal\n• alternate interior angles are equal\n• alternate exterior angles are equal\n• co-interior angles add up to 180°.\nidentifying angles on parallel lines\n1.\t Fill in the corresponding angles to those given.\n120° 60°\n80°\n100°\n45°\n135°\n2.\t Fill in the alternate exterior angles.\n120° 60°\n80°\n100°\n45°\n135°\n3.\t (a)\t Fill in the alternate interior angles.\n\t\n(b)\t Circle the two pairs of co-interior angles in each figure.\n120°\n60°\n80°\n100°\n45°\n135°\nMaths_LB_gr8_bookA.indb   223\n2014/07/03   10:54:30 AM\n\n224\t MATHEMATICS Grade 8: Term 2\n4.\t (a)\t Without measuring, fill in all the angles in the following figures that are equal to \n\t\nx and y.\n\t\n(b)\t Explain your reasons for each x and y that you filled in to your partner.\n\t\nA\t\nB\nx\ny\nx\ny\n5.\t Give the value of x and y below.\n75°\nx\ny\n165°\n12.5\t Finding unknown angles on parallel lines\nworking out unknown angles\nWork out the sizes of the unknown angles. Give reasons for your answers. (The first one \nhas been done as an example.)\n1.\t Find the sizes of x, y and z.\ny\nx\nz\nE\n74°\nB\nA\nD\nC\nF\n\t\nx = 74°\t\n[alt. ∠ with given 74°; AB // CD]\n\t\ny = 74°\t\n[corr. ∠ with x; AB // CD]\nor\t y = 74°\t\n[vert. opp. ∠ with given 74°]\n\t\nz = 106°\t\n[co-int. ∠ with x; AB // CD]\nor\t z = 106°\t\n[∠s on a straight line]\nMaths_LB_gr8_bookA.indb   224\n2014/07/03   10:54:30 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n225\n2.\t Work out the sizes of p, q and r.\nr\nq\np\nG\n133°\nL\nK\nN\nM\nH\n3.\t Find the sizes of a, b, c and d.\nb\na\nd\nH\n140°\nK\nS\nT\nc\n4.\t Find the sizes of all the angles in  \nthis figure.\nE\nB\nC\n130°\nA\nD\nF\n5.\t Find the sizes of all the angles.  \n(Can you see two transversals and \ntwo sets of parallel lines?)\nS\nY\nR\n95°\nJ\nP\nM\nX\nK\nJ\nK\nC\nA\nD\nB\nMaths_LB_gr8_bookA.indb   225\n2014/07/03   10:54:30 AM\n\n226\t MATHEMATICS Grade 8: Term 2\nEXTENSION\nTwo angles in the following diagram are given as x and y. Fill in all the angles that are equal  \nto x and y.\nx\ny\nsum of the angles in a quadrilateral\nThe diagram below is a section of the previous diagram.\nx\nx\nx\nx\nx\nx\nx\nx\ny\ny\ny\ny\ny\ny\ny\ny\n1.\t What kind of quadrilateral is in the diagram? Give a reason for your answer.\n2.\t Look at the top left intersection. Complete the following equation:\n\t\nAngles around a point = 360°\n\t\n∴\t x + y + \n + \n\t= 360°\n3.\t Look at the interior angles of the quadrilateral. Complete the following  \nequations:\n\t\nSum of angles in the quadrilateral = x + y + \n + \n\t\nFrom question 2:\t x + y + \n + \n\t= 360°\n\t\n∴\t\nSum of angles in a quadrilateral\t= \n°\nCan you think of another \nway to use the diagram \nabove to work out the sum of \nthe angles in a quadrilateral?\nMaths_LB_gr8_bookA.indb   226\n2014/07/03   10:54:31 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n227\n12.6\t Solving more geometric problems\nangle relationships on parallel lines\n1.\t Calculate the sizes of 1 to 7.\n1\n3\nC\nD\n7\n4\n5\n6\n154°\n2\nW\nZ\nY\nX\n2.\t Calculate the sizes of x, y and z.\ny\nz\nC\nD\n21°\nx\nE\nA\nB\n33°\n3.\t Calculate the sizes of a, b, c and d.\nb\nc\nT\nR\n154°\na\nQ\nP\nS\n48°\nU\nd\nMaths_LB_gr8_bookA.indb   227\n2014/07/03   10:54:31 AM\n\n228\t MATHEMATICS Grade 8: Term 2\n4.\t Calculate the size of x.\n3x\n2x\nX\nZ\nY\nW\n5.\t Calculate the size of x.\n3x\n124°\n6.\t Calculate the size of x.\nx + 50°\n3x – 10°\n7.\t Calculate the sizes of a and CEP.\nA\nC\nP\nE\nF\nQ\n5a + 40°\na + 100°\nD\nB\nMaths_LB_gr8_bookA.indb   228\n2014/07/03   10:54:31 AM\n\n\t\nCHAPTER 12: GEOMETRY OF STRAIGHT LINES\t\n229\nincluding properties of triangles and quadrilaterals\n1.\t Calculate the sizes of 1 to 6.\nK\nA\nB\nT\nD\n1\n2\n3\n4\n132°\n5\n6\n2.\t RSTU is a trapezium. Calculate the sizes of T and R.\nR\nS\nT\nU\n143°\n112°\n3.\t JKLM is a rhombus. Calculate the sizes of J ML, M\n2 and K\n1.\nK\nL\nM\nJ\n1\n102°\n2\n1\n2\n4.\t ABCD is a parallelogram. Calculate the sizes of ADB, ABD, C and DBC.\nA\nB\nC\nD\n20°\n82°\nMaths_LB_gr8_bookA.indb   229\n2014/07/03   10:54:32 AM\n\nWorksheet\nGEOMETRY OF STRAIGHT LINES\n1.\t Look at the drawing below. Name the items listed alongside.\nJ\nA\nK\nS\nL\nB\nM\nT\n(a)\ta pair of vertically opposite angles\n(b)\ta pair of corresponding angles\n(c)\ta pair of alternate interior angles\n(d)\ta pair of co-interior angles\n2.\t In the diagram, AB // CD. Calculate the sizes of FHG, F, C and D. Give reasons for  \n\t\nyour answers.\nD\nE\nA\nC\n85°\nF\nB\nG\nH\n53°\n110°\nI\n3.\t In the diagram, OK = ON, KN // LM, KL // MN and LKO = 160°. \n\t\nCalculate the value of x. Give reasons for your answers.\nM\nK\nL\nN\nO\nx\n140°\n160°\nMaths_LB_gr8_bookA.indb   230\n2014/07/03   10:54:33 AM\n\nTerm 2\nRevision and assessment",
      "chapter_id": "12"
    },
    {
      "title": "Term 2: Revision and assessment",
      "content": "TERM 2: REVISION AND ASSESSMENT\t\n231\nRevision........................................................................................................................... 232\n• Algebraic expressions 2.............................................................................................. 232\n• Algebraic equations 2................................................................................................. 235\n• Construction of geometric figures.............................................................................. 236\n• Geometry of 2D shapes.............................................................................................. 240\n• Geometry of straight lines.......................................................................................... 242\nAssessment...................................................................................................................... 244\nMaths_LB_gr8_bookA.indb   231\n2014/07/03   10:54:33 AM\n\n232\t MATHEMATICS Grade 8: Term 2\nRevision\nShow all your steps in your working.\nalgebraic expressions 2\n1.\t Simplify:\n\t\n(a)\t x2 + x2\n\t\n(b)\t m + m × m + m\n\t\n(c)\t 5ab – 7a2 – 2a2 + 11ba\n\t\n(d)\t (3ac2)(−4a2b)\n\t\n(e)\t (–4a2b3)3\n\t\n(f)\t\n−\n\n\n\n\n\n\n6\n3\n2\n4\n2\nx yz\nxyz\n\t\n(g)\t\n100\n81\n4\n64\nx\ny\n\t\n(h)\t\n16\n9\n2\n2\nc\nc\n+  \nMaths_LB_gr8_bookA.indb   232\n2014/07/03   10:54:34 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n233\n\t\n(i)\t (2x + 3x)3\n\t\n(j)\t 3x2(4x3 − 5) \n\t\n(k)\t (4a − 7a)(a2 − 2a − 5)\n\t\n(l)\t\n9\n3\n2\n3\n2\n2\n2\nc de\nc d e f\n\t\n(m)\t 6\n3\n2\n2\n2\n2\nb\nb\nb\n−\n\t\n(n)\t 10\n5\n1\n5\n2\nx\nx\n−\n+\n\t\n(o)\t 14\n21\n7\n2\n2\nx\nx\nx\n−\n2.\t Simplify the following expressions:\n\t\n(a)\t 3(a + 2b) − 4(b − 2a)\n\t\n(b)\t 3 − 2(5x2 + 6x − 2)\nMaths_LB_gr8_bookA.indb   233\n2014/07/03   10:54:35 AM\n\n234\t MATHEMATICS Grade 8: Term 2\n\t\n(c)\t 2x(x2 − x + 1) − 3(4 − x)\n\t\n(d)\t (2a + b − 4c) − (5a + b − c)\n\t\n(e)\t a{2a2[4 + 2(3a + 1)] − a}\n3.\t If a = 0, b = –2, and c = 3, determine the value of the following without using a \ncalculator. Show all working:\n\t\n(a)\t b2c\n\t\n(b)\t 2b – b(ab – 5bc)\n\t\n(c)\t 2\n10\n3\n2\nb\nc\na\nc\n−\n+\n4.\t If y = –2, find the value of 2y3 – 4y + 3\nMaths_LB_gr8_bookA.indb   234\n2014/07/03   10:54:35 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n235\nalgebraic equations 2\n1.\t Solve the following equations:\n\t\n(a)\t −x = −7\n\t\n(b)\t 2x = 24\n\t\n(c)\t 3x − 6 = 0\n\t\n(d)\t 2x + 5 = 3\n\t\n(e)\t 3(x − 4) = −3\n\t\n(f)\t 4(2x − 1) = 5(x − 2)\n2.\t Sello is x years old. Thlapo is 4 years older than Sello. The sum of their ages is 32. \n\t\n(a)\t Write this information in an equation using x as the variable.\n\t\n(b)\t Solve the equation to find Thlapo’s age.\nMaths_LB_gr8_bookA.indb   235\n2014/07/03   10:54:35 AM\n\n236\t MATHEMATICS Grade 8: Term 2\n3.\t The length of a rectangle is (2x + 8) cm and the width is 2 cm. The area of the \nrectangle is 12 cm2. \n\t\n(a)\t Write this information in an equation using x as the variable.\n\t\n(b)\t Solve the equation to determine the value of x.\n\t\n(c)\t How long is the rectangle?\n4.\t The area of a rectangle is (8x2 + 2x) cm2, and the length is 2x cm. Determine the  \nwidth of the rectangle in terms of x, in its simplest form.\nconstruction of geometric figures\nDo not erase any construction arcs in these questions.\n1.\t (a)\t Construct DEF = 56° with your ruler, pencil, and a protractor. Label the angle  \n\t\ncorrectly.\n\t\n(b)\t Bisect DEF using only a compass, ruler, and pencil (no protractor).\nMaths_LB_gr8_bookA.indb   236\n2014/07/03   10:54:35 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n237\n2.\t Here is a rough sketch of a quadrilateral (NOT drawn to scale):\n7,7 cm\n6 cm\n5,8 cm\n78°\n\t\nConstruct the quadrilateral accurately and full size below.\n3.\t Using only a compass, ruler and pencil, construct:\n\t\n(a)\t A line through C perpendicular to AB\n\t\n(b)\t A line through D perpendicular to AB\nD\nA\nC\nB\nMaths_LB_gr8_bookA.indb   237\n2014/07/03   10:54:36 AM\n\n238\t MATHEMATICS Grade 8: Term 2\n4.\t Construct and label the following triangles and quadrilaterals:\n\t\n(a)\t Triangle ABC, where AB = 8 cm; BC = 5,5 cm and AC = 4,9 cm\n\t\n(b)\t Rhombus GHJK, where GH = 6 cm and G = 50°\nMaths_LB_gr8_bookA.indb   238\n2014/07/03   10:54:36 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n239\n5.\t Here is a rough sketch of triangle FGH (NOT drawn to scale):\n7,1 cm\nH\n49 mm\n300°\nG\nF\n\t\nUsing a ruler, pencil, and protractor, construct and label the triangle accurately.  \n6.\t Construct an angle of 120° without using a protractor.\nMaths_LB_gr8_bookA.indb   239\n2014/07/03   10:54:36 AM\n\n240\t MATHEMATICS Grade 8: Term 2\ngeometry of 2D shapes\n1.\t True or false: all equilateral triangles, no matter what size they are, have angles that \nequal 60°.\n2.\t (a)\t In a triangle, two of the angles are 35° and 63°. Calculate the size of the third \n\t\nangle.\n\t\n(b)\t In a quadrilateral, one of the angles is a right angle, and another is 80°. If the  \n\t\nremaining two angles are equal to each other, what is the size of each?\n3.\t If triangle MNP has M = 40° and N = 90°, what is the size of P?\n4.\t Write definitions of the triangles in the table below.\nEquilateral triangle\nIsosceles triangle\nRight-angled triangle\nMaths_LB_gr8_bookA.indb   240\n2014/07/03   10:54:36 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n241\n5.\t The following list gives the properties of three quadrilaterals, A, B and C. \n\t\n(a)\t Give the special names of each of shapes A, B and C.\n\t\n\t\nQuadrilateral A: The opposite sides are equal and parallel. \n\t\n\t\nQuadrilateral B: The adjacent sides are equal, while the opposite sides are not equal.\n\t\n\t\nQuadrilateral C: All of the angles are right angles.\n\t\n(b)\t What property must Quadrilateral A also have to make it a rhombus?\n\t\n(c)\t What property must Quadrilateral A also have to make it a rectangle?\n6.\t Determine the size of V. Show all steps of your working and give reasons.\n7.\t Determine the size of x. Give reasons.\nU\n25°\n2x\n3x + 5°\nV\nT\n45°\nx\nMaths_LB_gr8_bookA.indb   241\n2014/07/03   10:54:36 AM\n\n242\t MATHEMATICS Grade 8: Term 2\ngeometry of straight lines\n1.\t Study the diagram alongside:\n\t\n(a)\t Name an angle that is vertically  \n\t\nopposite to EHG.\n\t\n(b)\t Name an angle that is corresponding  \n\t\nto EHG.\n\t\n(c)\t Name an angle that is co-interior  \n\t\nwith EHG.\n\t\n(d)\t Name an angle that is alternate to EHG.\n2.\t Determine the size of x in each of the following diagrams. Show all steps of working \nand give reasons.\n\t\n(a)\n\t\n(b)\nF\nK\nD\nJ\nH\nE\nL\nG\nL\nP\nM\nQ\nN\n4x\nx\nF\nG\nE\nL\nJ\nD\nK\n110°\nx + 90°\nH\nMaths_LB_gr8_bookA.indb   242\n2014/07/03   10:54:37 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n243\n\t\n(c)\t\n\t\n(d)\n\t\n(e)\t Are line segments AB and DE parallel?  \n\t\nProve your answer.\nB\nG\nE\nL\nJ\nD\nK\n60°\n2x\n130°\nx\nB\nA\nD\nC\nF\nB\nD\nC\nH\nE\nA\nG\n75°\n105°\nMaths_LB_gr8_bookA.indb   243\n2014/07/03   10:54:37 AM\n\n244\t MATHEMATICS Grade 8: Term 2\nAssessment\nIn this section, the numbers in brackets at the end of a question indicate the number \nof marks the question is worth. Use this information to help you determine how much \nworking is needed. The total number of marks allocated to the assessment is 75.\n1.\t Simplify the following expressions:\n\t\n(a)\t 5x2 – 6x2 + 10x2\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(1)\n\t\n(b)\t 4(3x – 7) – 3(2 + x)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n   \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n (2)\n\t\n(c)\t (–2a2bc3)2 ÷ 4abcd\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   (3)\n\t\n(d)\t 2\n3\n15\n3\nx\nx\nx\n(\n)\n−\n\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n  \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n  (3)\n\t\n(e)\t\n108\n4\n15\n6\n3\nd\nd\n÷ \n\t \t\n\t\n \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n    (3)\n\t\n(f)\t 2[3x2 – (4 – x2)] – [9 + (4x)2]\t\n\t\n \t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n(3)\nMaths_LB_gr8_bookA.indb   244\n2014/07/03   10:54:38 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n245\n2.\t Find the value of a if b = 3, c = –4 and d = 2:\n\t\n(a)\t a = b + c × d\t\t\n\t\n\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(2)\n\t\n(b)\t ab2 = 2c – d ÷ 2\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(3)\n3.\t Solve the following equations:\n\t\n(a)\t –7x = 56\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(2)\n\t\n(b)\t 4(x + 3) = 16\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(2)\n4.\t Sipho, Fundiswa and Ntosh are brothers. Sipho earns Rx per month; Fundiswa earns \nR1 000 more than Sipho per month, and Ntosh earns double what Sipho earns. If  \nyou add their salaries together you get a total of R27 000. \n\t\n(a)\t Write this information in an equation using x.\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n(2)\n\t\n(b)\t Solve the equation to find how much Fundiswa earns per month.\t\n\t\n   \t\n (2)\nMaths_LB_gr8_bookA.indb   245\n2014/07/03   10:54:38 AM\n\n246\t MATHEMATICS Grade 8: Term 2\n5.\t Construct the following figure using only a pencil, ruler and compass. Do not erase \nany construction arcs.\n\t\n(a)\t An angle of 60° \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(2)\n \n\t\n(b)\t The perpendicular bisector of line VW, where VW = 10 cm\t\n\t\n\t\n\t\n\t\n(3)\n\t\n(c)\t Triangle KLM, where KL = 8,3 cm; LM = 5,9 cm and KM = 7 cm\t     \t \t\n\t\n(4)\nMaths_LB_gr8_bookA.indb   246\n2014/07/03   10:54:38 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n247\n\t\n(d)\t Parallelogram EFGH, where E = 60°, EF = 4,2 cm and EH = 8 cm\t \t\n\t\n\t\n(4)\n6.\t (a)\t What is/are the property/properties that make a rhombus different to a  \n\t\nparallelogram?\t  \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n (1)\n\t\n(b)\t True or false: a rectangle is a special type of parallelogram.\t\n\t\n\t\n  \t\n\t\n (1)\n7.\t Determine the size of x in each figure. Show all the necessary steps and give reasons.\n\t\n(a)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   (3)\n54°\n2x\nMaths_LB_gr8_bookA.indb   247\n2014/07/03   10:54:38 AM\n\n248\t MATHEMATICS Grade 8: Term 2\n\t\n(b)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n   (4)\n\t\n(c)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n    \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(3)\n8.\t Study the following diagram. Then answer the questions that follow:\nx\ny\n\t\n(a)\t Write down the correct word to complete the sentence: x and y form a pair of\n\t\n\t\n angles.\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(1)\n\t\n(b)\t Write down an equation that shows the relationship between angles x and y.\t\n(1)\n80°\n75°\nx\n2x – 20°\n38°\nx\nMaths_LB_gr8_bookA.indb   248\n2014/07/03   10:54:38 AM\n\n\t\nTERM 2: REVISION AND ASSESSMENT\t\n249\n9.\t Determine the size of x, showing all necessary steps and giving reasons for all \nstatements that use geometrical theorems:\n\t\n(a)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(4)\n  \t\n(b)\t \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n  \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(5)\n\t\n(c)         \t \t\n\t\n\t\n\t\n\t\n\t\n\t\n   \t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n\t\n(3)\nx\n115º\n125°\nx + 20°\n2x\n72°\nx\nA\nB\nC\nD\nE\nF\nG\nMaths_LB_gr8_bookA.indb   249\n2014/07/03   10:54:38 AM\n\n250\t MATHEMATICS Grade 8: Term 2\n10.\tConsider the following diagram, in which it is given: DEI = 30°, DE = EI, DF // IG,  \nand GH = IH.\nE\nD\nI\nH\nG\nF\n\t\n(a)\t Determine, with reasons, the size of H.    \t\t\n\t\n\t\n\t\n\t\n\t\n\t\n  \t\n   \t\n(6)\n\t\n(b)\t Which of the following statements is correct? Explain your answer.\t \t\n\t\n(2)\n\t\n\t\n(i)\t  ∆DEI is similar to ∆GHI\n\t\n\t\n(ii)\t  ∆DEI is congruent to ∆GHI\n\t\n\t\n(iii) We cannot determine a relationship between ∆DEI and ∆GHI since there  \n\t\n\t\n is not enough information given.\n\t\nStatement \n is correct because \nMaths_LB_gr8_bookA.indb   250\n2014/07/03   10:54:39 AM",
      "chapter_id": "14"
    }
  ]
}