Chapter 7 - Pearson Schools

Chapter 7Expanding brackets and factorisingThis chapter will show you how to✔ expand and simplify expressions with brackets✔ solve equations and inequalities involving brackets✔ factorise by removing a common factor✔ expand two brackets7.1 Expanding bracketsYou will need to know how to● multiply positive and negative numbers● add and subtract negative numbers● collect like termsWhen multiplying algebraic terms remember thatx × 3 5 3 × x 5 3xy × y 5 y 2gh 5 g × hMore complicated multiplications can also be simplified.EXAMPLE 1Simplify 3f × 4g.3f × 4g 5 3 × f × 4 × g5 3 × 4 × f × g5 12 × fg5 12fgTo multiply algebraic terms,multiply the numbers then multiplythe letters.Multiplying a bracketYou can work out 6 × 34 by thinking of 34 as 30 1 4.6 × 34 5 6 × (30 1 4)5 6 × 30 1 6 × 45 180 1 245 20430 46 6 × 306 × 4Algebra 101

Expanding brackets and factorisingEXERCISE 7A1 Simplify these expressions.(a) 2 × 5k (b) 3 × 6b (c) 4a × 5(d) 3a × 2b (e) 4c × 3d (f) x × 5y2 Expand the brackets to find the value of these expressions.Check your answers by working out the brackets first.(a) 2(50 1 7) (b) 5(40 1 6) (c) 6(70 1 3)(d) 3(40 2 2) (e) 7(50 2 4) (f) 8(40 2 3)3 Remove the brackets from these.(a) 5(p 1 6) (b) 3(x 1 y) (c) 4(u 1 v 1 w)(d) 2(y 2 8) (e) 7(9 2 z) (f) 8(a 2 b 1 6)4 Expand the brackets in these expressions.(a) 3(2c 1 6) (b) 5(4t 1 3) (c) 2(5p 1 q)(d) 3(2a 2 b) (e) 6(3c 2 2d) (f) 7(2x 1 y 2 3)(g) 6(3a 2 4b 1 c) (h) 2(x 2 1 3x 1 2) (i) 4(y 2 2 3y 2 10)5 Write down the 6 pairs of cards which show equivalent expressions.Remember, you must multiply eachterm inside the brackets by the termoutside the bracket.A common mistake is to forget tomultiply the second term.Remember3 × 2c 5 3 × 2 × c 5 6c4(x 1 2y) 4x 1 2y 2(4x 1 y) 4(2x 2 y)A B C D8x 2 8y 4x 1 8y 8(x 2 y) 2x 2 8yE F G H8x 1 2y 2(x 2 4y) 2(2x 1 y) 8x 2 4yI J K LYou can use the same method for expressions that have an algebraicterm instead of a number term outside the bracket.EXAMPLE 3Expand the brackets in these expressions.(a) a(a 1 4) (b) x(2x 2 y) (c) 3t(t 2 1 1)(a) a(a 1 4) 5 a × a 1 a × 45 a 2 1 4a(b) x(2x 2 y) 5 x × 2x 2 x × y5 2x 2 2 xy(c) 3t(t 2 1 1) 5 3t × t 2 1 3t × 15 3t 3 1 3tRemember a × a 5 a 2x × 2x 5 x × 2 × x 5 2 × x × x5 2x 2Remember x × y 5 xy3t × t 2 5 3 × t × t × t 5 3t 3Algebra 103

EXERCISE 7BExpand the brackets in these expressions.1 b(b 1 4) 2 a(5 1 a) 3 k(k 2 6)4 m(9 2 m) 5 a(2a 1 3) 6 g(4g 1 1)7 p(2p 1 q) 8 t(t 1 5w) 9 m(m 1 3n)10 x(2x 2 y) 11 r(4r 2 t) 12 a(a 2 4b)13 2t(t 1 5) 14 3x(x 2 8) 15 5k(k 1 l)16 3a(2a 1 4) 17 2g(4g 1 h) 18 5p(3p 2 2q)19 3x(2y 1 5z) 20 r(r 2 1 1) 21 a(a 2 1 3)22 t(t 2 –7) 23 2p(p 2 1 3q) 24 4x(x 2 1 x)Remember3x × 4x 5 4 × 3 × x × x5 12x 2Adding and subtracting expressions with bracketsAddingTo add expressions with brackets, expand the brackets first, thencollect like terms to simplify your answer.Collecting like terms means addingall the terms in x, all the terms in yand so on.EXAMPLE 4Expand then simplify these expressions.(a) 3(a 1 4) 1 2a 1 10 (b) 3(2x 1 5) 1 2(x 2 4)(a) 3(a 1 4) 1 2a 1 10 5 3a 1 12 1 2a 1 105 3a 1 2a 1 12 1 105 5a 1 22(b) 3(2x 1 5) 1 2(x 2 4) 5 6x 1 15 1 2x 2 85 6x 1 2x 1 15 2 85 8x 1 7Expand the brackets first. Thencollect like terms.Expand both sets of brackets first.SubtractingIf you have an expression like 23(2x 2 5), multiply both terms in thebrackets by 23.23 × 2x 5 26x and 23 × 25 5 15So 23(2x 2 5) 5 23 × 2x 1 23 × 255 26x 1 15Multiplying × × × × 104 Algebra

Expanding brackets and factorisingEXAMPLE 5Expand these expressions.(a) 22(3t 1 4) (b) 23(4x 2 1)(a) 22(3t 1 4) 5 22 × 3t 1 22 × 45 26t 1 285 26t 2 8(b) 23(4x 2 1) 5 23 3 4x 1 23 3 215 212x 1 322 × 3 5 26 22 × 4 5 2823 × 4 5 212 23 × 21 5 13EXAMPLE 6Expand then simplify these expressions.(a) 3(2t 1 1) 2 2(2t 1 4) (b) 8(x 1 1) 2 3(2x 2 5)(a) 3(2t 1 1) 2 2(2t 1 4) 5 6t 1 3 2 4t 2 85 6t 2 4t 1 3 2 85 2t 2 5(b) 8(x 1 1) 2 3(2x 2 5) 5 8x 1 8 2 6x 1 155 8x 2 6x 1 8 1 155 2x 1 23Remember to multiply both termsin the second bracket by 22.Expand the brackets first.Remember that 23 × 25 5 115.Then collect like terms.EXERCISE 7CExpand these expressions.1 22(2k 1 4) 2 23(2x 1 6) 3 25(3n 2 1)4 24(3t 1 5) 5 23(4p 2 1) 6 22(3x 2 7)Expand then simplify these expressions.7 3(y 1 4) 1 2y 1 10 8 2(k 1 6) 1 3k 1 99 4(a 1 3) 2 2a 1 6 10 3(t 2 2) 1 4t 2 1011 3(2y 1 3) 1 2(y 1 5) 12 4(x 1 7) 1 3(x 1 4)13 3(2x 1 5) 1 2(x 2 4) 14 2(4n 1 5) 1 5(n 2 3)15 3(x 2 5) 1 2(x 2 3) 16 4(2x 2 1) 1 2(3x 2 2)17 3(2b 1 1) 2 2(2b 1 4) 18 4(2m 1 3) 2 2(2m 1 5)19 2(5t 1 3) 2 2(3t 1 1) 20 5(2k 1 2) 2 4(2k 1 6)21 8(a 1 1) 2 3(2a 2 5) 22 2(4p 1 1) 2 4(p 2 3)23 5(2g 2 4) 2 2(4g 2 6) 24 2(w 2 4) 2 3(2w 2 1)25 x(x 1 3) 1 4(x 1 2) 26 x(2x 1 1) 2 3(x 2 4)Algebra 105

7.2 Solving equations involving bracketsEquations sometimes involve brackets.When dealing with equations involving brackets, you usually expandthe brackets first.EXAMPLE 7Solve 4(c 1 3) 5 20.Method A4(c 1 3) 5 204c 1 12 5 204c 1 12 2 12 5 20 2 124c 5 84c4 5 8 4c 5 2Method B4(c 1 3) 5 20c 1 3 5 20 4c 1 3 5 5c 5 5 2 3c 5 2Expand the bracket by multiplyingboth terms inside the bracket by theterm outside the bracket.You must subtract 12 from bothsides before dividing both sidesby 4.Since 4 divides exactly into 20 youcan divide both sides by 4 first.EXAMPLE 8Solve 2(3p 2 4) 5 7.2(3p – 4) 5 76p – 8 5 76p – 8 1 8 5 7 1 86p 5 156p6 5 15 6p 5 2.5Expand the bracket.You must add 8 to both sides beforedividing both sides by 6.106 Algebra

Expanding brackets and factorisingEXERCISE 7D1 Solve these equations.(a) 4(g 1 6) 5 32 (b) 7(k 1 1) 5 21 (c) 5(s 1 10) 5 65(d) 2(n 2 4) 5 6 (e) 3(f 2 2) 5 24 (f) 6(v 2 3) 5 42(g) 4(m 2 3) 5 14 (h) 2(w 1 7) 5 192 Solve these equations.(a) 4(5t 1 2) 5 48 (b) 3(2r 1 4) 5 30 (c) 2(2b 1 2) 5 22(d) 2(3w 2 6) 5 27 (e) 3(4x 2 2) 5 24 (f) 5(2y 1 11) 5 40(g) 6(2k 2 1) 5 36 (h) 3(2a 2 13) 5 18When two brackets are involved, expand both brackets then collectlike terms before solving.Like terms are terms of the samekind. In Example 9 there are onlyterms in m and number terms.EXAMPLE 9Solve 2(2m 1 10) 5 12(m 2 1).2(2m 1 10) 5 12(m 2 1)4m 1 20 5 12m 2 1220 1 12 5 12m 2 4m32 5 8mm 5 4Expand the brackets on both sidesof the equation and collect liketerms.Collect terms in m on the RHSbecause 12m on the RHS is greaterthan 4m on the LHS. This keeps them term positive.When an equation involves fractions, it can be transformed into anequation without fractions by multiplying all terms by the LCM ofthe numbers in the denominators.LCM means Lowest CommonMultiple.EXAMPLE 10Solve the equation x 1 1745 x 1 2.x 1 175 x 1 244(x 1 17)5 4(x 1 2)4x 1 17 5 4x 1 817 2 8 5 4x 2 x9 5 3xx 5 3Multiply both sides by 4, collect liketerms and then finally divide by 3.Note the use of brackets.Collect the terms in x on the RHSand the numbers on the LHS.4x on the RHS is greater than x onthe LHS.Algebra 107

EXAMPLE 11Solve the equation x 2 63x 2 65 x 1 43 515(x 2 6)515(x 1 4)355(x – 6) 5 3(x 1 4)5x 2 30 5 3x 1 125x 2 3x 5 12 1 302x 5 42x 5 215 x 1 45 .Look at the denominators.3 and 5 have a LCM of 15 somultiply both sides of the equationby 15.Note the use of brackets. Alwaysput them in when you multiply inthis way.Then solve using the method shownin Example 9.EXAMPLE 12Solve the equation 2x 1 361 x 2 235 5 2 .6(2x 1 3)16(x 2 2)5 6(5)6 3 22x 1 3 1 2(x 2 2) 5 152x 1 3 1 2x 2 4 5 154x 2 1 5 154x 5 16x 5 4The LCM here is 6.Note the use of brackets.The most common mistake is toforget to multiply all terms by theLCM.This means the number on the RHSas well as the terms on the LHS.EXERCISE 7E1 Solve these equations.(a) 2a 1 4 5 5(a 2 1) (b) 3(d 2 2) 5 2d 2 1(c) 5(x 1 3) 5 11x 1 3 (d) 12p 1 3 5 3(p 1 7)(e) 4t 1 3 5 3(2t 2 3) (f) 3b 2 4 5 2(2b 2 7)(g) 8(3g 2 1) 5 15g 1 10 (h) 3(2k 1 6) 5 17k 1 7(i) 2(y 1 5) 5 3y 1 12 (j) 5r 1 3 5 4(2r 1 3)2 Solve the following equations by expanding both brackets.(a) 2(b 1 1) 5 8(2b 2 5) (b) 5(4a 1 7) 5 3(8a 1 9)(c) 6(x 2 2) 5 3(3x 2 8) (d) 5(2p 1 2) 5 6(p 1 5)(e) 9(3s 2 4) 5 5(4s 2 3) (f) 4(10t 2 7) 5 3(6t 2 2)(g) 4(2w 1 2) 5 2(5w 1 7) (h) 3(3y 2 2) 5 7(y 2 2)Use Example 9 to help.108 Algebra

Expanding brackets and factorising3 Solve these equations.(a) d 1 315(c) 3x 2 18(e) c 2 844 Solve these equations.(a) x 1 1 5 x 2 13 4(c) 3x 1 15(e) x 1 275 3 2 d (b)6y 2 555 x 2 2 (d) 6 1 a25 c 1 1 (f) 10 2 b35 2x 35 3x 1 655 Solve these equations.(a) x 1 1 1 x 1 22 5(c) 3x 1 25(e) x 2 341 x 1 232 x 1 33(b) 2x 2 13(d) x 1 32(f) 8 2 x25 3 (b) x 1 245 2 (d) 3x 2 155 1 (f) 2x 1 545 y 1 35 a 1 45 12 1 b5 x 25 x 2 355 2x 1 251 x 1 172 x 1 232 x 1 435 35 1 55 2Use Example 10 to help.Use Example 11 to help.Use Example 12 to help.7.3 Solving inequalities involvingbracketsInequalities can also involve brackets.Remember that there is usually more than one answer when you solve aninequality and you need to state all possible values of the solution set.EXAMPLE 13Solve these inequalities.(a) 9 < 3(y 2 1) (b) 3(2x 2 5) . 2(x 1 4)(a) 9 < 3(y 2 1) (b) 3(2x 2 5) . 2(x 1 4)9 < 3y 2 3 6x 2 15 . 2x 1 89 1 3 < 3y 6x 2 2x . 8 1 1512 < 3y 4x . 234 < y x . 23 4x . 5 3 4You must remember to keep theinequality sign in your answer.For example, if you leave (a) as4 5 y you will lose a mark becauseyou have not included all possiblevalues of y.If you are asked for integersolutions to (b) the final answer willbe x > 6.Algebra 109

EXAMPLE 14n is an integer.List the values of n such that 211 , 2(n 2 3) , 1.211 , 2(n – 3) , 1211 , 2n – 6 , 1211 1 6 , 2n , 1 1 625 , 2n , 722.5 , n , 3.5Values of n are 22, 21, 0, 1, 2, 3This is a double inequality.Expand the bracket.Add 6 throughout.Remember to list the integersolutions as you were asked to inthe question.Remember to include 0.EXERCISE 7F1 Solve these inequalities.(a) 2(x 2 7) < 8 (b) 7 , 2(m 1 5)(c) 4(3w 2 1) . 20 (d) 3(2y 1 1) < 215(e) 2(p 2 3) . 4 1 3p (f) 1 2 5k , 2(5 1 2k)(g) 5(x 2 1) > 3(x 1 2) (h) 2(n 1 5) < 3(2n 2 2)2 Solve these inequalities then list the integer solutions.(a) 24 < 2x < 8 (b) 26 , 3y , 15(c) 28 < 4n , 17 (d) 212 , 6m < 30(e) 25 , 2(t 1 1) , 7 (f) 23 , 3(x 2 4) , 6(g) 26 < 5(y 1 1) < 11 (h) 217 , 2(2x 2 3) < 107.4 Factorising by removing a commonfactorFactorising an algebraic expression is the opposite of expandingbrackets. To factorise an expression, look for a common factor –that is, a number that divides into all the terms in the expression.To factorise completely, use the HCF of the terms.For example,6x 1 10 can be written as 2(3x 1 5)because 6x 5 2 × 3xand 10 5 2 × 5HCF means highest common factor.2 is a factor of 6x.2 is also a factor of 10.So 2 is a common factor of 6xand 10.Notice that the common factor isthe term outside the bracket.110 Algebra

Expanding brackets and factorisingEXAMPLE 15Copy and complete these.(a) 3t 1 15 5 3(h 1 5) (b) 4n 1 12 5 h(n 1 3)(a) 3t 1 15 5 3(t 1 5)(b) 4n 1 12 5 4(n 1 3)Because 3 × t 5 3t (and 3 × 5 5 15)4 × n 5 4n and 4 × 3 5 12EXAMPLE 16Factorise these expressions.(a) 5a 1 20 (b) 4x 2 12 (c) x 2 1 7x (d) 6p 2 q 2 2 9pq 3(a) 5a 1 20 5 5 × a 1 5 × 45 5(a 1 4)5 5(a 1 4)Check 5(a 1 4) 5 5 × a 1 5 × 45 5a 1 20 3(b) 4x 2 12 5 4 × x 2 4 × 35 4(x 2 3)5 4(x 2 3)(c) x 2 1 7x 5 x × x 1 x × 75 x(x 1 7)5 x(x 1 7)(d) 6p 2 q 2 2 9pq 35 3 × 2 × p × p × q × q 2 3 × 3 × p × q × q × q5 3pq 2 (2p 2 3q)5 3pq 2 (2p 2 3q)5 is a factor of 5a 5a 5 5 × a5 is also a factor of 20 20 5 5 × 4So 5 is a common factor of 5a and20 and is the term outside thebracket.Check your answer by removing thebrackets.2 is a factor of 4x and 12.4 is also a common factor of 4x and12. Use 4 because it is the highestcommon factor (HCF) of 4x and 12.Always look for the HCF.x is a common factor of x 2 and 7x.The HCF is 3pq 2 .EXERCISE 7G1 Copy and complete these.(a) 3x 1 15 5 3(h 1 5) (b) 5a 1 10 5 5(h 1 2)(c) 2x 2 12 5 2(x 2 h) (d) 4m 2 16 5 4(m 2 h)(e) 4t 1 12 5 h(t 1 3) (f) 3n 1 18 5 h(n 1 6)(g) 2b 2 14 5 h(b 2 7) (h) 4t 2 20 5 h(t 2 5)Use Example 15 to help.Don’t forget to check your answersby removing the brackets.Algebra 111

2 Factorise these expressions.(a) 5p 1 20 (b) 2a 1 12 (c) 3y 1 15(d) 7b 1 21 (e) 4q 1 12p (f) 6k 1 24l3 Factorise these expressions.(a) 4t 2 12 (b) 3x 2 9 (c) 5n 2 20(d) 2b 2 8 (e) 6a 2 18b (f) 7k 2 74 Factorise these expressions.(a) y 2 1 7y (b) x 2 1 5x (c) n 2 1 n(d) x 2 2 7x (e) p 2 2 8p (f) a 2 2 ab5 Factorise these expressions.(a) 6p 1 4 (b) 4a 1 10 (c) 6 2 4t(d) 12m 2 8n (e) 25x 1 15y (f) 12y 2 9z6 Factorise completely.(a) 3x 2 2 6x (b) 8x 2 2 xy (c) 8a 1 4ab(d) p 3 2 5p 2 (e) 3t 3 1 6t 2 (f) 10yz 2 15y 2(g) 18a 2 1 12ab (h) 16p 2 2 12pq7 Factorise these expressions.(a) 4ab 2 1 6ab 3 (b) 10xy 2 5x 2(c) 3p 2 q 2 6p 3 q 2(d) 8mn 3 1 4n 2 2 6m 2 n(e) 6h 2 k 2 12hk 3 2 18h 2 k 2Use Example 16(a) to help.Remember 5 5 5 × 1Use Example 16(c) to help.Remember n 5 n × 1Remember 6p 5 2 × 3pUse Example 16(a) to help. Look forthe common factors in the terms.Expanding two bracketsYou can use a grid method to multiply two numbers.For example,34 × 57 34 × 57 5 (30 1 4) × (50 1 7)× 50 7 5 30 × 50 1 30 × 7 1 4 × 50 1 4 × 730 1500 210 5 1500 1 210 1 200 1 284 200 28 5 1938You can also use a grid method when you multiply two bracketstogether. You have to multiply each term in one bracket by eachterm in the other bracket.For example,To expand and simplify (x 1 2)(x 1 5)× x 5x x 2 5x2 2x 10(x 1 2)(x 1 5) 5 x × x 1 x × 5 1 2 × x 1 2 × 55 x 2 1 5x 1 2x 1 105 x 2 1 7x 1 10You simplify the final expression bycollecting the like terms.5x 1 2x 5 7x.112 Algebra

Expanding brackets and factorisingIt is like working out the area ofa rectangle of length x 1 5 andwidth x 1 2.Total area 5 (x 1 2)(x 1 5)5 x 2 1 5x 1 2x 1 105 x 2 1 7x 1 10x2x 5area = x × x= x 2 area = x × 5= 5xarea = 2 × x= 2xarea = 2 × 5= 10EXAMPLE 17Expand and simplify these.(a) (a 1 4)(a 1 10)(b) (t 1 6)(t 2 2)(a) × a 10a a 2 10a4 4a 40(a 1 4)(a 1 10) 5 a × a 1 a × 10 1 4 × a 1 4 × 105 a 2 1 10a 1 4a 1 405 a 2 1 14a 1 40(b) × t 22t t 2 22t6 6t 212(t 1 6)(t 2 2) 5 t × t 1 t × (22) 1 6 × t 1 6 × (22)5 t 2 2 2t 1 6t 2 125 t 2 1 4t 2 12Remember you can use a grid tohelp.Remember to multiply each term inthe first bracket by each term in thesecond bracket.Remember you are multiplyingby 22.1ve × 2 ve 5 2 ve.Look again at the last example.(t 6) (t 2)5 t 2 2 2t 1 6t 2 125 t 2 1 4t 2 12The First terms in each bracket multiply to give t 2 .The Outside pair of terms multiply to give 22t.The Inside pair of terms multiply to give 16t.The Last terms in each bracket multiply to give 212.This method is often known as FOILand is another way of expandingbrackets.Algebra 113

EXERCISE 7HExpand and simplify.1 (a 1 2)(a 1 7) 2 (x 1 3)(x 1 1) 3 (x 1 5)(x 1 5)4 (t 1 5)(t 2 2) 5 (x 1 7)(x 2 4) 6 (n 2 5)(n 1 8)7 (x 2 4)(x 1 5) 8 (p 2 4)(p 1 4) 9 (x 2 9)(x 2 4)10 (h 2 3)(h 2 8) 11 (y 2 3)(y 2 3) 12 (4 1 a)(a 1 7)13 (m 2 7)(8 1 m) 14 (6 1 q)(7 1 q) 15 (d 1 5)(4 2 d)16 (8 2 x)(3 2 x) 17 (x 2 12)(x 2 7) 18 (y 2 16)(y 1 6)Be careful when there are negativesigns 2 this is where a lot ofmistakes are made.Squaring an expressionYou can use the same method of expanding two brackets for examplesinvolving the square of an expression.To square an expression, write out the bracket twice and expand.EXAMPLE 18Expand and simplify (x 1 4) 2 .(x 4) (x 4)(x 1 4) 2 5 (x 1 4)(x 1 4)5 x 2 1 4x 1 4x 1 165 x 2 1 8x 1 16You need to multiply the expression(x 1 4) by itself so write down thebracket twice and expand as youdid in Example 17 or use FOIL as inthis example.Notice that you do not just squarethe x and the 4, there are two otherterms in the expansion.EXERCISE 7I1 Expand and simplify.(a) (x 1 5) 2 (b) (x 1 6) 2 (c) (x 2 3) 2(d) (x 1 1) 2 (e) (x 2 4) 2 (f) (x 2 5) 2(g) (x 1 7) 2 (h) (x 2 8) 2 (i) (3 1 x) 2(j) (2 1 x) 2 (k) (5 2 x) 2 (l) (x 1 a) 22 Copy and complete these by finding the correct number to go ineach box.(a) (x 1 h) 2 5 x 2 1 hx 1 36 (b) (x 2 h) 2 5 x 2 2 hx 1 49(c) (x 1 h) 2 5 x 2 1 18x 1 h (d) (x 2 h) 2 5 x 2 2 20x 1 hIn question 1, see if you can spotthe pattern between the terms inthe brackets and the finalexpression.114 Algebra

Expanding brackets and factorising3 Expand and simplify.(a) (x 1 4)(x 2 4) (b) (x 1 5)(x 2 5) (c) (x 1 2)(x 2 2)(d) (x 2 11)(x 1 11) (e) (x 2 3)(x 1 3) (f) (x 2 1)(x 1 1)(g) (x 1 9)(x 2 9) (h) (x 1 a)(x 2 a) (i) (t 1 x)(t 2 x)What happens to the x term whenyou multiply brackets of the form(x 1 a)(x 2 a)?EXAMPLE 19Remember to multiply each term inExpand and simplify (3x 2 y)(x 2 2y).5 3x 2 2 7xy 1 2y 2 2ve × 2 ve 5 1 ve.the first bracket by each term in thesecond bracket.× x 22y3x 3x 2 26xyBe careful when there are negative2y 2xy 2y 2signs. This is where a lot of mistakesare made.(3x 2 y)(x 2 2y) 5 3x 2 2 6xy 2 xy 1 2y1ve × 2 ve 5 2 ve.EXERCISE 7JExpand and simplify.1 (3a 1 2)(a 1 4) 2 (5x 1 3)(x 1 2) 3 (2t 1 3)(3t 1 5)4 (4y 1 1)(2y 1 7) 5 (6x 1 5)(2x 1 3) 6 (4x 1 3)(x 2 1)7 (2z 1 5)(3z 2 2) 8 (y 1 1)(7y 2 8) 9 (3n 2 5)(n 1 8)10 (3b 2 5)(2b 1 1) 11 (p 2 4)(7p 1 3) 12 (2z 2 3)(3z 2 4)13 (5x 2 9)(2x 2 1) 14 (2y 2 3)(2y 2 3) 15 (2 1 3a)(4a 1 5)16 (3x 1 4) 2 17 (2x 2 7) 2 18 (5 2 4x) 219 (2x 1 1)(2x 2 1) 20 (3y 1 2)(3y 2 2) 21 (5n 1 4)(5n 2 4)22 (3x 1 5)(3x 2 5) 23 (1 1 2x)(1 2 2x) 24 (3t 1 2x)(3t 2 2x)Can you see the connectionbetween questions 19–24 andquestion 3 in Exercise 7I ?Algebra 115

EXAMINATION QUESTIONS1 Solve the inequality 7 2 5x > 217, given that x is a positive integer. [3](CIE Paper 2, Nov 2000)2 Solve the inequality 25 2 3x , 7. [2](CIE Paper 2, Jun 2001)3 Solve the inequality 3(x 1 7) , 5x 2 9. [2](CIE Paper 2, Jun 2002)4 (a) Solve the inequality 5 2 2x 3 > 1 2 + x 4 . [3](b) List the positive integers which satisfy the inequality 5 2 2x 3 > 1 2 + x 4 . [1](CIE Paper 2, Nov 2002)5 Solve the equation x 2 8 5 22. [2]4(CIE Paper 2, Jun 2004)6 (a) Factorise completely 12x 2 2 3y 2 . [2](b) (i) Expand (x 2 3) 2 . [2](ii) x 2 2 6x 1 10 is to be written in the form (x 2 p) 2 1 q.Find the values of p and q. [2](CIE Paper 2, Jun 2004)7 Solve the equation 3x 2 25(CIE Paper 2, Nov 2004)5 8. [2]116 Algebra