Methods in Algebra -Cambridge University

CHAPTER ONE

Methods in Algebra

Fluency in algebra, and particularly in factoring, is absolutely vital for everythingin this course. Much of this chapter will be a review of earlier work, but severaltopics will probably be quite new, including:• the sum and difference of cubes in Section 1E• three simultaneous equations in three variables in Section 1H.

1 A Arithmetic with PronumeralsA pronumeral is a symbol that stands for a number. The pronumeral may standfor a known number, or for an unknown number, or it may be a variable, standingfor any one of a whole set of possible numbers. Pronumerals, being numbers, cantherefore take part in all the operations that are possible with numbers, such asaddition, subtraction, multiplication and division (except by zero).

Like and Unlike Terms: An algebraic expression consists of pronumerals, numbers andthe operations of arithmetic. Here is an example:

x2 + 2x + 3x2 − 4x − 3

This particular algebraic expression can be simplified by combining like terms.• The two like terms x2 and 3x2 can be combined to give 4x2.• Another pair of like terms 2x and −4x can be combined to give −2x.• This yields three unlike terms, 4x2, −2x and −3, which cannot be combined.

WORKED EXERCISE:Simplify each expression by combining like terms.(a) 7a + 15 − 2a − 20 (b) x2 + 2x + 3x2 − 4x − 3

SOLUTION:

(a) 7a + 15 − 2a − 20 = 5a − 5 (b) x2 + 2x + 3x2 − 4x − 3 = 4x2 − 2x − 3

Multiplying and Dividing: To simplify a product like 3y × (−6y), or a quotient like10x2y ÷ 5y, work systematically through the signs, then the numerals, and theneach pronumeral in turn.

WORKED EXERCISE:Simplify these products and quotients.(a) 3y × (−6y) (b) 4ab × 7bc (c) 10x2y ÷ 5y

SOLUTION:

(a) 3y × (−6y) = −18y2 (b) 4ab × 7bc = 28ab2c (c) 10x2y ÷ 5y = 2x2

ISBN: 9781107679573 Photocopying is restricted under law and this material must not be transferred to another party

© Bill Pender, David Sadler, Julia Shea, Derek Ward 2012 Cambridge University Press

� �2 CHAPTER 1: Methods in Algebra CAMBRIDGE MATHEMATICS 2 UNIT YEAR 11

Index Laws: Here are the standard laws for dealing with indices. They will be coveredin more detail in Chapter Seven.

1

THE INDEX LAWS:• To multiply powers of the same base, add the indices: ax ay = ax+y

• To divide powers of the same base, subtract the indices:ax

ay= ax−y

• To raise a power to a power, multiply the indices: (ax)n = axn

• The power of a product is the product of the powers: (ab)x = ax bx

• The power of a quotient is the quotient of the powers:(

a

b

)x

=ax

bx

In expressions with several factors, work systematically through the signs, thenthe numerals, and then each pronumeral in turn.

WORKED EXERCISE:Use the index laws above to simplify each expression.(a) 3x4 × 4x3

(b) (48x7y3) ÷ (16x5y3)

(c) (3a4)3

(d) (−5x2)3 × (2xy)4(e)

(2x

3y

)4

SOLUTION:

(a) 3x4 × 4x3 = 12x7 (multiplying powers of the same base)

(b) (48x7y3) ÷ (16x5y3) = 3x2 (dividing powers of the same base)

(c) (3a4)3 = 27a12 (raising a power to a power)

(d) (−5x2)3 × (2xy)4 = −125x6 × 16x4y4 (two powers of products)

= −2000x10y4 (multiplying powers of the same base)

(e)(

2x

3y

)4

=16x4

81y4 (a power of a quotient)

Exercise 1A1. Simplify:

(a) 5x + 3x (b) 5x − 3x (c) −5x + 3x (d) −5x − 3x

2. Simplify:(a) −2a + 3a + 4a (b) −2a − 3a + 4a (c) −2a − 3a − 4a (d) −2a + 3a − 4a

3. Simplify:(a) −2x + x

(b) 3y − y

(c) 3a − 7a

(d) −8b + 5b

(e) 4x − (−3x)(f) −2ab − ba

(g) −3pq + 7pq

(h) −5abc− (−2abc)

4. Simplify:(a) 6x + 3 − 5x

(b) −2 + 2y − 1(c) 3a − 7 − a + 4(d) 3x − 2y + 5x + 6y

(e) −8t + 12 − 2t − 17(f) 2a2 + 7a − 5a2 − 3a

(g) 9x2 − 7x + 4 − 14x2 − 5x − 7(h) 3a− 4b− 2c + 4a + 2b− c + 2a− b− 2c



� �CHAPTER 1: Methods in Algebra 1A Arithmetic with Pronumerals 3

5. Simplify:(a) −3a × 2 (b) −4a × (−3a) (c) a2 × a3 (d) (a2)3

6. Simplify:(a) −10a ÷ 5 (b) −24a ÷ (−8a) (c) a7 ÷ a3 (d) 7a2 ÷ 7a

7. Simplify:

(a)5x

x(b)

−7x3

x(c)

−12a2b

−ab(d)

−27x6y7z2

9x3y3z

8. Simplify:(a) t2 + t2 (b) t2 − t2 (c) t2 × t2 (d) t2 ÷ t2

9. Simplify:(a) −6x + 3x (b) −6x − 3x (c) −6x × 3x (d) −6x ÷ 3x

D E V E L O P M E N T

10. If a = −2, find the value of:(a) 3a + 2 (b) a3 − a2 (c) 3a2 − a + 4 (d) a4 + 3a3 + 2a2 − a

11. If x = 2 and y = −3, find the value of:(a) 3x + 2y (b) y2 − 5x (c) 8x2 − y3 (d) x2 − 3xy + 2y2

12. Subtract:(a) x from 3x (b) −x from 3x (c) 2a from −4a (d) −b from −5b

13. Multiply:(a) 5a by 2(b) 6x by −3

(c) −3a by a

(d) −2a2 by −3ab

(e) 4x2 by −2x3

(f) −3p2q by 2pq3

14. Divide:(a) −2x by x

(b) 3x3 by x2

(c) x3y2 by x2y

(d) a6x3 by −a2x3

(e) 14a5b4 by −2a4b

(f) −50a2b5c8 by −10ab3c2

15. Find the sum of:(a) x + y + z, 2x + 3y − 2z and 3x − 4y + z

(b) 2a − 3b + c, 15a − 21b − 8c and 24b + 7c + 3a

(c) 5ab + bc − 3ca, ab − bc + ca and −ab + 2ca + bc

(d) x3 − 3x2y + 3xy2 , −2x2y − xy2 − y3 and x3 + 4y3

16. From:(a) 7x2 − 5x + 6 take 5x2 − 3x + 2(b) 4a − 8b + c take a − 3b + 5c

(c) 3a + b − c − d take 6a − b + c − 3d

(d) ab − bc − cd take −ab + bc − 3cd

17. Subtract:(a) x3 − x2 + x + 1 from x3 + x2 − x + 1(b) 3xy2 − 3x2y + x3 − y3 from x3 + 3x2y + 3xy2 + y3

(c) b3 + c3 − 2abc from a3 + b3 − 3abc

(d) x4 + 5 + x − 3x3 from 5x4 − 8x3 − 2x2 + 7




18. Simplify:(a) 2a2b4 × 3a3b2 (b) −6ab5 × 4a3b3 (c) (−3a3)2 (d) (−2a4b)3

C H A L L E N G E

19. What must be added to 4x3 − 3x2 + 2 to give 3x3 + 7x − 6?

20. Simplify:

(a)3a × 3a × 3a

3a + 3a + 3a(b)

3c × 4c2 × 5c3

3c2 + 4c2 + 5c2 (c)ab2 × 2b2c3 × 3c3a4

a3b3 + 2a3b3 + 3a3b3

21. Simplify:

(a)(−2x2)3

−4x(b)

(3xy3)3

3x2y4 (c)(−ab)3 × (−ab2)2

−a5b3 (d)(−2a3b2)2 × 16a7b

(2a2b)5

22. Divide the product of (−3x7y5)4 and (−2xy6)3 by (−6x3y8)2.

1 B Expanding BracketsExpanding brackets is routine in arithmetic. For example, to calculate 7 × 61,

7 × (60 + 1) = 7 × 60 + 7 × 1,

which quickly gives the result 7 × 61 = 420 + 7 = 427. The algebraic version ofthis procedure can be written as:

2EXPANDING BRACKETS IN ALGEBRA:

a(x + y) = ax + ay

WORKED EXERCISE:Expand and simplify each expression.(a) 3x(4x − 7) (b) 5a(3 − b) − 3b(6 − 5a)

SOLUTION:

(a) 3x(4x − 7) = 12x2 − 21x (b) 5a(3 − b) − 3b(6 − 5a) = 15a − 5ab − 18b + 15ab

= 15a + 10ab − 18b

Expanding the Product of Two Bracketed Terms: Each pair of brackets should be ex-panded in turn and the reulting expression should then be simplified.

WORKED EXERCISE:Expand and simplify each expression.(a) (x + 3)(x − 5) (b) (3 + x)(9 + 3x + x2)

SOLUTION:

(a) (x + 3)(x − 5)= x(x − 5) + 3(x − 5)= x2 − 5x + 3x − 15= x2 − 2x − 15

(b) (3 + x)(9 + 3x + x2)= 3(9 + 3x + x2) + x(9 + 3x + x2)= 27 + 9x + 3x2 + 9x + 3x2 + x3

= 27 + 18x + 6x2 + x3



� �CHAPTER 1: Methods in Algebra 1B Expanding Brackets 5

Special Expansions: These three identities are important and must be memorised.Examples of these expansions occur very frequently, and knowing the formulaegreatly simplifies the working. They are proven in the exercises.

3

SQUARE OF A SUM: (A + B)2 = A2 + 2AB + B2

SQUARE OF A DIFFERENCE: (A − B)2 = A2 − 2AB + B2

DIFFERENCE OF SQUARES: (A + B)(A − B) = A2 − B2

WORKED EXERCISE: Use the three special expansions above to simplify:

(a) (4x + 1)2 (b) (s − 3t)2 (c) (x + 3y)(x − 3y)

SOLUTION:

(a) (4x + 1)2 = 16x2 + 8x + 1 (the square of a sum)

(b) (s − 3t)2 = s2 − 6st + 9t2 (the square of a difference)

(c) (x + 3y)(x − 3y) = x2 − 9y2 (the difference of squares)

Exercise 1B1. Expand:

(a) 3(x − 2)(b) 2(x − 3)(c) −3(x − 2)

(d) −2(x − 3)(e) −3(x + 2)(f) −2(x + 3)

(g) −(x − 2)(h) −(2 − x)(i) −(x + 3)

2. Expand:(a) 3(x + y)(b) −2(p − q)(c) 4(a + 2b)

(d) x(x − 7)(e) −x(x − 3)(f) −a(a + 4)

(g) 5(a + 3b − 2c)(h) −3(2x − 3y + 5z)(i) xy(2x − 3y)

3. Expand and simplify:(a) 2(x + 1) − x

(b) 3a + 5 + 4(a − 2)(c) 2 + 2(x − 3)(d) −3(a + 2) + 10

(e) 3 − (x + 1)(f) b + c − (b − c)(g) (2x − 3y) − (3x − 2y)(h) 3(x − 2) − 2(x − 5)

(i) −4(a − b) − 3(a + 2b)(j) 4(s − t) − 5(s + t)(k) 2x(x + 6y) − x(x − 5y)(l) −7(2a−3b+c)−6(−a+4b−2c)

4. Expand and simplify:(a) (x + 2)(x + 3)(b) (y + 4)(y + 7)(c) (t + 6)(t − 3)(d) (x − 4)(x + 2)(e) (t − 1)(t − 3)

(f) (2a + 3)(a + 5)(g) (u − 4)(3u + 2)(h) (4p + 5)(2p − 3)(i) (2b − 7)(b − 3)(j) (5a − 2)(3a + 1)

(k) (6 − c)(c − 3)(l) (2d − 3)(4 + d)

(m) (2x + 3)(y − 2)(n) (a − 2)(5b + 4)(o) (3 − 2m)(4 − 3n)


5. (a) By expanding (A+B)(A+B), prove the special expansion (A+B)2 = A2 +2AB+B2 .(b) Similarly, prove the special expansions:

(i) (A − B)2 = A2 − 2AB + B2 (ii) (A − B)(A + B) = A2 − B2




6. Use the special expansions to expand:(a) (x + y)2

(b) (x − y)2

(c) (x−y)(x+y)(d) (a + 3)2

(e) (b − 4)2

(f) (c + 5)2

(g) (d − 6)(d + 6)(h) (7 + e)(7 − e)(i) (8 + f)2

(j) (9 − g)2

(k) (h + 10)(h − 10)(l) (i + 11)2

(m) (2a + 1)2

(n) (2b − 3)2

(o) (3c + 2)2

(p) (2d + 3e)2

(q) (2f +3g)(2f −3g)(r) (3h − 2i)(3h + 2i)

(s) (5j + 4)2

(t) (4k − 5�)2

(u) (4 + 5m)(4 − 5m)(v) (5 − 3n)2

(w) (7p + 4q)2

(x) (8 − 3r)2

7. Expand:(a) −a(a2 − a − 1)(b) −2x(x3 − 2x2 − 3x + 1)

(c) 3xy(2x2y − 5x3)(d) −2a2b(a2b3 − 2a3b)

8. Simplify:(a) 14 − (

10 − (3x − 7) − 8x)

(b) 4(a − 2(b − c) − (

a − (b − 2)))

9. Expand and simplify:

(a)(

t +1t

)2

(b)(

t − 1t

)2(c)

(t +

1t

) (t − 1

t

)

C H A L L E N G E

10. Subtract a(b + c − a) from the sum of b(c + a − b) and c(a + b − c).

11. Multiply:(a) a − 2b by a + 2b

(b) 2 − 5x by 5 + 4x(c) 4x + 7 by itself(d) x2 + 3y by x2 − 4y

(e) a + b − c by a − b

(f) 9x2 − 3x + 1 by 3x + 1

12. Expand and simplify:(a) (a − b)(a + b) − a(a − 2b)(b) (x + 2)2 − (x + 1)2

(c) (a − 3)2 − (a − 3)(a + 3)

(d) (p + q)2 − (p − q)2

(e) (2x + 3)(x − 1) − (x − 2)(x + 1)(f) 3(a − 4)(a − 2) − 2(a − 3)(a − 5)

13. Use the special expansions to find the value of:(a) 1022 (b) 9992 (c) 203 × 197

1 C FactoringFactoring is the reverse process of expanding brackets, and will be needed on aroutine basis throughout the course. There are four basic methods, but in everysituation, common factors should always be taken out first.

4

THE FOUR BASIC METHODS OF FACTORING:

• HIGHEST COMMON FACTOR: Always try this first.

• DIFFERENCE OF SQUARES: This involves two terms.

• QUADRATICS: This involves three terms.

• GROUPING: This involves four or more terms.

Factoring should continue until each factor is irreducible, meaning that it cannotbe factored further.



� �CHAPTER 1: Methods in Algebra 1C Factoring 7

Factoring by Taking Out the Highest Common Factor: Always look first for any commonfactors of all the terms, and then take out the highest common factor.

WORKED EXERCISE:Factor each expression by taking out the highest common factor.(a) 4x3 + 3x2 (b) 18a2b3 − 30b3

SOLUTION:

(a) The highest common factor of 4x3 and 3x2 is x2,so 4x3 + 3x2 = x2(4x + 3).

(b) The highest common factor of 18a2b3 and 30b3 is 6b3,so 18a2b3 − 30b3 = 6b3(3a2 − 5).

Factoring by Difference of Squares: The expression must have two terms, both ofwhich are squares. Sometimes a common factor must be taken out first.

WORKED EXERCISE:Use the difference of squares to factor each expression.(a) a2 − 36 (b) 80x2 − 5y2

SOLUTION:

(a) a2 − 36 = (a + 6)(a − 6)

(b) 80x2 − 5y2 = 5(16x2 − y2) (Take out the highest common factor.)= 5(4x − y)(4x + y) (Use the difference of squares.)

Factoring Monic Quadratics: A quadratic is called monic if the coefficient of x2 is 1.Suppose that we want to factor the monic quadratic expression x2 − 13x + 36.We look for two numbers:• whose sum is −13 (the coefficient of x), and• whose product is +36 (the constant term).

WORKED EXERCISE:Factor these monic quadratics.(a) x2 − 13x + 36 (b) a2 + 12a − 28

SOLUTION:

(a) The numbers with sum −13 and product +36 are −9 and −4,so x2 − 13x + 36 = (x − 9)(x − 4).

(b) The numbers with sum +12 and product −28 are +14 and −2,so a2 + 12a − 28 = (a + 14)(a − 2).

Factoring Non-monic Quadratics: In a non-monic quadratic like 2x2 +11x+12, wherethe coefficient of x2 is not 1, we look for two numbers:• whose sum is 11 (the coefficient of x), and• whose product is 12 × 2 = 24 (the constant times the coefficient of x2).




WORKED EXERCISE:Factor these non-monic quadratics.(a) 2x2 + 11x + 12 (b) 6s2 − 11s − 10

SOLUTION:

(a) The numbers with sum 11 and product 12 × 2 = 24 are 8 and 3,so 2x2 + 11x + 12 = (2x2 + 8x) + (3x + 12) (Split 11x into 8x + 3x.)

= 2x(x + 4) + 3(x + 4) (Take out the HCF of each group.)= (2x + 3)(x + 4). (x + 4 is a common factor.)

(b) The numbers with sum −11 and product −10 × 6 = −60 are −15 and 4,so 6s2 − 11s − 10 = (6s2 − 15s) + (4s − 10) (Split −11s into −15s + 4s.)

= 3s(2s − 5) + 2(2s − 5) (Take out the HCF of each group.)= (3s + 2)(2s − 5). (2s − 5 is a common factor.)

Factoring by Grouping: When there are four or more terms, it is sometimes possibleto split the expression into groups, factor each group in turn, and then factor thewhole expression by taking out a common factor or by some other method.

WORKED EXERCISE:Factor each expression by grouping.(a) 12xy − 9x − 16y + 12 (b) s2 − t2 + s − t

SOLUTION:

(a) 12xy − 9x − 16y + 12 = 3x(4y − 3) − 4(4y − 3) (Take out the HCF of each pair.)= (3x − 4)(4y − 3) (4y − 3 is a common factor.)

(b) s2 − t2 + s − t = (s + t)(s − t) + (s − t) (Factor s2 − t2 using difference of squares.)= (s − t)(s + t + 1) (s − t is a common factor.)

Exercise 1C1. Factor, by taking out any common factors:

(a) 2x + 8(b) 6a − 15(c) ax − ay

(d) 20ab − 15ac

(e) x2 + 3x

(f) p2 + 2pq

(g) 3a2 − 6ab

(h) 12x2 + 18x

(i) 20cd − 32c

(j) a2b + b2a

(k) 6a2 + 2a3

(l) 7x3y − 14x2y2

2. Factor, by grouping in pairs:(a) mp + mq + np + nq

(b) ax − ay + bx − by

(c) ax + 3a + 2x + 6(d) a2 + ab + ac + bc

(e) z3 − z2 + z − 1

(f) ac + bc − ad − bd

(g) pu − qu − pv + qv

(h) x2 − 3x − xy + 3y

(i) 5p − 5q − px + qx

(j) 2ax − bx − 2ay + by

(k) ab + ac − b − c

(l) x3 + 4x2 − 3x − 12(m) a3 − 3a2 − 2a + 6(n) 2t3 + 5t2 − 10t − 25(o) 2x3 − 6x2 − ax + 3a

3. Factor, using the difference of squares:(a) a2 − 1(b) b2 − 4(c) c2 − 9(d) d2 − 100

(e) 25 − y2

(f) 1 − n2

(g) 49 − x2

(h) 144 − p2

(i) 4c2 − 9(j) 9u2 − 1(k) 25x2 − 16(l) 1 − 49k2

(m) x2 − 4y2

(n) 9a2 − b2

(o) 25m2 − 36n2

(p) 81a2b2 − 64



� �CHAPTER 1: Methods in Algebra 1D Algebraic Fractions 9

4. Factor each quadratic expression. They are all monic quadratics.(a) a2 + 3a + 2(b) k2 + 5k + 6(c) m2 + 7m + 6(d) x2 + 8x + 15(e) y2 + 9y + 20(f) t2 + 12t + 20

(g) x2 − 4x + 3(h) c2 − 7c + 10(i) a2 − 7a + 12(j) b2 − 8b + 12(k) t2 + t − 2(l) u2 − u − 2

(m) w2 − 2w − 8(n) a2 + 2a − 8(o) p2 − 2p − 15(p) y2 + 3y − 28(q) c2 − 12c + 27(r) u2 − 13u + 42

(s) x2 − x − 90(t) x2 + 3x − 40(u) t2 − 4t − 32(v) p2 + 9p − 36(w) u2 − 16u − 80(x) t2 + 23t − 50


5. Factor each quadratic expression. They are all non-monic quadratics.(a) 3x2 + 4x + 1(b) 2x2 + 5x + 2(c) 3x2 + 16x + 5(d) 3x2 + 8x + 4(e) 2x2 − 3x + 1(f) 5x2 − 13x + 6

(g) 5x2 − 11x + 6(h) 6x2 − 11x + 3(i) 2x2 − x − 3(j) 2x2 + 3x − 5(k) 3x2 + 2x − 5(l) 3x2 + 14x − 5

(m) 2x2 − 7x − 15(n) 2x2 + x − 15(o) 6x2 + 17x − 3(p) 6x2 − 7x − 3(q) 6x2 + 5x − 6(r) 5x2 + 23x + 12

(s) 5x2 + 4x − 12(t) 5x2 − 19x + 12(u) 5x2 − 11x − 12(v) 5x2 + 28x − 12(w) 9x2 − 6x − 8(x) 3x2 + 13x − 30

6. Use the techniques of the previous questions to factor each expression.(a) a2 − 25(b) b2 − 25b

(c) c2 − 25c + 100(d) 2d2 + 25d + 50(e) e3 + 5e2 + 5e + 25(f) 16 − f 2

(g) 16g2 − g3

(h) h2 + 16h + 64

(i) i2 − 16i − 36(j) 5j2 + 16j − 16(k) 4k2 − 16k − 9(l) 2k3 − 16k2 − 3k + 24

(m) 2a2 + ab − 4a − 2b

(n) 6m3n4 + 9m2n5

(o) 49p2 − 121q2

(p) t2 − 14t + 40

(q) 3t2 + 2t − 40(r) 5t2 + 54t + 40(s) 5t2 + 33t + 40(t) 5t3 + 10t2 + 15t

(u) u2 + 15u − 54(v) 3x3 − 2x2y − 15x + 10y

(w) 1 − 36a2

(x) 4a2 − 12a + 9

C H A L L E N G E

7. Write each expression as a product of three factors. (Take out any common factors first.)(a) 3a2 − 12(b) x4 − y4

(c) x3 − x

(d) 5x2 − 5x − 30

(e) 25y − y3

(f) 16 − a4

(g) 4x2 + 14x − 30(h) a4 + a3 + a2 + a

(i) c3 + 9c2 − c − 9(j) x3 − 8x2 + 7x

(k) x4 − 3x2 − 4(l) ax2 − a − 2x2 + 2

1 D Algebraic FractionsAn algebraic fraction is a fraction containing pronumerals. They are manipulatedin the same way as arithmetic fractions, and factoring may play a major role.

Adding and Subtracting Algebraic Fractions: A common denominator is needed. Find-ing the lowest common denominator may involve factoring each denominator.

5

ADDING AND SUBTRACTING ALGEBRAIC FRACTIONS:

• First factor every denominator.

• Then work with the lowest common denominator.




WORKED EXERCISE:Use a common denominator to simplify each algebraic fraction.

(a)5x

6+

11x

4(b)

23x

− 35x

(c)1

x − 4− 1

x(d)

2 + x

x2 − x− 5

x − 1

SOLUTION:

(a)5x

6+

11x

4=

10x

12+

33x

12

=43x

12

(b)23x

− 35x

=1015x

− 915x

=1

15x

(c)1

x − 4− 1

x=

x − (x − 4)x(x − 4)

=4

x(x − 4)

(d)2 + x

x2 − x− 5

x − 1

=2 + x

x(x − 1)− 5

x − 1

=2 + x − 5x

x(x − 1)

=2 − 4x

x(x − 1)

Cancelling Algebraic Fractions: The key step here is to factor the numerator anddenominator completely before cancelling factors.

6

CANCELLING ALGEBRAIC FRACTIONS:

• First factor the numerator and denominator.

• Then cancel out all common factors.

WORKED EXERCISE:Simplify each algebraic fraction.

(a)6x + 8

6 (b)x2 − x

x2 − 1

SOLUTION:

(a)6x + 8

6=

2(3x + 4)6

=3x + 4

3(which could be written as x + 4

3 ).

(b)x2 − x

x2 − 1=

x(x − 1)(x + 1)(x − 1)

=x

x + 1

Multiplying and Dividing Algebraic Fractions: These processes are done exactly as forarithmetic fractions.

7

MULTIPLYING AND DIVIDING ALGEBRAIC FRACTIONS:

• First factor all numerators and denominators completely.

• Then cancel common factors.

To divide by an algebraic fraction, multiply by its reciprocal.



� �CHAPTER 1: Methods in Algebra 1D Algebraic Fractions 11

WORKED EXERCISE:Simplify these products and quotients of algebraic fractions.

(a)2a

a2 − 9× a − 3

5a(b)

12x

x + 1÷ 6x

x2 + 2x + 1

SOLUTION:

(a)2a

a2 − 9× a − 3

5a=

2a

(a − 3)(a + 3)× a − 3

5a(Factor a2 − 9.)

=2

5(a + 3)(Cancel a − 3 and a.)

(b)12x

x + 1÷ 6x

x2 + 2x + 1=

12x

x + 1× x2 + 2x + 1

6x(Multiply by the reciprocal.)

=12x

x + 1× (x + 1)2

6x(Factor x2 + 2x + 1.)

= 2(x + 1) (Cancel x + 1 and 6x.)

Simplifying Compound Fractions: A compound fraction is a fraction in which eitherthe numerator or the denominator is itself a fraction.

8

SIMPLIFYING COMPOUND FRACTIONS:

• Find the lowest common multiple of the denominators on the top and bottom.

• Multiply top and bottom by this lowest common multiple.

This will clear all the fractions from the top and bottom together.

WORKED EXERCISE:Simplify each compound fraction.

(a)12 − 1

314 + 1

6

(b)1t + 11t − 1

SOLUTION:

(a)12 − 1

314 + 1

6

=12 − 1

314 + 1

6

× 1212

=6 − 43 + 2

= 25

(b)1t + 11t − 1

=1t + 11t − 1

× t

t

=1 + t

1 − t

Exercise 1D1. Simplify:

(a)x

x(b)

2x

x(c)

x

2x(d)

a

a2 (e)3x2

9xy(f)

12ab

4a2b

2. Simplify:

(a)x

3× 3

x

(b)a

4÷ a

2

(c) x2 × 3x

(d)12b

× b2

(e)3x

4× 2

x2

(f)5a÷ 10

(g)2ab

3× 6

ab2

(h)8ab

5÷ 4ab

15




3. Write as a single fraction:

(a)x

2+

x

5

(b)a

3− a

6

(c)x

8− y

12

(d)2a

3+

3a

2

(e)7b

10− 19b

30

(f)xy

30− xy

12

(g)1x

+12x

(h)34x

+43x

(i)1a− 1

b

(j) x +1x

(k) a +b

a

(l)1x− 1

x2


4. Simplify:

(a)x + 1

2+

x + 23

(b)2x − 1

5+

2x + 34

(c)x + 3

6+

x − 312

(d)x + 2

2− x + 3

3

(e)2x + 1

4− 2x − 3

5

(f)2x − 1

3− 2x + 1

6

(g)2x + 1

3− x − 5

6+

x + 44

(h)3x − 7

5+

4x + 32

− 2x − 510

(i)x − 53x

− x − 35x

5. Factor where possible and then simplify:

(a)2p + 2q

p + q

(b)3t − 122t − 8

(c)x2 + 3x

3x + 9

(d)a

ax + ay

(e)3a2 − 6ab

2a2b − 4ab2

(f)x2 + 2x

x2 − 4

(g)a2 − 9

a2 + a − 12

(h)x2 + 2x + 1

x2 − 1

(i)x2 + 10x + 25x2 + 9x + 20

(j)ac + ad + bc + bd

a2 + ab

(k)y2 − 8y + 152y2 − 5y − 3

(l)9ax + 6bx − 6ay − 4by

9x2 − 4y2

6. Simplify:

(a)1x

+1

x + 1

(b)1x− 1

x + 1

(c)1

x + 1+

1x − 1

(d)2

x − 3+

3x − 2

(e)3

x + 1− 2

x − 1

(f)2

x − 2− 2

x + 3

(g)x

x + y+

y

x − y

(h)a

x + a− b

x + b

(i)x

x − 1− x

x + 1

7. Simplify:

(a)8a3b

5÷ 4ab

15

(b)2a

3b× 5c2

2a2b× 3b2

2c

(c)12x2yz

8xy3 × 24xy2

36yz2

(d)3a2b

4b3c× 2c2

8a3 ÷ 6ac

16b2

C H A L L E N G E

8. Simplify:

(a)3x + 3

2x× x2

x2 − 1

(b)a2 + a − 2

a + 2× a2 − 3a

a2 − 4a + 3

(c)c2 + 5c + 6

c2 − 16÷ c + 3

c − 4

(d)x2 − x − 20

x2 − 25× x2 − x − 2

x2 + 2x − 8÷ x + 1

x2 + 5x

(e)ax + bx − 2a − 2b

3x2 − 5x − 2× 9x2 − 1

a2 + 2ab + b2

(f)2x2 + x − 15x2 + 3x − 28

÷x2 + 6x + 9x2 − 4x

÷6x2 − 15x

x2 − 49



� �CHAPTER 1: Methods in Algebra 1E Factoring the Sum and Difference of Cubes 13

9. Simplify:

(a)1

x2 + x+

1x2 − x

(b)1

x2 − 4+

1x2 − 4x + 4

(c)1

x − y+

2x − y

x2 − y2

(d)3

x2 + 2x − 8− 2

x2 + x − 6

(e)x

a2 − b2 − x

a2 + ab

(f)1

x2 − 4x + 3+

1x2 − 5x + 6

− 1x2 − 3x + 2

10. Simplify:

(a)b − a

a − b

(b)v2 − u2

u − v

(c)x2 − 5x + 6

2 − x

(d)1

a − b− 1

b − a

(e)m

m − n+

n

n − m

(f)x − y

y2 + xy − 2x2

11. Study the worked exercise on compound fractions and then simplify:

(a)1 − 1

2

1 + 12

(b)2 + 1

3

5 − 23

(c)12 − 1

5

1 + 110

(d)1720 − 3

445 − 3

10

(e)1x

1 + 2x

(f)t − 1

t

t + 1t

(g)1

1b + 1

a

(h)xy + y

xxy − y

x

(i)1 − 1

x+11x + 1

x+1

(j)3

x+2 − 2x+1

5x+2 − 4

x+1

1 E Factoring the Sum and Difference of CubesThe factoring of the difference of cubes is similar to the factoring of the differenceof squares. The sum of cubes can also be factored, whereas the sum of squarescannot be factored.

9DIFFERENCE OF CUBES: A3 − B3 = (A − B)(A2 + AB + B2)SUM OF CUBES: A3 + B3 = (A + B)(A2 − AB + B2)

The proofs of these identities are left to the first question in the following exercise.

WORKED EXERCISE:Factor each expression.(a) x3 − 8 (b) 27a3 + 1

SOLUTION:

(a) x3 − 8 = (x − 2)(x2 + 2x + 4) (Use the difference of cubes x3 − 23.)

(b) 27a3 + 1 = (3a + 1)(9a2 − 3a + 1) (Use the sum of cubes (3a)3 + 13.)

WORKED EXERCISE:

(a) Simplifya3 + 1a + 1

. (b) Factor a3 − b3 + a − b.

SOLUTION:

(a)a3 + 1a + 1

=(a + 1)(a2 − a + 1)

a + 1= a2 − a + 1

(b) a3 − b3 + a − b

= (a − b)(a2 + ab + b2) + (a − b)= (a − b)(a2 + ab + b2 + 1)




Exercise 1E1. (a) Prove the identity A3 − B3 = (A − B)(A2 + AB + B2) by expanding the RHS.

(b) Similarly, prove the identity A3 + B3 = (A + B)(A2 − AB + B2).

2. Factor each expression.(a) x3 + y3

(b) a3 − b3

(c) y3 + 1

(d) k3 − 1(e) a3 + 8(f) b3 − 8

(g) 27 − t3

(h) 27 + u3

(i) x3 + 64

(j) y3 − 64(k) 125 + a3

(l) 125 − b3


3. Factor each expression.(a) 8p3 + 1(b) 8q3 − 1(c) u3 − 64v3

(d) t3 + 64u3

(e) 27c3 + 8(f) 27d3 − 8(g) 64m3 − 125n3

(h) 64p3 + 125q3

(i) 216e3 − 343f 3

(j) 216g3 + 343h3

(k) a3b3c3 + 1000(l) 729x3 − 1331y3

4. Factor each expression fully. (Remember to take out any common factors first.)(a) 5x3 − 5(b) 2x3 + 16(c) a4 − ab3

(d) 24t3 + 81(e) x3y − 125y

(f) 250p3 − 432q3

(g) 27x4 + 1000xy3

(h) 5x3y3 − 5(i) x6 + x3y3

C H A L L E N G E

5. Simplify each expression.

(a)x3 − 1x2 − 1

(b)a2 − 3a − 10

a3 + 8

(c)a3 + 16a2 × 3a

a2 + a

(d)x2 − 9

x4 − 27x÷ x + 3

x2 + 3x + 9

6. Simplify:

(a)3

a − 2− 3a

a2 + 2a + 4

(b)1

x3 − 1+

x + 1x2 + x + 1

(c)1

x2 − 2x − 8− 1

x3 + 8

(d)a2

a3 + b3 +a − b

a2 − ab + b2 +1

a + b

1 F Solving Linear EquationsThe first principle in solving any equation is to simplify the equation by doingthe same things to both sides. Linear equations can be solved completely byfollowing this principle

10SOLVING LINEAR EQUATIONS:• Any number can be added to or subtracted from both sides.• Both sides can be multiplied or divided by any non-zero number.

An equation involving algebraic fractions can often be reduced to a linear equationby following these steps.



� �CHAPTER 1: Methods in Algebra 1F Solving Linear Equations 15

WORKED EXERCISE: Solve each equation.(a) 6x + 5 = 4x − 9 (b)

4 − 7x

4x − 7= 1

SOLUTION:

(a) 6x + 5 = 4x − 9

− 4x 2x + 5 = −9

− 5 2x = −14

÷ 2 x = −7

(b)4 − 7x

4x − 7= 1

× (4x − 7) 4 − 7x = 4x − 7

+ 7x 4 = 11x − 7

+ 7 11 = 11x

÷ 11 x = 1

Changing the Subject of a Formula: Similar sequences of operations allow the subjectof a formula to be changed from one pronumeral to another.

WORKED EXERCISE:Change the subject of each formula to x.

(a) y = 4x − 3 (b) y =x + 1x + 2

SOLUTION:

(a) y = 4x − 3

+ 3 y + 3 = 4x

÷ 4y + 3

4= x

x =y + 3

4

(b) y =x + 1x + 2

× (x + 2) xy + 2y = x + 1

Rearranging, xy − x = 1 − 2y

Factoring, x(y − 1) = 1 − 2y

÷ (y − 1) x =1 − 2y

y − 1

Exercise 1F1. Solve:

(a) a − 10 = 5

(b) t + 3 = 1

(c) 5c = −35

(d)n

6= −3

(e) −2x = −20

(f) 3x = 2

(g) −a = 5

(h)x

−4= −1

(i) −1 − x = 0

(j) 0·1y = 5

(k) 2t = t

(l) − 12 x = 8

2. Solve:(a) 2x + 1 = 7(b) 5p − 2 = −2

(c)a

2− 1 = 3

(d) 3 − w = 4(e) 3x − 5 = 22

(f) 4x + 7 = −13

(g) 1 − 2x = 9(h) 6x = 3x − 21

(i) −2 = 4 +t

5

(j) −13 = 5a − 6(k) 19 = 3 − 7y

(l) 23 − u

3= 7

3. Solve:(a) 3n − 1 = 2n + 3(b) 4b + 3 = 2b + 1(c) 5x − 2 = 2x + 10(d) 5 − x = 27 + x

(e) 16 + 9a = 10 − 3a

(f) 13y − 21 = 20y − 35(g) 13 − 12x = 6 − 3x

(h) 3(x + 7) = −2(x − 9)

(i) 8 + 4(2 − x) = 3 − 2(5 − x)(j) 7x − (3x + 11) = 6 − (15 − 9x)(k) 4(x + 2) = 4x + 9(l) 3(x − 1) = 2(x + 1) + x − 5





4. Solve:

(a)x

8=

12

(b)a

12=

23

(c)y

20=

45

(d)1x

= 3

(e)2a

= 5

(f) 3 =92y

(g)2x + 1

5= −3

(h)5a

3= a + 1

(i)7 − 4x

6= 1

(j)5 + a

a= −3

(k)9 − 2t

t= 13

(l) 6 − c

3= c

(m)1a

+ 4 = 1 − 2a

(n)4

x − 1= −5

(o)3x

1 − 2x= 7

(p)11t

8t + 13= −2

5. Solve:(a) (x − 3)(x + 6) = (x − 4)(x − 5)(b) (1 + 2x)(4 + 3x) = (2 − x)(5 − 6x)

(c) (x + 3)2 = (x − 1)2

(d) (2x − 5)(2x + 5) = (2x − 3)2

6. (a) If v = u + at, find a when t = 4, v = 20 and u = 8.(b) Given that v2 = u2 + 2as, find the value of s when u = 6, v = 10 and a = 2.

(c) Suppose that1u

+1v

=1t. Find v, given that u = −1 and t = 2.

(d) If S = −15, n = 10 and a = −24, find �, given that S = n2 (a + �).

(e) The formula F = 95 C + 32 relates temperatures in degrees Fahrenheit and Celsius.

Find the value of C that corresponds to F = 95.

(f) Suppose that c and d are related by the formula3

c + 1=

5d − 1

. Find c when d = −2.

7. Solve each problem by forming, and then solving, a linear equation.(a) Three less than four times a certain number is equal to 21. Find the number.(b) Five more than twice a certain number is one more than the number itself. What is

the number?(c) Bill and Derek collect Batman cards. Bill has three times as many cards as Derek,

and altogether they have 68 cards. How many cards does Derek have?(d) If I paid $1.45 for an apple and an orange, and the apple cost 15 cents more than the

orange, how much did the orange cost?

8. Solve:

(a)x

3− x

5= 2

(b) y +y

2= 1

(c)a

10− a

6= 1

(d)x

6+

23

=x

2− 5

6

(e)x

3− 2 =

x

2− 3

(f)1x− 3 =

12x

(g)12x

− 23

= 1 − 13x

(h)x − 2

3=

x + 44

(i)3

x − 2=

42x + 5

(j)x + 1x + 2

=x − 3x + 1

(k)(3x − 2)(3x + 2)

(3x − 1)2 = 1

(l)a + 5

2− a − 1

3= 1

9. Rearrange each formula so that the pronumeral written in square brackets is the subject.

(a) a = bc − d [b]

(b) t = a + (n − 1)d [n]

(c)p

q + r= t [r]

(d) u = 1 +3v

[v]



� �CHAPTER 1: Methods in Algebra 1G Solving Quadratic Equations 17

C H A L L E N G E

10. Solve:

(a)34− x + 1

12=

23− x − 1

6

(b)x + 1

2− x − 1

3=

x + 13

− x − 12

(c)2x

5+

2 − 3x

4=

310

− 3 − 5x

2

(d) 34 (x − 1) − 1

2 (3x + 2) = 0

(e)4x + 1

6− 2x − 1

15=

3x − 55

− 6x + 110

(f)7(1 − x)

12− 3 + 2x

9=

5(2 + x)6

− 4 − 5x

1811. Solve each problem by forming, and then solving, a linear equation.

(a) My father is 40 years older than me and he is three times my age. How old am I?(b) The fuel tank in my new car was 40% full. I added 28 litres and then found that it

was 75% full. How much fuel does the tank hold?(c) A basketballer has scored 312 points in 15 games. How many points must he average

per game in his next 3 games to take his overall average to 20 points per game?(d) A cyclist rides for 5 hours at a certain speed and then for 4 hours at a speed 6 km/h

greater than his original speed. If he rides 294 km altogether, what was his first speed?

12. Rearrange each formula so that the pronumeral written in square brackets is the subject.

(a)a

2− b

3= a [a]

(b)1f

+2g

=5h

[g]

(c) x =y

y + 2[y]

(d) a =b + 5b − 4

[b]

(e) c =7 + 2d

5 − 3d[d]

(f) u =v + w − 1v − w + 1

[v]

1 G Solving Quadratic EquationsThere are three approaches to solving a quadratic equation:• factoring• completing the square• using the quadratic formula.

This section reviews factoring and the quadratic formula. Completing the squarewill be reviewed in Section 1I.

Solving a Quadratic by Factoring: This method is the simplest, but does not always work.

11SOLVING A QUADRATIC BY FACTORING:1. Get all the terms on the left, then factor the left-hand side.2. Use the principle that if AB = 0, then A = 0 or B = 0.

WORKED EXERCISE:

Solve the quadratic equation 5x2 + 34x − 7 = 0 by factoring.

SOLUTION:

5x2 + 34x − 7 = 05x2 + 35x − x − 7 = 0 (35 and −1 have sum 34 and product −7 × 5 = −35.)

5x(x + 7) − (x + 7) = 0(5x − 1)(x + 7) = 0 (The LHS is now factored.)

5x − 1 = 0 or x + 7 = 0 (One of the factors must be zero.)x = 1

5 or x = −7 (There are two solutions.)




Solving a Quadratic by the Formula: This method works whether the solutions arerational numbers or involve surds. It will be proven in Chapter Ten.

12

THE QUADRATIC FORMULA:• The solutions of ax2 + bx + c = 0 are:

x =−b +

√b2 − 4ac

2aor x =

−b −√b2 − 4ac

2a.

• Always calculate b2 − 4ac first.(Later, this quantity will be called the discriminant and given the symbol Δ.)

WORKED EXERCISE:Solve each quadratic equation using the quadratic formula.(a) 5x2 + 2x − 7 = 0 (b) 3x2 + 4x − 1 = 0

SOLUTION:

(a) For 5x2 + 2x − 7 = 0,

a = 5, b = 2 and c = −7.Hence b2 − 4ac = 22 + 140

= 144= 122 ,

so x =−2 + 12

10or

−2 − 1210

= 1 or −125 .

(b) For 3x2 + 4x − 1 = 0,

a = 3, b = 4 and c = −1.Hence b2 − 4ac = 42 + 12

= 28= 4 × 7,

so x =−4 + 2

√7

6or

−4 − 2√

76

=−2 +

√7

3or

−2 −√7

3.

Exercise 1G1. Solve:

(a) x2 = 9(b) y2 = 25(c) a2 − 4 = 0

(d) c2 − 36 = 0(e) 1 − t2 = 0(f) x2 = 9

4

(g) 4x2 − 1 = 0(h) 9a2 − 64 = 0(i) 25y2 = 16

2. Solve, by factoring:(a) x2 − 5x = 0(b) y2 + y = 0(c) c2 + 2c = 0(d) k2 − 7k = 0

(e) t2 = t

(f) 3a = a2

(g) 2b2 − b = 0(h) 3u2 + u = 0

(i) 4x2 + 3x = 0(j) 2a2 = 5a

(k) 3y2 = 2y

(l) 12h + 5h2 = 0

3. Solve, by factoring:(a) x2 + 4x + 3 = 0(b) x2 − 3x + 2 = 0(c) x2 + 6x + 8 = 0(d) a2 − 7a + 10 = 0(e) t2 − 4t − 12 = 0(f) c2 − 10c + 25 = 0

(g) n2 − 9n + 8 = 0(h) p2 + 2p − 15 = 0(i) a2 − 10a − 24 = 0(j) y2 + 4y = 5(k) p2 = p + 6(l) a2 = a + 132

(m) c2 + 18 = 9c(n) 8t + 20 = t2

(o) u2 + u = 56(p) k2 = 24 + 2k

(q) 50 + 27h + h2 = 0(r) α2 + 20α = 44



� �CHAPTER 1: Methods in Algebra 1G Solving Quadratic Equations 19


4. Solve, by factoring:(a) 2x2 + 3x + 1 = 0(b) 3a2 − 7a + 2 = 0(c) 4y2 − 5y + 1 = 0(d) 2x2 + 11x + 5 = 0(e) 2x2 + x − 3 = 0(f) 3n2 − 2n − 5 = 0

(g) 3b2 − 4b − 4 = 0(h) 2a2 + 7a − 15 = 0(i) 2y2 − y − 15 = 0(j) 3y2 + 10y = 8(k) 5x2 − 26x + 5 = 0(l) 4t2 + 9 = 15t

(m) 13t + 6 = 5t2

(n) 10u2 + 3u − 4 = 0(o) 25x2 + 1 = 10x(p) 6x2 + 13x + 6 = 0(q) 12b2 + 3 + 20b = 0(r) 6k2 + 13k = 8

5. Solve each equation, using the quadratic formula. Give exact answers, followed by ap-proximations to four significant figures where appropriate.(a) x2 − x − 1 = 0(b) y2 + y = 3(c) a2 + 12 = 7a

(d) u2 + 2u − 2 = 0

(e) c2 − 6c + 2 = 0(f) 4x2 + 4x + 1 = 0(g) 2a2 + 1 = 4a(h) 5x2 + 13x − 6 = 0

(i) 2b2 + 3b = 1(j) 3c2 = 4c + 3(k) 4t2 = 2t + 1(l) x2 + x + 1 = 0

6. Solve, by factoring:

(a) x =x + 2

x

(b) a +10a

= 7

(c) y +2y

=92

(d) (5b − 3)(3b + 1) = 1

(e)5k + 7k − 1

= 3k + 2

(f)u + 32u − 7

=2u − 1u − 3

7. Find the exact solutions of:

(a) x =1x

+ 2

(b)4x − 1

x= x

(c) a =a + 4a − 1

(d)5m

2= 2 +

1m

(e)y + 1y + 2

=3 − y

y − 4

(f) 2(k − 1) =4 − 5k

k + 1

8. (a) If y = px − ap2, find p, given that a = 2, x = 3 and y = 1.(b) Given that (x − a)(x − b) = c, find x when a = −2, b = 4 and c = 7.

(c) Suppose that S =n

2(2a + (n − 1)d

). Find the positive value of n that gives S = 80

when a = 4 and d = 6.

9. Solve each problem by forming and then solving a suitable quadratic equation.(a) Find a positive integer that, when increased by 30, is 12

less than its square.(b) Two positive numbers differ by 3 and the sum of their

squares is 117. Find the numbers.( − 7)x cm

( + 2)x cmx cm

(c) Find the value of x in the diagram opposite.

C H A L L E N G E

10. Solve each equation.

(a)2

a + 3+

a + 32

=103

(b)k + 10k − 5

− 10k

=116

(c)3t

t2 − 6=

√3

(d)3m + 13m − 1

− 3m − 13m + 1

= 2




11. Solve each problem by constructing and then solving a quadratic equation.(a) A rectangular area can be completely tiled with 200 square tiles. If the side length of

each tile was increased by 1 cm, it would take only 128 tiles to tile the area. Find theside length of each tile.

(b) The numerator of a certain fraction is 3 less than its denominator. If 6 is added tothe numerator and 5 to the denominator, the value of the fraction is doubled. Findthe fraction.

(c) A photograph is 18 cm by 12 cm. It is to be surrounded by a frame of uniform widthwhose area is equal to that of the photograph. Find the width of the frame.

(d) A certain tank can be filled by two pipes in 80 minutes. The larger pipe by itself canfill the tank in 2 hours less than the smaller pipe by itself. How long does each pipetake to fill the tank on its own?

(e) Two trains each make a journey of 330km. One of the trains travels 5 km/h fasterthan the other and takes 30minutes less time. Find the speeds of the trains.

1 H Solving Simultaneous EquationsThis section reviews the two algebraic approaches to solving simultaneous equa-tions — substitution and elimination. Both linear and non-linear simultaneousequations are reviewed. The methods are extended to systems of three equationsin three unknowns, which will be new for most readers.

Solution by Substitution: This method can be applied whenever one of the equationscan be solved for one of the variables.

13SOLVING SIMULTANEOUS EQUATIONS BY SUBSTITUTION:• Solve one of the equations for one of the variables.• Then substitute it into the other equation.

WORKED EXERCISE:

Solve each pair of simultaneous equations by substitution.(a) 3x − 2y = 29 (1)

4x + y = 24 (2)(b) y = x2 (1)

y = x + 2 (2)

SOLUTION:

(a) Solving (2) for y, y = 24 − 4x. (2A)Substituting (2A) into (1), 3x − 2(24 − 4x) = 29

x = 7.

Substituting x = 7 into (1), 21 − 2y = 29y = −4.

Hence x = 7 and y = −4. (This should be checked in the original equations.)

(b) Substituting (1) into (2), x2 = x + 2x2 − x − 2 = 0

(x − 2)(x + 1) = 0x = 2 or −1.

From (1), when x = 2, y = 4, and when x = −1, y = 1.Hence x = 2 and y = 4, or x = −1 and y = 1. (Check in the original equations.)



� �CHAPTER 1: Methods in Algebra 1H Solving Simultaneous Equations 21

Solution by Elimination: This method, when it can be used, is more elegant, and caninvolve less algebraic manipulation.

14SOLVING SIMULTANEOUS EQUATIONS BY ELIMINATION:

Take suitable multiples of the equations so that one variable is eliminated whenthe equations are added or subtracted.

WORKED EXERCISE:

Solve each pair of simultaneous equations by elimination.(a) 3x − 2y = 29 (1)

4x + 5y = 8 (2)(b) x2 + y2 = 53 (1)

x2 − y2 = 45 (2)SOLUTION:

(a) Taking 4 × (1) and 3 × (2),12x − 8y = 116 (1A)

12x + 15y = 24. (2A)Subtracting (1A) from (2A),

23y = −92

÷ 23 y = −4.

Substituting into (1),3x + 8 = 29

x = 7.

Hence x = 7 and y = −4.

(b) Adding (1) and (2),2x2 = 98x2 = 49.

Subtracting (2) from (1),2y2 = 8y2 = 4.

Hence x = 7 and y = 2,

or x = 7 and y = −2,

or x = −7 and y = 2,

or x = −7 and y = −2.

Systems of Three Equations in Three Variables: The key step here is to reduce thesystem to two equations in two variables.

15SOLVING THREE SIMULTANEOUS EQUATIONS:

Using either substitution or elimination, produce two simultaneous equationsin two of the variables.

WORKED EXERCISE:

Solve simultaneously: 3x − 2y − z = −8 (1)5x + y + 3z = 23 (2)4x + y − 5z = −18 (3)

SOLUTION:

Subtracting (3) from (2), x + 8z = 41. (4)Doubling (3), 8x + 2y − 10z = −36 (3A)and adding (1) and (3A), 11x − 11z = −44

x − z = −4. (5)Equations (4) and (5) are now two equations in two unknowns.Subtracting (5) from (4), 9z = 45

z = 5.

Substituting z = 5 into (5), x = 1and substituting into (2), y = 3.

Hence x = 1, y = 3 and z = 5. (This should be checked in the original equations.)




Exercise 1H1. Solve, by substituting the first equation into the second:

(a) y = x and 2x + y = 9(b) y = 2x and 3x − y = 2(c) y = x − 1 and 2x + y = 5

(d) a = 2b + 1 and a − 3b = 3(e) p = 2 − q and p − q = 4(f) v = 1 − 3u and 2u + v = 0

2. Solve, by either adding or subtracting the two equations:(a) x + y = 5 and x − y = 1(b) 3x − 2y = 7 and x + 2y = −3(c) 2x + y = 9 and x + y = 5

(d) a + 3b = 8 and a + 2b = 5(e) 4c − d = 6 and 2c − d = 2(f) p − 2q = 4 and 3p − 2q = 0

3. Solve, by substitution:(a) y = 2x and 3x + 2y = 14(b) y = −3x and 2x + 5y = 13(c) y = 4 − x and x + 3y = 8(d) x = 5y + 4 and 3x − y = 26

(e) 2x + y = 10 and 7x + 8y = 53(f) 2x − y = 9 and 3x − 7y = 19(g) 4x − 5y = 2 and x + 10y = 41(h) 2x + 3y = 47 and 4x − y = 45

4. Solve, by elimination:(a) 2x + y = 1 and x − y = −4(b) 2x + 3y = 16 and 2x + 7y = 24(c) 3x + 2y = −6 and x − 2y = −10(d) 5x − 3y = 28 and 2x − 3y = 22(e) 3x + 2y = 7 and 5x + y = 7(f) 3x + 2y = 0 and 2x − y = 56

(g) 15x + 2y = 27 and 3x + 7y = 45(h) 7x − 3y = 41 and 3x − y = 17(i) 2x + 3y = 28 and 3x + 2y = 27(j) 3x − 2y = 11 and 4x + 3y = 43(k) 4x + 6y = 11 and 17x − 5y = 1(l) 8x = 5y and 13x = 8y + 1


5. Solve, by substitution:(a) y = 2 − x and y = x2

(b) y = 2x − 3 and y = x2 − 4x + 5(c) y = 3x2 and y = 4x − x2

(d) x − y = 5 and y = x2 − 11

(e) x − y = 2 and xy = 15(f) 3x + y = 9 and xy = 6(g) x2 − y2 = 16 and x2 + y2 = 34(h) x2 + y2 = 117 and 2x2 − 3y2 = 54

6. Solve each problem by constructing and then solving a pair of simultaneous equations.(a) Find two numbers that differ by 16 and have a sum of 90.(b) I paid 75 cents for a pen and a pencil. If the pen cost four times as much as the pencil,

find the cost of each item.(c) If 7 apples and 2 oranges cost $4, while 5 apples and 4 oranges cost $4·40, find the

cost of each apple and orange.(d) Twice as many adults as children attended a certain concert. If adult tickets cost $8

each, child tickets cost $3 each and the total takings were $418, find the numbers ofadults and children who attended.

(e) A man is 3 times as old as his son. In 12 years time he will be twice as old as his son.How old is each of them now?

(f) At a meeting of the members of a certain club, a proposal was voted on. If 357members voted and the proposal was carried by a majority of 21, how many voted forand how many voted against?



� �CHAPTER 1: Methods in Algebra 1I Completing the Square 23

7. Solve simultaneously:

(a)y

4− x

3= 1 and

x

2+

y

5= 10 (b) 4x +

y − 23

= 12 and 3y − x − 35

= 6

C H A L L E N G E

8. Solve simultaneously:(a) x = 2y

y = 3z

x + y + z = 10(b) x + 2y − z = −3

3x − 4y + z = 132x + 5y = −1

(c) 2a − b + c = 10a − b + 2c = 9

3a − 4c = 1(d) p + q + r = 6

2p − q + r = 1p + q − 2r = −9

(e) 2x − y − z = 17x + 3y + 4z = −20

5x − 2y + 3z = 19(f) 3u + v − 4w = −4

u − 2v + 7w = −74u + 3v − w = 9

9. Solve simultaneously:(a) x + y = 15 and x2 + y2 = 125(b) x − y = 3 and x2 + y2 = 185(c) 2x + y = 5 and 4x2 + y2 = 17

(d) x + y = 9 and x2 + xy + y2 = 61(e) x + 2y = 5 and 2xy − x2 = 3(f) 3x + 2y = 16 and xy = 10

10. Set up a pair of simultaneous equations to solve each problem.(a) The value of a certain fraction becomes 1

5 if one is added to its numerator. If one istaken from its denominator, its value becomes 1

7 . Find the fraction.(b) Kathy paid $320 in cash for a CD player. If she paid in $20 notes and $10 notes and

there were 23 notes altogether, how many of each type were there?(c) A certain integer is between 10 and 100. Its value is 8 times the sum of its digits, and

if it is reduced by 45, its digits are reversed. Find the integer.(d) Two people are 16 km apart on a straight road. They start walking at the same time.

If they walk towards each other, they will meet in 2 hours, but if they walk in thesame direction (so that the distance between them is decreasing), they will meet in8 hours. Find their walking speeds.

1 I Completing the SquareCompleting the square can be done in all situations, whereas factoring is notalways possible. In particular, the quadratic formula that was reviewed in Sec-tion 1G will proven in Chapter Ten by completing the square.

The review in this section is mostly restricted to monic quadratics, in which thecoefficient of x2 is 1. Chapter Ten will deal with non-monic quadratics.

Perfect Squares: The expansion of the quadratic (x + α)2 is

(x + α)2 = x2 + 2αx + α2 .

Notice that the coefficient of x is twice α, and the constant is the square of α.

Reversing the process, the constant term in a perfect square can be found bytaking half the coefficient of x and squaring the result.

16 COMPLETING THE SQUARE IN AN EXPRESSION x2 + bx + · · · :Halve the coefficient b of x and square the result.




WORKED EXERCISE:Complete the square in each expression.(a) x2 + 16x + · · · (b) x2 − 3x + · · ·SOLUTION:

(a) The coefficient of x is 16, half of 16 is 8, and 82 = 64,so x2 + 16x + 64 = (x + 8)2 .

(b) The coefficient of x is −3, half of −3 is −112 , and (−11

2 )2 = 214 ,

so x2 − 3x + 214 = (x − 11

2 )2 .

Solving Quadratic Equations by Completing the Square: This process always works.

17SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE:

Complete the square in the quadratic by adding the same to both sides.

WORKED EXERCISE:Solve each quadratic equation by completing the square.(a) t2 + 8t = 20 (b) x2 − x − 1 = 0

SOLUTION:

(a) t2 + 8t = 20

+ 16 t2 + 8t + 16 = 36

(t + 4)2 = 36

t + 4 = 6 or t + 4 = −6

t = 2 or −10

(b) x2 − x − 1 = 0

+ 1 x2 − x = 1

+ 14 x2 − x + 1

4 = 114

(x − 12 )2 = 5

4

x − 12 = 1

2

√5 or x − 1

2 = − 12

√5

x = 12 + 1

2

√5 or 1

2 − 12

√5

The Word ‘Algebra’: Al-Khwarizmi was a famous and influential mathematician whoworked in Baghdad during the early ninth century when the Arabs excelled inscience and mathematics. The Arabic word ‘algebra’ comes from the title of hismost important work and means ‘the restoration of broken parts’ — a referenceto the balancing of terms on both sides of an equation. Al-Khwarizmi’s own namecame into modern European languages as ‘algorithm’.

Exercise 1I1. What constant must be added to each expression in order to create a perfect square?

(a) x2 + 2x

(b) y2 − 6y

(c) a2 + 10a

(d) m2 − 18m

(e) c2 + 3c

(f) x2 − x

(g) b2 + 5b

(h) t2 − 9t

2. Factor:(a) x2 + 4x + 4(b) y2 + 2y + 1

(c) p2 + 14p + 49(d) m2 − 12m + 36

(e) t2 − 16t + 64(f) x2 + 20x + 100

(g) u2 − 40u + 400(h) a2 − 24a + 144

3. Copy and complete:(a) x2 + 6x + · · · = (x + · · · )2

(b) y2 + 8y + · · · = (y + · · · )2

(c) a2 − 20a + · · · = (a − · · · )2

(d) b2 − 100b + · · · = (b − · · · )2

(e) u2 + u + · · · = (u + · · · )2

(f) t2 − 7t + · · · = (t − · · · )2

(g) m2 + 50m + · · · = (m + · · · )2

(h) c2 − 13c + · · · = (c − · · · )2



� �CHAPTER 1: Methods in Algebra 1J Chapter Review Exercise 25


4. Solve each quadratic equation by completing the square.(a) x2 − 2x = 3(b) x2 − 6x = 0(c) a2 + 6a + 8 = 0

(d) y2 + 3y = 10(e) b2 − 5b − 14 = 0(f) x2 + 4x + 1 = 0

(g) x2 − 10x + 20 = 0(h) y2 − y + 2 = 0(i) a2 + 7a + 7 = 0

C H A L L E N G E

5. Solve, by dividing both sides by the coefficient of x2 and then completing the square:(a) 3x2 − 15x + 18 = 0(b) 2x2 − 4x − 1 = 0(c) 3x2 + 6x + 5 = 0

(d) 2x2 + 8x + 3 = 0(e) 4x2 + 4x − 3 = 0(f) 4x2 − 2x − 1 = 0

(g) 3x2 − 8x − 3 = 0(h) 2x2 + x − 15 = 0(i) 2x2 − 10x + 7 = 0

6. (a) If x2 + y2 + 4x − 2y + 1 = 0, show that (x + 2)2 + (y − 1)2 = 4.(b) Show that the equation x2 + y2 − 6x − 8y = 0 can be written in the form

(x − a)2 + (y − b)2 = c,

where a, b and c are constants. Hence find a, b and c.(c) If x2 + 1 = 10x + 12y, show that (x − 5)2 = 12(y + 2).(d) Find values for A, B and C if y2 − 6x + 16y + 94 = (y + C)2 − B(x + A).

1J Chapter Review Exercise

1. Simplify:(a) −8y + 2y (b) −8y − 2y (c) −8y × 2y (d) −8y ÷ 2y

2. Simplify:(a) −2a2 − a2 (b) −2a2 − (−a2) (c) −2a2 × (−a2) (d) −2a2 ÷ (−a2)

3. Simplify:(a) 3t − 1 − t

(b) −6p + 3q + 10p(c) 7x − 4y − 6x + 2y

(d) 2a2 + 8a − 13 + 3a2 − 11a − 5

4. Simplify:(a) −6k6 × 3k3 (b) −6k6 ÷ 3k3 (c) (−6k6)2 (d) (3k3)3

5. Expand and simplify:(a) 4(x + 3) + 5(2x − 3)(b) 8(a − 2b) − 6(2a − 3b)(c) −(a − b) − (a + b)(d) −4x2(x + 3) − 2x2(x − 1)

(e) (n + 7)(2n − 3)(f) (r + 3)2

(g) (y − 5)(y + 5)(h) (3x − 5)(2x − 3)

(i) (t − 8)2

(j) (2c + 7)(2c − 7)(k) (4p + 1)2

(l) (3u − 2)2

6. Factor:(a) 18a + 36(b) 20b − 36(c) 9c2 + 36c

(d) d2 − 36(e) e2 + 13e + 36(f) f 2 − 12f + 36

(g) 36 − 25g2

(h) h2 − 9h − 36(i) i2 + 5i − 36(j) 2j2 + 11j + 12(k) 3k2 − 7k − 6(l) 5�2 − 14� + 8

(m) 4m2 + 4m − 15(n) n3 + 8(o) p3 − 27(p) p3 + 9p2 + 4p + 36(q) qt − rt − 5q + 5r

(r) u2w + vw − u2x − vx




7. Simplify:

(a)x

2+

x

4

(b)x

2− x

4

(c)x

2× x

4

(d)x

2÷ x

4

(e)3a

2b+

2a

3b

(f)3a

2b− 2a

3b

(g)3a

2b× 2a

3b

(h)3a

2b÷ 2a

3b

(i)x

y+

y

x

(j)x

y− y

x

(k)x

y× y

x

(l)x

y÷ y

x

8. Simplify:

(a)x + 4

5+

x − 53

(b)5

x + 4+

3x − 5

(c)x + 1

2− x − 4

5

(d)2

x + 1− 5

x − 4

(e)x

2− x + 3

4

(f)2x− 4

x + 39. Factor where possible, then simplify:

(a)6a + 3b

10a + 5b

(b)2x − 2y

x2 − y2

(c)x2 + 2x − 3x2 − 5x + 4

(d)2x2 + 3x + 1

2x3 + x2 + 2x + 1

(e)a + b

a2 + 2ab + b2

(f)3x2 − 19x − 14

9x2 − 4

(g)x3 − 8x2 − 4

(h)a2 − 2a − 3

a3 + 110. Solve each linear equation.

(a) 3x + 5 = 17(b) 3(x + 5) = 17

(c)x + 5

3= 17

(d)x

3+ 5 = 17

(e) 7a − 4 = 2a + 11(f) 7(a − 4) = 2(a + 11)

(g)a − 4

7=

a + 112

(h)a

7− 4 =

a

2+ 11

11. Solve each quadratic equation by factoring the left-hand side.(a) a2 − 49 = 0(b) b2 + 7b = 0(c) c2 + 7c + 6 = 0(d) d2 + 6d − 7 = 0

(e) e2 − 5e + 6 = 0(f) 2f 2 − f − 6 = 0(g) 2g2 − 13g + 6 = 0(h) 3h2 + 2h − 8 = 0

12. Solve, using the quadratic formula. Write the solutions in simplest exact form.(a) x2 − 4x + 1 = 0(b) y2 + 3y − 3 = 0(c) t2 + 6t + 4 = 0

(d) 3x2 − 2x − 2 = 0(e) 2a2 + 5a − 5 = 0(f) 4k2 − 6k − 1 = 0

13. Solve each quadratic by completing the square on the left-hand side.(a) x2 + 4x = 6(b) y2 − 6y + 3 = 0

(c) x2 − 2x = 12(d) y2 + 10y + 7 = 0

Online Multiple Choice Quiz



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Methods in Algebra -Cambridge University