algebraic manipulation sample - jschs...

35

2.1 Adding and subtracting like termsA pronumeral (letter) represents a number. It may stand for an unknown value or series of values that change. For example, in the equation x + 5 = 8, x is a pronumeral that represents a value. Its value can be determined because we know 3 + 5 = 8, so x = 3.

When a term has a pronumeral and a number, the number is written before the pronumeral and is called the coeffi cient. For example, the term 3xy has a coeffi cient of 3 and its pronumerals are written after the coeffi cient in alphabetical order.

Like termsTerms that have exactly the same pronumerals such as 2a and 5a are called like terms. Only like terms can be added and subtracted. It involves adding and subtracting the coeffi cients. Adding and subtracting like terms simplifi es the algebraic expression. It is often called collecting the like terms.

C H A P T E R

2Algebraic manipulation

Syllabus topic — AM1 Algebraic manipulation Add and subtract like terms

Multiply and divide algebraic terms

Expand and factorise algebraic expressions

Evaluate the subject of the formula through substitution

Solve linear equations involving up to 3 steps

Solve equations following substitution

9781107627291c02_p35-72.indd 35 8/29/12 9:25 PM

Uncorrected sample pages • Cambridge University Press © Powers 2012 • 978-1-107-62729-1 • Ph 03 8671 1400

SAMPLE

36 Preliminary Mathematics General

Adding and subtracting like terms

1 Find the like terms or the terms that have exactly the same pronumerals. 2 Only like terms can be added or subtracted; unlike terms cannot.3 Add or subtract the coeffi cients or the numbers before the pronumeral of the like terms.

Example 1 Adding and subtracting like terms

Simplify 2ab + 3 + 5ab - 7.

Solution1 Rewrite the expression by grouping

the like terms.2 Add and subtract the coefficients.

2ab + 3 + 5ab - 7 = (2ab + 5ab) + 3 - 7 = 7ab - 4

Example 2 Adding and subtracting like terms

Simplify 4y + 6y2 - 3y - 5y2.

Solution1 Rewrite the expression by

grouping the like terms.2 Add and subtract the coefficients.

4y + 6y2 - 3y - 5y2 = 6y2 - 5y2 + 4y - 3y = y2 + y

Adding and subtracting algebraic fractions To add and subtract algebraic fractions rewrite each fraction as an equivalent fraction with a common denominator, then add or subtract the numerators. A common denominator is always found my multiplying the denominators of both fractions together.

Example 3 Adding and subtracting algebraic fractions

Simplify x x

6 4+ .

Solution

1 Find a common denominator for 6 and 4. Both 6 and 4 divide into 12. Alternatively, multiply 6 by 4 and use 24.

2 Multiply the first fraction by 2 (6 × 2 = 12) and the second fraction by 3 (4 × 3 = 12).

3 Write the equivalent fractions.4 Add the numerators of the equivalent fractions.

+ =+ = × + ×

= += +

=

x x x x×x x×

x x

x

6 4

2x x2x x

6 2×6 2×3

4 3×4 3×2

12

3x x3x x

125

12

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SAMPLE

37Chapter 2 — Algebraic manipulation

Exercise 2A 1 Choose the like terms out of each of the following.

a 4r, 6p, r, 7 b 5x, 3xy, 2x c -a, 5a, -8b, 7d xy, 4xy, xy2, 3yx e 3, 2m, mn, 9m f c, cd, cde, dc, ce

2 Simplify by collecting like terms.

a 4y + 3y b 3 17p p3 1p p3 17p p73 1+3 13 1p p3 1+3 1p p3 1 c 7 6h h7 6h h7 67 6h h7 6−7 6h h7 6

d 3x xx x−x x e d dd d+d dd d4d d f 6 12y y6 1y y6 12y y26 1y y6 1−6 1y y6 1

g − +d d− +d d− + ( )d d( )d d−d d−( )−d d−d d( )d d4d d( )d d h 33t tt t+ −t t( )11( )11t t( )t t11t t11( )11t t11t t+ −t t( )t t+ −t t i 11 f ff f− −f f( )f f( )f f5f f5( )5f f5f f− −f f( )f f− −f f

j 4 6hg4 6hg4 6gh4 6+4 6 k 5 2ab5 2ab5 2ba5 2+5 2 l xyz xyz+ 3


a 5 4 2c c5 4c c5 4 2c c2+ +5 4+ +5 4c c+ +c c5 4c c5 4+ +5 4c c5 4 b 4 4 7f f4 4f f4 4+ −f f+ −f f4 4f f4 4+ −4 4f f4 4 c 8 5 12+ +8 5+ +8 5r r12r r12+ +r r+ +

d 6 4 3x y6 4x y6 4 x+ −6 4+ −6 4x y+ −x y6 4x y6 4+ −6 4x y6 4 e 3 7 2b a3 7b a3 7 a+ −b a+ −b a3 7b a3 7+ −3 7b a3 7 f h d hh d+ −h d2 6h d2 6h d+ −2 6+ −h d+ −h d2 6h d+ −h d

g 4 2de4 2de4 2ed4 2ed4 2de4 2+ −4 24 2ed4 2+ −4 2ed4 2 h 7 2a b7 2a b7 2a b2a b27 2+ +7 27 2a b7 2+ +7 2a b7 2a b−a b i xy yx xy+ +2 3yx2 3yx+ +2 3+ +yx+ +yx2 3yx+ +yx

j 6 2ba6 2ba6 2b a− +6 2− +6 2b a− +b a( )b a( )b ab( )bb a−b a( )b a−b a k 7 2a b7 2a b7 2a b2a b27 2a b7 2+ −7 2a b7 2+ −7 2+ −7 2a b+ −a b( )7 2( )7 2a b( )a b7 2a b7 2( )7 2a b7 27 2a b7 2+ −7 2a b7 2( )7 2a b7 2+ −7 2a b7 2 l 5 8g h5 8g h5 8g h5 8g h5 85 8+ +5 85 8g h5 8+ +5 8g h5 85 8g h5 8− +5 8g h5 8( )5 8( )5 8g h( )g h5 8g h5 8( )5 8g h5 85 8− +5 8( )5 8− +5 85 8g h5 8− +5 8g h5 8( )5 8g h5 8− +5 8g h5 8


a 8 3 42 28 32 28 3x x8 3x x8 32 2x x2 28 32 28 3x x8 32 28 3 x2 2x2 2− −x x− −x x8 3x x8 3− −8 3x x8 3 + b 4 32 24 32 24 34 3a a4 3a a2 2a a2 24 3+ −4 34 32 24 3+ −4 32 24 34 3a a4 3+ −4 3a a4 3 +a a+a a c 7 8 6 72 26 72 26 7t t7 8t t7 8 t t6 7t t6 72 2t t2 26 72 26 7t t6 72 26 7+ −2 2+ −2 2t t+ −t t7 8t t7 8+ −7 8t t7 8 6 7t t6 7−6 7t t6 7

d 3 8 42 23 82 23 8 42 24m m3 8m m3 83 82 23 8m m3 82 23 8 m m2 2m m2 2+ −2 2+ −2 23 82 23 8+ −3 82 23 8m m+ −m m3 8m m3 8+ −3 8m m3 82 2m m2 2+ −2 2m m2 23 82 23 8m m3 82 23 8+ −3 82 23 8m m3 82 23 8 m m−m m e e e e e2 2e e2 2e e e e2 2e e2 222 2+ +e e+ +e e2 2+ +2 2e e2 2e e+ +e e2 2e ee e+ +e e2 2+ +2 2e e2 2e e+ +e e2 2e ee e2e e+ +e e2e e2 222 2+ +2 222 2e e2 2e e2e e2 2e e+ +e e2 2e e2e e2 2e e e e−e e f d d d d+ −d d+ −d d d d+d d2 2d d2 2d d+ −2 2+ − d d+d d2 2d d+d d52 252 2

g 2 522 522 52 5w w2 5 w2 5+ +2 52 522 5+ +2 522 52 5w w2 5+ +2 5w w2 5 + h 6 426 426 46 4− +6 46 426 4− +6 426 46 4−6 46 4v v6 46 4− +6 4v v6 4− +6 46 426 4− +6 426 4v v6 426 4− +6 426 4 i 8 7 7 3 2r r8 7r r8 7 7 3r r7 3rr r− −r r8 7r r8 7− −8 7r r8 7 7 3−7 3

5 Add or subtract the algebraic fractions.

a a a

3 3+ b

3

5

2

5

x x2x x2− c 2

4

3

4

m m3m m3+

d 3

7 7

x x− e d d

11

d d2d d

11+ f

6

15

2

15

y y2y y2−

g 4

3

9

3

s s9s s9+ h 9

8 8

f f4f f4− i y y

2 4+y y+y y

j 7

6 3

e e− k g g

2

4g g4g g

6+g g+g g

l r r

2

4r r4r r

10−

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SAMPLE


Development

6 Which of the following is equivalent to m m m n n+ + +m m m n+ + +m m m n + ?

a 2 2m m2 2m m2 2n2 2+ +2 22 2m m2 2+ +2 2m m2 2 b m n3 2m n3 2m nm n+m n3 2+3 2m n3 2m n+m n3 2m n

c 5 4 3m m5 4m m5 4 n n− +5 4− +5 4m m− +m m5 4m m5 4− +5 4m m5 4 n n−n n d 7 4 2m m n7 4m m n7 4 2m m n2 n− +7 4− +7 4m m n− +m m n7 4m m n7 4− +7 4m m n7 4 −

e 3 23 2n n m3 2m3 2− +3 23 2n n m3 2− +3 2n n m3 23 2− −3 2 f − + − +2 5− +2 5− + 2 3− +2 3− +m m2 5m m2 5− +2 5− +m m− +2 5− + n n2 3n n2 3− +2 3− +n n− +2 3− +

7 Copy and add like terms where possible to complete the table.

+ x 3x x + y

3x

7y

x - y

2y

8 Matteo spent $y shopping. He spent $x for a pair of jeans, $3x for a shirt and $2x for a belt. Write an expression in simplified form for how many dollars he has left.

9 The perimeter of a plane shape is the distance around the boundary of the shape. The plane shape opposite is a rectangle with a length l and a breadth b. Write an expression for the perimeter of this rectangle by collecting like terms.

10 The isosceles triangle opposite has three sides whose lengths are 3x + y, 3x + y and x + 2y. Write an expression in simplified form for the perimeter of this triangle by collecting like terms.

11 Add or subtract the algebraic fractions.

a w w

4 3+ b

a a

4 5− c

x x

7

2x x2x x

3+

d z z

3 5− e

3

8 6

h h+ f 5

12 8

r r−

g u u

10

4u u4u u

15+ h

3

4 10

e e− i w w w

2 4 6+ ++ +

j 3

5 4 2

a a a− +− + k 7

10 6 3

x x x− +− + l dd d+ −+ −2 10

l

b

3x + y

3x + y

x + 2y

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SAMPLE


2.2 Multiplication and division of algebraic termsAlgebraic terms are multiplied and divided to form a single algebraic expression. Terms usually contain a coeffi cient before a pronumeral. The multiplication sign between the coeffi cient and the pronumeral is omitted. For example, the algebraic term 4x can be written in expanded form as 4 × x. After an algebraic expression is written in expanded form, the coeffi cients can be multiplied or divided and the pronumerals can be multiplied or divided. Index notation should be used to write expressions in shorter way such as a × a = a2. If the algebraic terms contain fractions it is easier to cancel any common factors in the numerator or denominator. This makes the calculations easier.

Multiplication and division of algebraic terms

1 Write in expanded form. 2 If the algebraic term is a fraction cancel any common factors.3 Multiply and divide the coeffi cients. 4 Multiply and divide the pronumerals.5 Write the coeffi cient before the pronumerals.6 Write the pronumerals in alphabetical order and express in index notation.

Example 4 Multiplying algebraic terms

Simplify the following.a 2cd de× (−3 ) b x x x2x x2x x× 3x x× 3x x × 4

Solution

1 Write in expanded form.2 Multiply the coefficients (2 × -3 = -6).3 Write the pronumerals in alphabetical

order.4 Express answer using index notation

(d × d = d 2).5 Write in expanded form.6 Multiply the coefficients (1 × 3 × 4 = -12).7 Write the coefficient before the

pronumerals.8 Write the pronumerals using index

notation.9 Express the answer using index notation

(x2 × x × x = x4).

a 2 2 36

6 2

cd2 2cd2 2de2 2de2 2 c d d ec d d e

cd e

× (2 2× (2 2−32 2−32 2)2 2)2 2= ×2 2= ×2 2 × ×c d× ×c d − ×3− ×3 d e×d e= − × × × ×d d e× × ×d d e

= −

b x x x x x x

x x x

x

x

2 2x x2 2x x x x2 2x x2 1x x2 1x x 1

4

4

x x12x x2 2122 2

12

12

12

× 3x x× 3x x2 2× 32 2x x2 2x x× 3x x2 2x x × 42 2× 42 2x x= ×x x2 2= ×2 2x x2 2x x= ×x x2 2x xx x12x x= ×x x12x x2 2122 2= ×2 2122 2x x2 2x x12x x2 2x x= ×x x2 2x x12x x2 2x x × ×x x× ×x x

= ×12= ×12 × ×x x× ×x x2 1× ×2 1x x2 1x x× ×x x2 1x x

= ×12= ×12

=

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SAMPLE


Example 5 Dividing algebraic terms

Simplify 1818 662a ba b2a b22a b2 a÷ .

Solution1 Write in fraction form.2 Write in expanded form.3 Divide the coefficients (18 ÷ 6 = 3).4 Cancel the pronumeral a in both the

numerator and denominator (common factor of a, a

aa a= ÷a a= ÷a a = 1).

5 Write the coefficient before the pronumerals.6 Write the pronumerals in alphabetical order.

18 618

618

63

3

22

1

1

a b2a b2 aa b2a b2

aa a b

aa aa a b

aaa

÷ =6÷ =6a÷ =a

= × × ×a a× × ×a a

×

= × ×a a× ×a a ×

= ×3= ×3 ×××=

bab3

Example 6 Dividing algebraic terms with fractions

Simplify

x

y

xy

12

8

20× .

Solution1 Write fractions in expanded form.2 Determine any common factors in

the numerator and the denominator.

3 Cancel out the common factors (4 is a common factor of 8 and 12, 2 is a common factor of 2 and 4).

4 Cancel the pronumeral y in both the numerator and denominator as it is a common factor.

5 Multiply the numerators together.6 Multiply the denominators

together.7 Express the answer using index

notation (x × x = x2).8 Write the coefficient before the

pronumerals. However, it is acceptable to the leave answer

as x2

30.

x

y

xy x

y

x y

x

yy

x yx y

12

8

20 12

8

20

4 34 3

4 24 24 2

1 11 1y1 1yy1 1y4 31 14 3

1 11 14 21 14 24 21 14 2

× =× =xy× =xy

×× × ×x y× ×x y

=× ×4 3× ×4 31 1× ×1 14 31 14 3× ×4 31 14 3

×× ×4 2× ×4 24 2× ×4 21 1× ×1 14 21 14 2× ×4 21 14 24 21 14 2× ×4 21 14 2 ×x y×x y 111

2

22

4 54 524 52

3 2 5

30

1

30

4 5×4 5

= ×× ×3 2× ×3 2

=

x x×x x×

xxor

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SAMPLE


Exercise 2B 1 Multiply these algebraic terms.

a 6g × 4 b 7 × 5m c 2 42 4×2 4d × 7d a a×a a×a a7a a7a a e 20 5x x5x x5×x x×x x f 5 65 6s s5 65 6× ×5 65 6s s5 6× ×5 6s s5 6

g 3 23 2e e3 23 2× 73 2× 73 2× 73 23 2e e3 2× 73 2e e3 23 2×3 2 h 4 )× (4 )× (−44 )−4w w4 )w w× (w w× (4 )× (w w× (−4w w−44 )−4w w−4 i −3 1f f3 1f f3 15f f53 1f f3 1×3 1f f3 1

2 Multiply these algebraic terms.a mn mn× 4 b 15 3yz y× c 5pq pr× (−2 )

d 3 2x x2x x2 × 2x x× 2x x e 2 3 2r r3 2r r3 23 2× 43 2r r× 4r r3 2r r3 2× 43 2r r3 2 f 24 2 2q q2 2q q2 22 2× 22 2q q× 2q q2 2q q2 2× 22 2q q2 2

g 3 3st st× 5 h −de2 2× −2 2× −2 2( )d e( )d e2 2( )2 2d e2 2d e( )d e2 2d e5( )52 252 2( )2 252 2× −( )× −2 2× −2 2( )2 2× −2 2 i z z z2z z2z z× 4z z× 4z z × 5

3 What is the product of 4uw2 and uw?

4 Simplify these algebraic terms.

a 12

6

y b 16

4

a c 6

18

mn

d 14

2

b

b e

12

12

3m

m f

2

26

xyz

x

g 8

64

2x

x h 5

15

2r

rs i 10

2

pq

q

5 Write in fraction form and simplify.a 20 4z ÷ b 22 22w ÷ c 27 2h ÷ −( )3( )3÷ −( )÷ −

d 33 11y y11y y11÷y y÷y y e 12 4ab a÷ f 25 52kj kj÷

g 4 283 24 23 24 283 28x x4 2x x4 28x x83 2x x3 24 23 24 2x x4 23 24 283 28x x83 284 23 24 2÷4 23 24 24 2x x4 2÷4 2x x4 24 23 24 2x x4 23 24 2÷4 23 24 2x x4 23 24 2 h 2 8 2s s2 8s s2 82 8÷2 82 8s s2 8÷2 8s s2 8 i 7 2 3b m7 2b m7 21b m17 2b m7 2÷7 2b m7 2

6 Express 16

12

2xy

x in simplest form.

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SAMPLE


Development

7 Multiply the algebraic terms in the first column by each algebraic term given to complete the table.

× p 2p2 pq

5p

6pq

2q2

8 Simplify the following

a 8

34

yy× b 6

2

5d

d× c 5

92

ww×

d 5

4

12x

x× e

2 3

6a

a× f 5

7

2

9

w w2w w2×

g 10 3

4

a

b

b× h m

n

mn

9

6

15× i d

e

c

d×

3

9 Divide the algebraic terms in the first column by each algebraic term given to complete the table.

÷ r 2r 4rt

8r

12rt

4r2

10 The plane shape opposite is a rectangle with a length of 52x and a

breadth of 3x2. Write an expression in simplest form for the area of this rectangle.

11 Express 8

16

3 2

2 3

x y3 2x y3 2

x y2 3x y2 3 in simplest form.

12 Simplify the following.

a 15

3

3

4

2 332 33h

k

k2 3k2 3

× b 21

15

5

7

2 352 35v

uv

u2 3u2 3

× c 18

12

6

9 2

2 362 36a

b

b2 3b2 3

9 2a9 2

a× ×× ×

d 9 6

3

3 29 63 29 62

m m9 6m m9 63 2m m3 29 63 29 6m m9 63 29 6 n3 2n3 2

m n2m n2

9 63 29 6×9 63 29 69 6m m9 6×9 6m m9 69 63 29 6m m9 63 29 6×9 63 29 6m m9 63 29 6 e

3 1

8

4

2 1

3( )3 1( )3 1

( )2 1( )2 1

3 1( )3 1m3 1( )3 1

m

m

( )m( )2 1( )2 1m2 1( )2 1

3 1( )3 1+3 1( )3 1 ×2 1( )2 1+2 1( )2 1

f 7

10

5 2

21

4

6

y

y( )2( )2y( )y

( )5 2( )5 2y( )y5 2y5 2( )5 2y5 2

( )−( )× 5 2( )5 2−5 2( )5 2

5x

3x2

2

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SAMPLE


2.3 Expanding algebraic expressionsGrouping symbols in algebraic expressions indicate the order of operations. The two most commonly used grouping symbols are parentheses ( ) and brackets [ ]. They are removed by using the distributive law or a × (b + c) = ab + ac. This is illustrated below.

Using order of operations2 3 1 2 4

82 3× +2 3 = ×1 2= ×1 2

=( )2 3( )2 3 1 2( )1 2× +( )× +2 3× +2 3( )2 3× +2 3

Using distributive law

2 3 1 2 3 2 16 28

2 3× +2 3 = ×1 2= ×1 2 3 2 1+ ×3 2 1= +6 2= +6 2=

( )2 3( )2 3 1 2( )1 2× +( )× +2 3× +2 3( )2 3× +2 3

To expand an algebraic expression using the distributive law, multiply the number or terms inside the grouping symbols by the number or term outside the grouping symbols. The resulting algebraic expression is simplifi ed by collecting the like terms.

Make sure you remember to multiply all the terms inside the grouping symbol by the number or term outside the grouping symbols.

Expanding algebraic expressions

1 Multiply the number or term outside the grouping symbol by the a fi rst term inside the grouping symbol.b second term inside the grouping symbol.

2 Simplify and collect like terms if required. a b a b a c

ab ac( )a b( )a b c( )c+ =( )+ =( )c( )c+ =c( )c a b a× +a b a ×

= +ab= +ab a b a b a c

ab ac( )a b( )a b c( )c− =( )− =( )c( )c− =c( )c a b a× −a b a ×

= −ab= −ab

Example 7 Expanding algebraic expressions

Expand 5 2( )5 2( )5 2 3( )3 .( )y( )( )−( )

Solution1 Multiply the first term inside the parenthesis (2y) by

the number outside the parenthesis (5).2 Multiply the second term inside the parenthesis (-3)

by the number outside the parenthesis (5).3 Write in simplest form.

5 2 3 5 2 5 310 15

( )5 2( )5 2 3 5( )3 5y y3 5y y3 5 2 5y y2 5( )y y( )3 5( )3 5y y3 5( )3 5y

3 5y y3 5− =3 5y y3 5( )y y( )− =( )y y( )3 5( )3 5y y3 5( )3 5− =3 5( )3 5y y3 5( )3 5 × −2 5× −2 5y y× −y y2 5y y2 5× −2 5y y2 5 ×= −10= −10y= −y

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SAMPLE


Example 8 Expanding algebraic expressions

Expand -(m - 5).

Solution1 Multiply the first term inside the parenthesis (m)

by the number outside the parenthesis (-1).2 Multiply the second term inside the parenthesis

(-5) by the number outside the parenthesis (-1).3 Write in simplest form.

− − × −= − × −= − +

( )− −( )− − ( )× −( )× −m m× −m m× −( )m m( )− −( )− −m m− −( )− − ( )m m( )× −( )× −m m× −( )× −m

m

5 1( )5 1( )m m5 1m m= −m m= −5 1= −m m= −( )m m( )5 1( )m m( ) ( )5( )1 1× −1 1× −m1 1m× −m× −1 1× −m× − 5

5

Example 9 Expanding and simplify algebraic expressions

Remove the grouping symbols for 2(3x + 4) + 3(x - 1) and simplify if possible.

Solution1 Multiply the first term inside the parenthesis

(3x) by the number outside the parenthesis (2).2 Multiply the second term inside the parenthesis

(+4) by the number outside the parenthesis (2).3 Repeat the first two steps for the second

parenthesis.4 Simplify by collecting the like terms.

2 3 4 32 3 2 4 3 3 16 8 3 3

( )2 3( )2 3 4 3( )4 3( )1( )1x x4 3x x4 3( )x x( )4 3( )4 3x x4 3( )4 3( )x x( )x x2 4x x2 4 3 3x x3 3

x x6 8x x6 8 3 3x x3 3

4 3+ +4 3( )+ +( )4 3( )4 3+ +4 3( )4 34 3x x4 3+ +4 3x x4 3( )x x( )+ +( )x x( )4 3( )4 3x x4 3( )4 3+ +4 3( )4 3x x4 3( )4 3( )−( )= ×2 3= ×2 3 + ×2 4+ ×2 4x x+ ×x x2 4x x2 4+ ×2 4x x2 4 + ×3 3+ ×3 3x x+ ×x x3 3x x3 3+ ×3 3x x3 3+ ×3 3+ ×3 3 −= +6 8= +6 86 8x x6 8= +6 8x x6 8 + −3 3+ −3 3x x+ −x x3 3x x3 3+ −3 3x x3 3= 9 599 599 5x9 59 5+9 5

Example 10 Expanding and simplifying algebraic expressions

Expand a(3a + 2) - a(a - 1) and simplify if possible.

Solution1 Multiply the first term inside the parenthesis

(3a) by the term outside the parenthesis (a).2 Multiply the second term inside the

parenthesis (+2) by the term outside the parenthesis (a).

3 Repeat the first two steps for the second parenthesis.

4 Simplify by collecting the like terms.

a a a aa a a aa a a a

( )a a( )a a ( )a a( )a a( )a a( )a a ( )a a( )a a

( )3 2( )a a( )a a3 2a a( )a a ( )1( )( )3 2( )a a( )a a3 2a a( )a a ( )1( )3 2a a3 2a a a a3 2a a

+ −( )+ −( )( )3 2( )+ −( )3 2( ) ( )−( )= ×a a= ×a a + −( )+ −( )( )3 2( )+ −( )3 2( ) × −a a× −a a( )× −( )a a( )a a× −a a( )a a= ×a a= ×a a3 2+ ×3 2a a3 2a a+ ×a a3 2a a− ×a a− ×a a a aaa aa

a a a a

a a

− ×a a− ×a a −= + − +a a− +a a

= +

1

3 2a a3 2a a= +3 2= +a a= +a a3 2a a= +a a

2 3a a2 3a a= +2 3= +a a= +a a2 3a a= +a a

2 2a a2 2a a − +2 2− +a a− +a a2 2a a− +a a3 22 23 2a a3 2a a2 2a a3 2a a= +3 2= +2 2= +3 2= +a a= +a a3 2a a= +a a2 2a a= +a a3 2a a= +a a22 322 3= +2 3= +2= +2 3= +a a= +a a2 3a a= +a a2a a= +a a2 3a a= +a a

9781107627291c02_p35-72.indd 44 8/29/12 9:26 PM


SAMPLE


Exercise 2C 1 Ryan was required to remove the grouping symbols. This was his solution.

3 3 2 9 2( )3 3( )3 3 2 9( )2 9x x2 9x x2 9( )x x( )2 9( )2 9x x2 9( )2 92 9x x2 9− =2 9x x2 9( )x x( )− =( )x x( )2 9( )2 9x x2 9( )2 9− =2 9( )2 9x x2 9( )2 9 − Where is the error in Ryan’s working?

2 Expand each of the following.

a 3 2( )3 2( )3 23 2( )3 2a3 2( )3 23 2( )3 2+3 2( )3 2 b 2 1( )2 1( )2 1( )d( )2 1( )2 1d2 1( )2 12 1( )2 1+2 1( )2 1 c 7 2( )7 2( )7 2( )b( )7 2( )7 2b7 2( )7 27 2( )7 2−7 2( )7 2

d 2 3( )2 3( )2 3 4( )4( )x( )( )+( ) e 2 5( )2 5( )2 5 7( )7( )x( )( )−( ) f 4 9( )4 9( )4 9 1( )1( )b( )( )+( )

g 4 5( )4 5( )4 5 2( )2( )+( )( )t( ) h 6 1( )6 1( )6 1 2( )2( )−( )( )w( ) i 5 3( )5 3( )5 3 9( )9( )+( )( )d( )

j 8 5( )8 5( )8 5( )e d( )2( )2e d2( )2( )e d( )−( )e d( ) k 5 4( )5 4( )5 4 9( )9( )a b( )9( )9a b9( )9( )a b( )+( )a b( ) l 7 2( )7 2( )7 2( )h g( )8( )8h g8( )8( )h g( )+( )h g( )


a − +4 3− +4 3− +( )4 3( )4 3− +4 3− +( )− +4 3− +4 3( )4 3x4 3( )4 3− +4 3− +( )− +4 3− +x− +4 3− +( )− +4 3− + b − +3 5− +3 5− +( )3 5( )3 5− +3 5− +( )− +3 5− +( )y( )3 5( )3 5y3 5( )3 5− +3 5− +( )− +3 5− +y− +3 5− +( )− +3 5− + c − +( )− +( )− +( )b( )− +( )− +b− +( )− +( )8( )

d − −7 2− −7 2− −( )7 2( )7 2− −7 2− −( )− −7 2− −( )k( )7 2( )7 2k7 2( )7 2− −7 2− −( )− −7 2− −k− −7 2− −( )− −7 2− − e − −6 1− −6 1− −( )6 1( )6 1− −6 1− −( )− −6 1− −6 1( )6 1w6 1( )6 1− −6 1− −( )− −6 1− −w− −6 1− −( )− −6 1− − f − −2 1− −2 1− −( )2 1( )2 1− −2 1− −( )− −2 1− − 3( )3( )x( )2 1( )2 1x2 1( )2 1− −2 1− −( )− −2 1− −x− −2 1− −( )− −2 1− −

g − +2 4− +2 4− +( )− +( )− +2 4( )2 4− +2 4− +( )− +2 4− + 2( )2( )q( ) h − −5 3− −5 3− −( )− −( )− −5 3( )5 3− −5 3− −( )− −5 3− − 4( )4( )r( ) i − −7 8− −7 8− −( )− −( )− −7 8( )7 8− −7 8− −( )− −7 8− − 2( )2( )s( )


a y y( )y y( )y y( )+( )( )1( ) b v v( )v v( )v v( )+( )( )4( ) c n n( )n n( )n n( )+( )( )10( )

d x x( )x x( )x x( )2 3( )x x( )x x2 3x x( )x x( )2 3( )−( )2 3( ) f e e( )e e( )e e( )3 5( )e e( )e e3 5e e( )e e( )3 5( )+( )3 5( ) g d d( )d d( )d d( )6 2( )d d( )d d6 2d d( )d d( )6 2( )−( )6 2( )

h z e( )z e( )z e f( )f( )7 3( )z e( )z e7 3z e( )z e f( )f7 3f( )f( )7 3( )+( )7 3( ) i a b( )a b( )a b c( )c( )2 3( )a b( )a b2 3a b( )a b( )2 3( )−( )2 3( ) j c d( )c d( )c d e( )e( )+( )( )4( )

5 Remove the grouping symbols and simplify if possible.

a 2 1 4( )2 1( )2 1g g4g g4( )g g( )2 1( )2 1g g2 1( )2 1+ +( )+ +( )2 1( )2 1+ +2 1( )2 1g g+ +g g( )g g( )+ +( )g g( )2 1( )2 1g g2 1( )2 1+ +2 1( )2 1g g2 1( )2 1 b 7 2( )7 2( )7 2s s( )s s( )7 2( )7 2s s7 2( )7 2+ +( )+ +( )7 2( )7 2+ +7 2( )7 2s s+ +s s( )s s( )+ +( )s s( )7 2( )7 2s s7 2( )7 2+ +7 2( )7 2s s7 2( )7 2

c 3 9 2( )3 9( )3 9y y2y y2( )y y( )3 9( )3 9y y3 9( )3 9y y− −y y( )y y( )− −( )y y( )3 9( )3 9y y3 9( )3 9− −3 9( )3 9y y3 9( )3 9 d 5 4x x5 4x x5 4x x− −x x5 4x x5 4− −5 4x x5 4( )2( )2x x( )x x− −( )− −x x− −x x( )x x− −x x

e 6 26 2z z6 2+ −6 2+ −6 2z z+ −z z6 2z z6 2+ −6 2z z6 2( )1( )1+ −( )+ −z z+ −z z( )z z+ −z z f 3 7q q3 7q q3 7q q− −q q3 7q q3 7− −3 7q q3 7( )5( )5q q( )q q− −( )− −q q− −q q( )q q− −q q


a 4 1 2 5( )4 1( )4 1x x2 5x x2 5( )x x( )4 1( )4 1x x4 1( )4 1− +( )− +( )4 1( )4 1− +4 1( )4 1x x− +x x( )x x( )− +( )x x( )4 1( )4 1x x4 1( )4 1− +4 1( )4 1x x4 1( )4 1 2 5+2 5 b 7 3 2 4 2( )7 3( )7 3 2 4( )2 4y y2 4y y2 4( )y y( )2 4− +2 4( )− +( )2 4( )2 4− +2 4( )2 42 4y y2 4− +2 4y y2 4( )y y( )− +( )y y( )2 4( )2 4y y2 4( )2 4− +2 4( )2 4y y2 4( )2 4 −

c 2 5 2 8( )2 5( )2 5 2 8( )2 8b b2 8b b2 8( )b b( )2 8( )2 8b b2 8( )2 82 8b b2 8+ −2 8b b2 8( )b b( )+ −( )b b( )2 8( )2 8b b2 8( )2 8+ −2 8( )2 8b b2 8( )2 82 8−2 8 d 4 17 5r r4 1r r4 17 5r r7 5+ +4 1+ +4 17 5+ +7 5r r+ +r r4 1r r4 1+ +4 1r r4 17 5r r7 5+ +7 5r r7 5( )3( )3r r( )r r −( )−

e 2 8 3 2n n2 8n n2 8 3 2n n3 2− +2 8− +2 8n n− +n n2 8n n2 8− +2 8n n2 8 ( )3 2( )3 23 2n n3 2( )3 2n n3 23 2+3 2( )3 2+3 2 f 5 2q q5 2q q5 2+ −5 2+ −5 2q q+ −q q5 2q q5 2+ −5 2q q5 2 ( )9( )9q q( )q q +( )+

9781107627291c02_p35-72.indd 45 8/29/12 9:26 PM


SAMPLE


Development


a 2 1 5 1( )2 1( )2 1 ( )5 1( )5 1x x5 1x x5 1( )x x( )2 1( )2 1x x2 1( )2 1 5 1( )5 1x x5 1( )5 1+ +( )+ +( )2 1( )2 1+ +2 1( )2 1x x+ +x x( )x x( )+ +( )x x( )2 1( )2 1x x2 1( )2 1+ +2 1( )2 1x x2 1( )2 1 5 1( )5 1−5 1( )5 1 b 3 2 2 1( )3 2( )3 2 ( )2 1( )2 1y y2 1y y2 1( )y y( )3 2( )3 2y y3 2( )3 2 ( )y y( )2 1( )2 1y y2 1( )2 1+ +( )+ +( )3 2( )3 2+ +3 2( )3 2y y+ +y y( )y y( )+ +( )y y( )3 2( )3 2y y3 2( )3 2+ +3 2( )3 2y y3 2( )3 2 2 1( )2 1+2 1( )2 1

c 5 2 3 4( )5 2( )5 2 ( )3 4( )3 4a a3 4a a3 4( )a a( )5 2( )5 2a a5 2( )5 2 3 4( )3 4a a3 4( )3 4+ +( )+ +( )5 2( )5 2+ +5 2( )5 2a a+ +a a( )a a( )+ +( )a a( )5 2( )5 2a a5 2( )5 2+ +5 2( )5 2a a5 2( )5 2 3 4( )3 4+3 4( )3 4 d 8 3 5 3( )8 3( )8 3 ( )5 3( )5 3c c5 3c c5 3( )c c( )8 3( )8 3c c8 3( )8 3 5 3( )5 3c c5 3( )5 3− +( )− +( )8 3( )8 3− +8 3( )8 3c c− +c c( )c c( )− +( )c c( )8 3( )8 3c c8 3( )8 3− +8 3( )8 3c c8 3( )8 3 5 3( )5 3+5 3( )5 3

e 6 6 2 2( )6 6( )6 6 ( )2 2( )2 2 1( )1s s2 2s s2 2( )s s( )6 6( )6 6s s6 6( )6 6 ( )s s( )2 2( )2 2s s2 2( )2 2+ +( )+ +( )6 6( )6 6+ +6 6( )6 6s s+ +s s( )s s( )+ +( )s s( )6 6( )6 6s s6 6( )6 6+ +6 6( )6 6s s6 6( )6 6 ( )−( ) f 5 7( )5 7( )5 7 ( )7( )7h h2 2h h2 2( )h h( )5 7( )5 7h h5 7( )5 7 ( )h h( )2 2( )2 2h h2 2( )2 2h h+ +h h( )h h( )+ +( )h h( )5 7( )5 7h h5 7( )5 7+ +5 7( )5 7h h5 7( )5 7 ( )−( )

g 4 3 1 2( )4 3( )4 3 1 2( )1 2( )2( )2x x1 2x x1 2( )x x( )1 2( )1 2x x1 2( )1 2( )x x( )1 2x x1 2− −1 2x x1 2( )x x( )− −( )x x( )1 2( )1 2x x1 2( )1 2− −1 2( )1 2x x1 2( )1 2( )−( ) h 9 5 7 2( )9 5( )9 5 ( )7 2( )7 2z z7 2z z7 29 5( )9 5z z9 5( )9 5 7 2( )7 2z z7 2( )7 2+ −( )+ −( )9 5( )9 5+ −9 5( )9 5z z+ −z z( )z z( )+ −( )z z( )9 5( )9 5z z9 5( )9 5+ −9 5( )9 5z z9 5( )9 5 7 2( )7 2−7 2( )7 2

i 5 2 4 3( )5 2( )5 2 4 3( )4 3( )7( )7c c4 3c c4 3( )c c( )4 3( )4 3c c4 3( )4 3( )c c( )4 3c c4 3− −4 3c c4 3( )c c( )− −( )c c( )4 3( )4 3c c4 3( )4 3− −4 3( )4 3c c4 3( )4 3( )+( ) j 5 5 1 4( )5 5( )5 5 1 4( )1 4( )2( )2g g1 4g g1 4( )g g( )1 4( )1 4g g1 4( )1 4( )g g( )1 4g g1 4− −1 4g g1 4( )g g( )− −( )g g( )1 4( )1 4g g1 4( )1 4− −1 4( )1 4g g1 4( )1 4( )−( )

k 7 2 3 3( )7 2( )7 2 3 3( )3 3( )3 3( )3 33 3u u3 3( )u u( )3 3( )3 3u u3 3( )3 33 3( )3 3u u3 3( )3 33 3u u3 3− −3 3u u3 3( )u u( )− −( )u u( )3 3( )3 3u u3 3( )3 3− −3 3( )3 3u u3 3( )3 33 3( )3 3−3 3( )3 3 l − −( )− −( )− − ( )( )4 1( )− −( )− −4 1− −( )− − 3 3( )3 3( )− −( )− −3 3− −( )− −d d− −d d− −( )d d( ) ( )d d( )( )4 1( )d d( )4 1( )− −( )− −4 1− −( )− −d d− −( )− −4 1− −( )− − 3 3d d3 3− −3 3− −d d− −3 3− −( )3 3( )d d( )3 3( )− −( )− −3 3− −( )− −d d− −( )− −3 3− −( )− −

8 Taylah was required to remove the grouping symbols and simplify. This was her solution.

6 3 2 2 3 18 12 2 616 6

( )6 3( )6 3 2 2( )2 2( )3 1( )3 1x x2 2x x2 2( )x x( )2 2( )2 2x x2 2( )2 2( )x x( ) x x8 1x x8 12 2x x2 2x

2 2x x2 2− −2 2x x2 2( )x x( )− −( )x x( )2 2( )2 2x x2 2( )2 2− −2 2( )2 2x x2 2( )2 2 3 1+ =3 1( )+ =( )3 1( )3 1+ =3 1( )3 1 x x− −x x8 1x x8 1− −8 1x x8 12 2x x2 2− −2 2x x2 2 += −16= −16x= −x

Where is the error in Taylah’s working?


a x x( )x x( )x x ( )( )− +( )5 2x x5 2x x( )5 2( ) ( )5 2( )x x( )x x5 2x x( )x x− +5 2− +( )− +( )5 2( )− +( ) ( )+( )5 2( )+( ) b b b b b( )b b( )b b ( )b b( )b b( )+ +( )3 1b b3 1b b( )3 1( ) ( )3 1( )b b( )b b3 1b b( )b b+ +3 1+ +( )+ +( )3 1( )+ +( ) ( )+( )3 1( )+( )

c y y y y( )y y( )y y ( )y y( )y y( )− +( )3 8y y3 8y y( )3 8( ) ( )3 8( )y y( )y y3 8y y( )y y− +3 8− +( )− +( )3 8( )− +( ) ( )+( )3 8( )+( ) d g g g g( )g g( )g g ( )g g( )g g( )2 3( )g g( )g g2 3g g( )g g ( )3( )+ −( )+ −( )( )2 3( )+ −( )2 3( ) ( )+( )

e v v( )v v( )v v ( )v v( )v v( )− +( ) ( )+( )7 6v v7 6v v( )7 6( ) ( )7 6( )v v( )v v7 6v v( )v v− +7 6− +( )− +( )7 6( )− +( ) ( )4( ) f b b b b( )b b( )b b ( )b b( )b b( )5 1( )b b( )b b5 1b b( )b b b b( )b b2 4b b( )b b− −( )− −( )( )5 1( )− −( )5 1( ) b b( )b b2 4b b( )b b+b b( )b b2 4b b( )b b

g 2 2u u2 2u u2 2 u u( )2 2( )2 2u u( )u u2 2u u2 2( )2 2u u2 2 ( )9( )9u u( )u u− +( )− +( )2 2( )2 2− +2 2( )2 2 ( )+( ) h 4 64 6n n4 6 n n( )4 6( )4 64 6n n4 6( )4 6n n4 6 ( )1( )1n n( )n n− −( )− −( )4 6( )4 6− −4 6( )4 6 ( )+( )

i 3 7d d3 7d d3 7 d d( )3 7( )3 7d d( )d d3 7d d3 7( )3 7d d3 7 ( )2 5( )2 5d d( )d d2 5d d2 5( )2 5d d2 5+ +( )+ +( )3 7( )3 7+ +3 7( )3 7 2 5( )2 5+2 5( )2 5 j e e e e( )e e( )e e ( )e e( )e e( )+ −( ) ( )−( )2 7( )2 7( )+ −2 7+ −( )+ −( )2 7( )+ −( ) ( )9( )

k 6 3k k6 3k k6 3 k k( )6 3( )6 3k k( )k k6 3k k6 3( )6 3k k6 3 ( )3( )3k k( )k k− +( )− +( )6 3( )6 3− +6 3( )6 3 ( )+( ) l t t( )t t( )t t ( )t t( )t t( )5 3( )t t( )t t5 3t t( )t t 7 2t t7 2t t( )7 2( )t t( )t t7 2t t( )t t− +( )− +( )t t( )t t− +t t( )t tt t( )t t5 3t t( )t t− +t t( )t t5 3t t( )t t t t( )t t−t t( )t t

10 Expand and simplify the algebraic expression 2 3ab2 3ab2 3 ab( )2 3( )2 3ab( )ab2 3ab2 3( )2 3ab2 3 ( )1( )1ab( )ab− −( )− −( )2 3( )2 3− −2 3( )2 3 ( )−( ).


a x x2x x2x x 2 1( )x x( )x x2 3( )2 3x x2 3x x( )x x2 3x x ( )x( )x2 1( )2 1x2 1x( )x2 1x+ −( )+ −( )2 3( )2 3+ −2 3( )2 3 2 1( )2 1+2 1( )2 1 b a a2a a2a a 2 4( )a a( )a a 2 4( )2 4( )a( )a 3( )32 4+ −2 4( )+ −( )2 4( )2 4+ −2 4( )2 4( )+( )

c y y2y y2y y 3 7( )y y( )y y5 2( )5 2y y5 2y y( )y y5 2y y ( )y( )y3 7( )3 7y3 7y( )y3 7y+ −( )+ −( )5 2( )5 2+ −5 2( )5 2 3 7( )3 7+3 7( )3 7 d b b b b( )b b( )b b ( )b b( )b b( )+ −( ) ( )+( )7 3b b7 3b b( )7 3( ) ( )7 3( )b b( )b b7 3b b( )b b+ −7 3+ −( )+ −( )7 3( )+ −( ) ( )2( )2b b2b bb b7 3b b2b b7 3b b

e z z z z( )z z( )z z ( )z z( )z z( )3 1( )z z( )z z3 1z z( )z z ( )5( )2z z2z z− +( )− +( )( )3 1( )− +( )3 1( ) ( )−( ) f e e( )e e( )e e ( )e e( )e e7 2e e7 2e e( )7 2( )e e( )e e7 2e e( )e e ( )7 2( )e e( )e e7 2e e( )e e( )6( )27 227 2e e7 2e e2e e7 2e e7 2− −7 2( )7 2( )− −( )7 2( )e e( )e e7 2e e( )e e− −e e( )e e7 2e e( )e e ( )+( )

g x x( )x x( )x x ( )2 2( )2 2( ) ( )2 2( )7 2x x7 2x x( )7 2( ) ( )7 2( )x x( )x x7 2x x( )x x2 27 22 2x x2 2x x7 2x x2 2x x( )2 2( )7 2( )2 2( ) ( )2 2( )7 2( )2 2( )x x( )x x2 2x x( )x x7 2x x( )x x2 2x x( )x x( )+ −( )( )2 2( )+ −( )2 2( )7 2+ −7 2( )7 2( )+ −( )7 2( )2 27 22 2+ −2 27 22 2( )2 2( )7 2( )2 2( )+ −( )2 2( )7 2( )2 2( ) ( )7 2( )+( )7 2( ) h a a a a( )a a( )a a ( )a a( )a a( )2 1( )a a( )a a2 1a a( )a a ( )4( )2 2a a2 2a a( )2 2( ) ( )2 2( )a a( )a a2 2a a( )a a( )2 1( )2 2( )2 1( )− +( )− +( )( )2 1( )− +( )2 1( )2 2− +2 2( )2 2( )− +( )2 2( )( )2 1( )2 2( )2 1( )− +( )2 1( )2 2( )2 1( ) ( )+( )

i v v v v( )v v( )v v ( )v v( )v v2 1v v2 1v v( )2 1( )v v( )v v2 1v v( )v v v v( )v v2 1v v( )v v2 2( )2 2( ) ( )2 2( )v v( )v v2 2v v( )v v2 12 22 1v v2 1v v2 2v v2 1v v( )2 1( )2 2( )2 1( ) ( )2 1( )2 2( )2 1( )v v( )v v2 1v v( )v v2 2v v( )v v2 1v v( )v v2 1− −2 1( )2 1( )− −( )2 1( )v v( )v v2 1v v( )v v− −v v( )v v2 1v v( )v v v v( )v v−v v( )v v j a a b b2a a2a a( )a a( )a a b b( )b b( )a b( )a bb b+ −b b( )+ −( )b b( )b b+ −b b( )b b a b( )a b+a b( )a b

k x x y x2 2x x2 2x x( )x x( )x x y x( )y x2 2( )2 2x x2 2x x( )x x2 2x x ( )x y( )x y3( )3x y3x y( )x y3x y+ −y x+ −y x( )+ −( )y x( )y x+ −y x( )y x( )+( )x y( )x y+x y( )x y l y y z y2 2y y2 2y y 4 1z y4 1z y2 24 12 2z y2 2z y4 1z y2 2z y( )y y( )y y z y( )z y2 2( )2 2y y2 2y y( )y y2 2y y2 24 12 2( )2 24 12 2( )2 2( )2 24 1( )4 1z4 1z( )z4 1z2 24 12 2( )2 24 12 2z2 2z4 1z2 2z( )z2 2z4 1z2 2zz y4 1z y+ −z y4 1z y2 24 12 2+ −2 24 12 2z y2 2z y4 1z y2 2z y+ −z y2 2z y4 1z y2 2z y( )+ −( )2 2( )2 2+ −2 2( )2 24 1( )4 1+ −4 1( )4 1z y4 1z y( )z y4 1z y+ −z y4 1z y( )z y4 1z y2 24 12 2( )2 24 12 2+ −2 24 12 2( )2 24 12 2z y2 2z y4 1z y2 2z y( )z y2 2z y4 1z y2 2z y+ −z y2 2z y4 1z y2 2z y( )z y2 2z y4 1z y2 2z y4 1( )4 1+4 1( )4 1

12 Expand and simplify ( ) ( ) .( )n r( ) ( )n n( )r n( )r n( )( )+ −( )( )n r( )+ −( )n r( ) ( )+( )4 7( )4 7( ) ( )4 7( )( )n r( )4 7( )n r( )n n4 7n n( )n n( )4 7( )n n( )+ −4 7+ −( )+ −( )4 7( )+ −( )( )n r( )+ −( )n r( )4 7( )n r( )+ −( )n r( )n n+ −n n4 7n n+ −n n2 2( )2 2( )( )n n( )2 2( )n n( )r n2 2r n( )r n( )2 2( )r n( )( )+( )2 2( )+( )4 72 24 7( )4 7( )2 2( )4 7( )n n4 7n n2 2n n4 7n n( )n n( )4 7( )n n( )2 2( )n n( )4 7( )n n( )+ −4 7+ −2 2+ −4 7+ −n n+ −n n4 7n n+ −n n2 2n n+ −n n4 7n n+ −n n

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SAMPLE


2.4 Factorising algebraic expressionsFactorising is the reverse process to expanding. For example, expanding the expression 5(2y - 3) produces 10y - 15, whereas factorising the expression 10y - 15 produces 5(2y - 3). The fi rst step in factorising an expression is to fi nd the largest factor of each term or the highest common factor (HCF). In this case the HCF of 10y and 15 is 5. The HCF is written outside the grouping symbol and terms inside are found by dividing the HCF into each term.

Factorising algebraic expressions

1 Find the largest factor of each term or the HCF.2 Write the HCF outside the grouping symbol.3 Divide the HCF into each term to fi nd the terms inside the grouping symbols.4 Check the factorisation by expanding the expression.

Example 11 Factorising algebraic expressions

Factorise 3p - 6.

Solution1 Find the largest factor of each term (HCF is 3).2 Write the HCF or 3 outside the grouping symbol.3 Divide the HCF or 3 into each term to find the terms

inside the grouping symbols.4 Check by expanding the expression.

3 6 3 3 23 23 2

p p3 6p p3 6 3 3p p3 3p p− =p p3 6p p3 6− =3 6p p3 6 3 3× −3 33 3p p3 3× −3 3p p3 3 ×= ×3 2= ×3 2= −3 2= −3 2

( )3 2( )3 2p( )p3 2p3 2( )3 2p3 23 2−3 2( )3 2−3 2( )3 2( )3 23 2p3 2( )3 2p3 2p( )p3 2p3 2( )3 2p3 23 2= −3 2( )3 2= −3 23 2= −3 2( )3 2= −3 23 2p3 2= −3 2p3 2( )3 2p3 2= −3 2p3 2

Example 12 Factorising algebraic expressions

Factorise 2x2 + 6x.

Solution1 Find the largest factor of each term (HCF is 2x).2 Write the HCF or 2x outside the grouping symbol.3 Divide the HCF or 2x into each term to find the

terms inside the grouping symbols.4 Check by expanding the expression.

2 6 2 2 32 32 3

22 622 6x x2 6x x2 62 622 6x x2 622 6 x x2 2x x2 2x2 3x x2 32 3x x2 3

+ =2 6+ =2 6x x+ =x x2 6x x2 6+ =2 6x x2 6 2 2× +2 22 2x x2 2× +2 2x x2 2 ×= ×2 3= ×2 32 3x x2 3= ×2 3x x2 3= +2 3= +2 32 3x x2 3= +2 3x x2 3

( )2 3( )2 32 3x x2 3( )2 3x x2 32 3+2 3( )2 3+2 3( )2 3( )2 32 3x x2 3( )2 3x x2 32 3= +2 3( )2 3= +2 32 3x x2 3= +2 3x x2 3( )2 3x x2 3= +2 3x x2 3

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SAMPLE


Exercise 2D 1 Find the HCF to complete the factorisations.

a 3 9m mm m3 9m m3 9+ =3 9+ =3 9m m+ =m m3 9m m3 9+ =3 9m m3 9 ( )3( )3m m( )m m +( )+ b 4 16 46 4x x4 1x x4 16 4x x6 46 4x x6 4x x− =x x4 1x x4 1− =4 1x x4 16 4x x6 4− =6 4x x6 4( )6 4( )6 46 4x x6 4( )6 4x x6 46 4−6 4( )6 4−6 4

c 10 20y yy y20y y20+ =20+ =20y y+ =y y20y y20+ =20y y20 ( )2( )2y y( )y y +( )+ d 5 30 60 6x x5 3x x5 30 6x x0 60 6x x0 6x x− =x x5 3x x5 3− =5 3x x5 30 6x x0 6− =0 6x x0 6( )0 6( )0 60 6x x0 6( )0 6x x0 60 6−0 6( )0 6−0 6

e 14 56a aa aa a56a a56+ =56+ =56a a+ =a a56a a56+ =56a a56 ( )2 8( )2 8a a( )a a2 8a a2 8( )2 8a a2 82 8+2 8( )2 8+2 8 f 6 22 32 32 32 3w w6 2w w6 22 3w w2 32 3w w2 3w w− =w w6 2w w6 2− =6 2w w6 22 3w w2 3− =2 3w w2 3( )2 3( )2 3 11( )11w w( )w w2 3w w2 3( )2 3w w2 3 −( )−

g 8 12 22 2e ee e8 1e e8 12 2e e2 22 2e e2 2+ =8 1+ =8 12 2+ =2 2e e+ =e e8 1e e8 1+ =8 1e e8 12 2e e2 2+ =2 2e e2 2( )2 2( )2 2 3( )3e e( )e e2 2e e2 2( )2 2e e2 2 +( )+ h 9 12 32 32 32 3n n9 1n n9 12 3n n2 32 3n n2 3n n− =n n9 1n n9 1− =9 1n n9 12 3n n2 3− =2 3n n2 3( )2 3( )2 3 4( )4n n( )n n2 3n n2 3( )2 3n n2 3 −( )−

2 Complete the following factorisations.a 2 8 2y2 8y2 8+ =2 8+ =2 8 +( ) b 5 20 55 2a5 2+ =5 2+ =5 20 5+ =0 5 +( )c 7 7 77 7x7 7− =7 7− =7 7 −( ) d 8 24 8h8 2h8 2− =8 2− =8 24 8− =4 8 −( )e 14 70 7v + =70+ =70 +( ) f 15 45 5s − =45− =45 −( )g 6 9 3d6 9d6 9− =6 9− =6 9 −( ) h 10 5 5+ =5 5+ =5 5 +( )5 5x5 55 5+ =5 5x5 5+ =5 5

3 Jakob was required to factorise an algebraic expression. This was his solution.

2 6 2 6x x2 6x x2 6 2 6x x2 6x x− =x x2 6x x2 6− =2 6x x2 6 ( )2 6( )2 62 6x x2 6( )2 6x x2 62 6−2 6( )2 6−2 6

Where is the error in Jakob’s working?

4 Factorise each of the following.a 5 205 2a5 25 2+5 2 b 3 183 1x3 13 1+3 1 c 8 56p8 5p8 58 5−8 5d 7 217 2z7 27 2+7 2 e 32 4+ d f 27 9− tg 4 10n4 1n4 14 1+4 1 h 16 36c + i 15 20f +j 24 16− x k 28 8g − l 9 219 2w9 29 2−9 2

5 Anh’s working for a factorisation question is shown below.

6 18 6y y6 1y y6 18 6y y8 6+ =6 1+ =6 18 6+ =8 6y y+ =y y6 1y y6 1+ =6 1y y6 18 6y y8 6+ =8 6y y8 6( )3( )3y y( )y y +( )+

How can she check that the factorisation is correct?

6 Factorise each of the following by taking out a negative factor.a − −2 1− −2 1− − 2a2 1a2 1− −2 1− −a− −2 1− − b − −3 1− −3 1− − 5q3 1q3 1− −3 1− −q− −3 1− − c − −5 2− −5 2− − 05 2m5 2− −5 2− −m− −5 2− −d − +9 9− +9 9− + 09 9s9 9− +9 9− +s− +9 9− + e − −16− −16− − 4 y f − −32− −32− − 8h

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SAMPLE


Development

7 Complete the following factorisations.

a x x x2x x2x x4+ =x x+ =x x+ =x x+ =x x4+ =4x x4x x+ =x x4x x ( ) b 6 2y y y− =2− =2y y− =y y ( )

c − − = −2 3− −2 3− − 3e e2 3e e2 3− −2 3− −e e− −2 3− − e( ) d − − = −6 7− −6 7− − 4k k6 7k k6 7− −6 7− −k k− −6 7− − k( )

e 10 14 22r r14r r142r r2 r− =r r− =r r14r r14− =14r r14 ( ) f 9 3 329 329 3x x9 3x x9 39 329 3x x9 329 3 x+ =9 3+ =9 3x x+ =x x9 3x x9 3+ =9 3x x9 3 ( )

g q q q4 2q q4 2q q 24 254 2q q5q q+ =q q+ =q q4 2+ =4 2q q4 2q q+ =q q4 2q qq q+ =q q4 2+ =4 2q q4 2q q+ =q q4 2q qq q5q q+ =q q5q q4 254 2+ =4 254 2q q4 2q q5q q4 2q q+ =q q4 2q q5q q4 2q q ( ) h − −w w− −w w− −2 3w w2 3w w 22 (= −2 (= −w w2 (w w w2 (w2 32 (2 3w w2 3w w2 (w w2 3w w 22 (2 )

i 6 8 236 836 8n n6 8n n6 86 836 8n n6 836 8 n+ =6 8+ =6 8n n+ =n n6 8n n6 8+ =6 8n n6 8 ( ) j 6 9 356 956 9t t6 9t t6 96 956 9t t6 956 9 t+ =t t+ =t t6 9t t6 9+ =6 9t t6 9 ( )

8 Fully factorise the following expressions.

a y y2y y2y y5y y5y yy y−y y b 10 52m m5m m52m m2 +m m+m m c 4 1624 124 1x x4 1x x4 16x x64 124 1x x4 124 14 1x x4 1−4 1x x4 1

d 3 2v v+v v+v v e 8 4 3g g8 4g g8 48 4+8 48 4g g8 4+8 4g g8 4 f 4 12 4d d4 1d d4 12d d22 4d d2 44 12 44 1d d4 12 44 122 42d d22 424 1d d4 1+4 1d d4 14 12 44 1d d4 12 44 1+4 12 44 1d d4 12 44 1

g 2 4ab2 4ab2 4a2 4+2 4 h 5 1025 125 1xy5 1xy5 1 xy5 1+5 1 i 6 152 26 12 26 152 25b c6 1b c6 12 2b c2 26 12 26 1b c6 12 26 1 b2 2b2 26 1−6 1

j def edef ede ff e−f ef e2f e2 k 5 32 25 32 25 3x y5 3x y5 35 32 25 3x y5 32 25 3xy2 2xy2 25 3+5 35 32 25 3+5 32 25 3 l r t r t3 2r t3 2r t 2 2r t2 2r t−

9 Stacey was required to completely factorise an algebraic expression. This was her solution.

12 8 42 28 42 28 4x x8 4x x8 4+ =8 4+ =8 48 42 28 4+ =8 42 28 4x x+ =x x8 4x x8 4+ =8 4x x8 4( )3 2( )3 22 2( )2 23 22 23 2( )3 22 23 2x x( )x x3 2x x3 2( )3 2x x3 22 2x x2 2( )2 2x x2 23 22 23 2x x3 22 23 2( )3 22 23 2x x3 22 23 23 22 23 2+3 22 23 2( )3 22 23 2+3 22 23 23 2x x3 2+3 2x x3 2( )3 2x x3 2+3 2x x3 23 22 23 2x x3 22 23 2+3 22 23 2x x3 22 23 2( )3 22 23 2x x3 22 23 2+3 22 23 2x x3 22 23 2

Where is the error in Stacey’s working?

10 Factorise each of the following by taking out the highest common factor.a 4 12 8x y4 1x y4 12 8x y2 8+ +4 1+ +4 12 8+ +2 8x y+ +x y4 1x y4 1+ +4 1x y4 12 8x y2 8+ +2 8x y2 8 b 10 5 15x y5 1x y5 1 z− +5 1− +5 1x y− +x y5 1x y5 1− +5 1x y5 1 c 2 4 6a b c2 4a b c2 4 6a b c6a b c+ −a b c2 4a b c2 4+ −2 4a b c2 4

d 9 6 1229 629 6y y9 6y y9 69 629 6y y9 629 6+ −9 6+ −9 6y y+ −y y9 6y y9 6+ −9 6y y9 6 e 8 4 6 2− +8 4− +8 4r r6r r6− +r r− + f 3 6 92+ +3 6+ +3 6h h9h h92h h2+ +h h+ +2+ +2h h2+ +2

g 10 15 252v v15v v15− +15− +15 2− +2v v− +v v15v v15− +15v v15 h 8 4 12m m8 4m m8 4 n− +8 4− +8 4m m− +m m8 4m m8 4− +8 4m m8 4 n− +n i 9 15 18c c9 1c c9 15 1c c5 1d d5 1d d5 18d d8+ +9 1+ +9 15 1+ +5 1c c+ +c c9 1c c9 1+ +9 1c c9 15 1c c5 1+ +5 1c c5 15 1d d5 1+ +5 1d d5 1

11 Factorise by using the HCF for each of the following expressions.

a 2ab ac ag+ +ac+ +ac b 4x xy xx xy xx x zy xzy x+ +x x+ +x xy x+ +y xx xy xx x+ +x xy xx x c 7 1 2g g7 1g g7 1h g7 1h g7 14h g47 1− +7 17 1g g7 1− +7 1g g7 17 1h g7 1− +7 1h g7 1

d d d de2d d2d dd d3d d3d d3d d− +d d− +d dd d3d d− +d d3d d e b bc b2b b2b b 5c b5c b+ −b b+ −b bc b+ −c b f k k h khh khh kk k− +k k4 2h k4 2h k4 2k k4 2k k− +4 2− +k k− +k k4 2k k− +k k h k− +h k4 2h k− +h k24 224 2− +4 2− +2− +4 2− +

g xyz xy x y+ −5 2xy5 2xy+ −5 2+ −xy+ −xy5 2xy+ −xy2 2x y2 2x y5 22 25 2+ −5 2+ −2 2+ −5 2+ − h 2 5 32 2 2abc2 5abc2 5a b2 2a b2 2c a3c a3 b+ −2 5+ −2 5 2 2+ −2 2 2+ −2a b+ −a b2 2a b2 2+ −2 2a b2 2c a+ −c a2c a2+ −2c a2 i mn m n mn− +m n− +m n2− +2− +2 2mn2 2mn− +2 2− +m n− +m n2 2m n− +m n

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SAMPLE


2.5 SubstitutionSubstitution involves replacing the pronumeral in an algebraic expression with one or more numbers. The resulting numerical expression is evaluated and expressed to the specifi ed level of accuracy.

Substitution of values

1 Write the algebraic expression.2 Replace the variables in the expression with the numbers given in the question.3 Evaluate using the calculator.4 Write the answer to the specifi ed level of accuracy and correct units if necessary.

Example 13 Substituting values

Evaluate 3a - 4b + c given a = 2, b = 5 and c = -10.

Solution1 Write the algebraic expression.2 Substitute the values for a, b and c into

the algebraic expression.3 Evaluate.

3 4 3a b3 4a b3 4 c− ×3− ×3a b− ×a b3 4a b3 4− ×3 4a b3 4 2 − 4 × 5 + −10= 6 − 20 − 10= −24

+ =c+ =c− ×+ =− ×c− ×c+ =c− ×c

Example 14 Substituting values

Evaluate the following, given a = 2.a ( )( )2 5( )( )2 5( )a( )2 5( )( )2 5( )+( )2 5( ) b 3a3

Solution1 Write the algebraic expression.2 Substitute the value for a into


a ( )( )2 5( ) 2

93

( )2 5( )a( )2 5( )+ =( )+ =( )( )2 5( )+ =( )2 5( )

==

× 2 + 5

4 Write the algebraic expression.5 Substitute the value for a into


b 3a3 = 3 × 23

= 3 × 8 = 24

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SAMPLE


Exercise 2E 1 Evaluate these expressions given a = 3, b = 4 and c = 8.

a 5a ba b+a b b a ba b+a ba b4a b c a b c− +a b− +a b

d a b2 2a b2 2a ba b+a ba b2 2a b+a b2 2a b e 4b a× (b a× (b ab a−2b a) f 2ab

g abc

2 h b c2b c2b cb c÷b c i

2ab

c

2 Find the value of 3a + 2b if:a a = 5 and b = 5 b a = 6 and b = -4 c a = 0 and b = 0

d a = -7 and b = -2 e a = 2

3 and b = 1 f a = 2 and b = 1

2

3 Calculate the value of e2 - 3 if:a e = 1 b e = 3 c e = 10d e = 2 e e = -1 f e = -2

g e = 1

2 h e = 3.1 i e = 1

5

4 Find ( )( )( )w( )( )−( )( )7( ) if:a w = 100 b w = 144 c w = 256d w = 25 e w = 49 f w = 400

g w = 9 h w = 1 i w = 1

4

5 Determine the value of 2

4

m nm n−m n if:

a m = 4 and n = 4 b m = 10 and n = 2c m = 20 and n = 8 d m = 1 and n = -6e m = 0.5 and n = -11 f m = 2 and n = 0.25

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SAMPLE


Development

6 Evaluate these expressions given x = 7, y = -5 and z = 21.

a x z y2 2x z2 2x z+ +x z+ +x z2 2+ +2 2x z2 2x z+ +x z2 2x z b y x3y x3y x4y x4y xy x−y x

c 4 1x z4 1x z4 14 1+ −4 14 1x z4 1+ −4 1x z4 1 d z y+z y+z y4z y4z y2

e 3 2xy

z f

6

5

2y

zx

7 The cross-sectional area of this solid is an annulus. It is evaluated using π ( )( )R r( )2 2( )2 2( )( )R r( )2 2( )R r( )( )R r( )−( )R r( ) where R is the radius of the outer circle and r the radius of the inner circle. Find the area of an annulus if R is 8 cm and r is 4 cm. Answer correct to one decimal place.

8 Determine the value of 23

2p q2p q2 given that p = 4 and q = 6.

9 Evaluate 2 33 y2 3y2 32 3+2 3 if y = 12.

10 Find the value of 2π lg

when l = 2.6 and g = 9.8. Give your answer correct to two decimal places.

11 Find the value of u as2u a2u a2u a2u a+u a+u a if u = 6, a = 7 and s = 2.

12 Find the value of 1

2π fcπ fcπ if f = 10 and c = 2. Give your answer correct to three decimal

places.

13 Find the value of 3Rr

R rR r+R r when R = 8.2 and r = 4.9. Give your answer correct to two

decimal places.

14 What is the value of yA

( )y( )y( )+( )( )12( ) when y = 9 and A = 15. Give your answer correct to the nearest whole number.

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SAMPLE


2.6 Linear equationsAn equation is a mathematical statement that says that two things are equal. It has an equal sign. For example, these are all equations:

x + =4 7+ =4 7+ = 2 18p2 1p2 12 1=2 1 6 1 23y6 1y6 1− =6 1− =6 1

Linear equations have all their variables raised to the power of 1. The above three equations are linear equations. An equation such as x2 = 9 is not a linear equation as the variable is raised to the power of 2.

Solving an equationThe process of fi nding the unknown value for the variable is called solving the equation.

When solving an equation look to perform the opposite operation:• + is opposite to -• × is opposite to ÷• x2 is opposite to √x

Make sure the equation remains balanced like a set of scales. The same operation needs to be done on both sides of the equal sign to keep the balance.

Solving an equation

1 Look to perform the opposite operation (+ is opposite to -, × is opposite to ÷).2 Add or subtract the same number to both sides of the equation OR3 Multiply or divide both sides of the equation by the same number.4 To solve two or three step equations repeat the above steps as required. It is often

easier to fi rstly add or subtract the same number to both sides of the equation.

When a solution has been reached it can be checked. The solution of the equation must satisfy the equation. Always check your solution by substituting your answer into the original equation. The left-hand side of the equation must equal the right-hand side.

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SAMPLE


Example 15 Solving an equation

Solve the equation x + 4 = 7.

Solution1 Write the equation.2 The opposite operation to adding 4 is subtracting 4.

Subtract 4 from both sides of the equation.3 Check that the solution is correct by substituting

x = 3 into the original equation.

x

xx

+ =

+ ==

− −4 7+ =4 7+ =

4 7+ =4 7+ =3

4 4− −4 4− −

Example 16 Solving an equation

Solve the equation 2p = 18.

Solution1 Write the equation.2 The opposite operation to multiplying by 2 is

dividing by 2. Divide both sides of the equation by 2.3 Check that the solution is correct by substituting

p = 9 into the original equation.

2 182

2

18

29

p2 1p2 1p

p

2 1=2 1

=

=

Example 17 Solving two-step equations

Solve the equation 4a + 5 = -1. Express your answer as a simplest fraction.


Subtract 5 from both sides of the equation.3 The opposite operation to multiplying by 4 is dividing

by 4. Divide both sides of the equation by 4.

4 Simplify the improper fraction −( )64

by writing it as a

mixed number in simplified form −( )1 12

.

5 Check that the solution is correct by substituting

a = −1 12

into the original equation.

4 5 1

4 5 14 64

4

6

4

12

4

11

2

5 5

4 5a4 5

4 5a4 54 6a4 6a

a

+ =4 5+ =4 5 −

+ =4 5+ =4 5 −4 6= −4 6

=−

= −

= −

− −5 5− −5 5

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SAMPLE


Example 18 Solving three-step equations

Solve the equation 2 4 36 3y y2 4y y2 4 36y y36 3y y3y y− =y y2 4y y2 4− =2 4y y2 4y y−y y .

Solution1 Write the equation.

2 The opposite operation to subtracting 4 is adding 4. Add 4 to both sides of the equation.

3 The opposite operation to subtracting -3y is adding +3y. Add +3y to both sides of the equation.

4 The opposite operation to multiplying by 5 is dividing by 5. Divide both sides of the equation by 5.

5 Check that the solution is correct by substituting y = 8 into the original equation.

=

=

+ +

+ +

y y− =y y− = −y y−

y y− =y y− = −y y−

y y= −y y= −

y y= −y y= −

y

y

y

2 4y y2 4y y− =y y− =2 4− =y y− = 36y y36y y3y y3y y

2 4y y2 4y y− =y y− =2 4− =y y− = 36y y36y y3y y3y y

2 4y y2 4y y= −y y= −2 4= −y y= −0 3y y0 3y y= −y y= −0 3= −y y= −

2 4y y2 4y y= −y y= −2 4= −y y= −0 3y y0 3y y= −y y= −0 3= −y y= −

5 4=5 4=y5 4y 0

5

5

40

5

8

y y

4 4+ +4 4+ +

3 3+ +3 3+ +y y3 3y y+ +y y+ +3 3+ +y y+ +

Example 19 Solving a practical problem involving a linear equation

Seven is added to three times a number and the result is 22.a Write and equation using x to represent the number.b Solve the equation for x.

Solution1 Use mathematical symbols to replace the words in the

statement. 7 for seven+ for added3 for three× for times= for result

2 The opposite operation to adding 7 is subtracting 7. Subtract 7 from both sides of the equation.

3 The opposite operation to multiplying by 3 is dividing by 3. Divide both sides of the equation by 3.

4 Check that the solution is correct by substituting x = 5 into the original equation.

a 7 3 22+ =7 3+ =7 3x+ =x+ =

b 7 3 22

7 3 22

3 15

3

3

15

3

5

7 7

+ =7 3+ =7 3

+ =7 3+ =7 3

3 1=3 1

=

=

− −7 7− −7 7

x+ =x+ =

x+ =x+ =

3 1x3 1

x

x

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SAMPLE


Solving an equation using a graphics calculatorGraphics calculators have an equation mode that may be used to solve any type of linear equation. The red letters on the keypad are used as the variable. After the equation is entered set the required variable to zero and choose the ‘solver’ key. The calculator will show the answer and the values of the left-hand side and right-hand side of the equation.

Example 20 Solving a linear equation using a graphics calculator

Solve the equation 80 - 10y = 100.

Solution1 Select the EQUA menu.2 Select SOLVER (F3).3 Enter the equation. To place the variable (y) use

the ALPHA key to access the letters in red above the keys.

4 Highlight the required variable or Y = 0.5 Press SOLV and the calculator will show the

answer. 6 The values of the left (Lft) and right (Rgt) sides of

the formula are shown and should be equal.

Example 21 Solving a linear equation using a graphics calculator

Solve the equation 6x + 5 = 7 + 5x.

Solution1 Select the EQUA menu.2 Select SOLVER (F3).3 Enter the equation. To place the variable (x) use the

ALPHA key to access the letters in red above the keys.

4 Highlight the required variable or X = 0.5 Press SOLV and the calculator will show the answer. The values of the left (Lft) and right (Rgt) sides of


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SAMPLE


Exercise 2F 1 Solve the following linear equations.

a y + =8 1+ =8 1+ = 7 b x + =13+ =13+ = 28 c a + =7 1+ =7 1+ = 2d c + =7 4+ =7 4+ = −7 4− e m + =9 4+ =9 4+ = f 4 54 5+ =4 54 5−4 5d4 5d4 54 5+ =4 5d4 5+ =4 5g 8 118 1+ =8 1h8 1h8 18 1+ =8 1h8 1+ =8 1 h 9 49 4+ =9 49 4+ =9 4r9 4+ =9 4 i 10 1+ = −q+ =q+ =j 5 = x + 2 k 12 = m + 7 l -3 = g + 5

2 Solve the following linear equations.a a - 7 = 3 b k - 5 = 5 c d - 9 = 14d s - 5 = - 4 e z - 12 = -7 f k - 7 = -9g 11 - v = 4 h 7 - x = 3 i 6 - j = 7j 9 = h - 5 k 11 = f - 4 l -4 = c - (-1)

3 Solve the following linear equations.a 4x = 12 b 5w = 45 c 7v = 28d 2t = -12 e 6h = -30 f -4a = 40g 2w = 13 h 3c = -23 i 7e = -8j 17 = 8k k -75 = 7d l -14 = -3e

4 Solve the following linear equations.

a y

24= b d

78= c w

64=

d f

−=

35 e

a

−=

75 f

g

92=

g d

129= − h s

−= −

113 i x

−= −

54

j 62

=−y

k 102

= m l 9

9=

−w

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SAMPLE


5 Solve the following linear equations. All solutions are integers.a 2x + 4 = 8 b 5y + 3 = 33c 4q + 6 = 14 d 6d - 1 = 59e 7v - 3 = 25 f 4m - 2 = 10g 10g - 5 = -25 h 9x - 4 = -13i 11p - 6 = -50 j 45 = 3v - 3k 22 = 7n + 1 l 69 = 5b - 11

6 Solve the following linear equations. All solutions are integers.a 3 + 2y = 11 b 2 + 3c = 5c 5 + 3a = 23 d 31 = 7y + 10e 5 = 7 + 2m f 33 = 18 - 3dg 28 - 4q = 16 h 90 - 10r = 100i 48 - 16e = -16 j 14 = 8 + 2hk 36 = 8 + 7d l 78 = 18 - 10w

7 Solve the following linear equations. Express your answer as a simplest fraction.a 4m + 1 = 26 b 3c + 16 = 27c 2y + 13 = 16 d 28 = 16 + 5be 37 = 6 + 4x f 11 = 28 + 8dg 2z - 3 = 4 h 5h - 7 = 14i 6e - 2 = 13 j 21 = 2q + 10k 32 = 3y - 5 l 17 = 9x + 9

8 Rajiv was required to solve the following equation for homework. This was his solution.

7 6 87 14

2

7 6x7 67 1x7 1

x

+ =7 6+ =7 67 1=7 1

=

Where is the error in Rajiv’s working?

9 Two is added to a number and the result is 7.a Write and equation using x to represent the number.b Solve the equation for x.

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SAMPLE


Development

10 Solve the following linear equations.a 3x + 2 = 2x + 7 b 3y + 5 = 2y + 7 c 5v + 7 = 4v - 8d 9b - 4 = 2 + 7b e 6d + 1 = 16 + 3d f 8z - 5 = 9 + zg 8 + a = 3a - 10 h 11 + 2w = 6w + 3 i 4 + 3e = 9 - 2e

11 Matthew spent $93 on six show bags at the Royal Easter show, each costing the same price, $p.a Using p as the cost of one show

bag, write an equation showing the cost of the six show bags.

b Use the equation to find the cost of one show bag.

12 If 12 is added to a certain number, the result is three times the number. Find the number.

13 Solve the equation 2 4 1 5 2 17( )2 4( )2 4 1 5( )1 5( )2 1( )2 1 .y y1 5y y1 5( )y y( )1 5( )1 5y y1 5( )1 5( )y y( )1 5y y1 5− −1 5y y1 5( )y y( )− −( )y y( )1 5( )1 5y y1 5( )1 5− −1 5( )1 5y y1 5( )1 5 2 1− =2 1( )− =( )2 1( )2 1− =2 1( )2 1

14 Isabella was required to solve the following equation for homework. This was her solution. Where is the error in Isabella’s working?

4 3 11 62 14

7

x x4 3x x4 3 11x x11 6x x6x2 1x2 1x

x x− =x x4 3x x4 3− =4 3x x4 3 +x x+x x2 1=2 1

=

15 Ten is added to three times a number and the result is 19.a Write an equation using x to represent the number.b Solve the equation for x.

16 A number is increased by 4 and then this amount is doubled. The result is 20.a Write an equation for this information.b Find the number.

17 Four is added to twice a number and the result is 36. a Write an equation using n to represent the number.b Solve the equation for n.

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SAMPLE


2.7 Equations with fractionsEquations with fractions are solved in exactly the same way as other equations. Look to perform the opposite operation (+ is opposite to -, × is opposite to ÷) to both sides of the equation. Check your solution by substituting your answer into the original equation.

Example 22 Solving an equation with a fraction

Solve the equation e

34 10+ =4 1+ =4 1 .


Subtract 4 from both sides of the equation.3 The opposite operation to dividing by 3 is multiplying

by 3. Multiply both sides by 3.4 Check that the solution is correct by substituting

e = 18 into the original equation.

e

e

e

e

34 10

36

33

6 3

18

4 4

+ =4 1+ =4 1

=

× =× = 6 3×6 3

=

− −4 4− −4 4

Example 23 Solving an equation with a fraction

Solve the equation 3 5

26

3 5x3 53 5−3 5 = .

Solution1 Write the equation.2 The opposite operation to dividing by 2 is multiplying

by 2. Multiply both sides of the equation by 2.3 The opposite operation to subtracting 5 is adding 5.

Add 5 to both sides of the equation.4 The opposite operation to multiplying by 3 is dividing

by 3. Divide both sides of the equation by 3.5 Check that the solution is correct by substituting x = 5 2

3

into the original equation.

3 5

26

3 5 123 173

3

17

317

3

52

3

5 5

3 5x3 5

3 5x3 53 1x3 1x

x

3 5−3 5 =

− =3 5− =3 53 1=3 1

=

=

=

+ +5 5+ +5 5

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SAMPLE


Exercise 2G 1 Solve these equations.

a x + =+ =2+ =2+ =1

211 b y − =− =− =3− =3− =1

310

2

3 c m − =− =− =4− =4− =2

51

1

3

d 3 51

63 5a3 53 5=3 5 e 6 2

2

56 2s6 26 2=6 2 f 5 2

9

105 2v5 25 2=5 2

g x

23

1

5= h

r

51

1

4= i

h

24

2

7=

2 Solve these equations.

a k

1010= b

d

47= − c

v

124=

d 3

46

d = e 7

214

a = f 2

54

z =

g 83

= k h 20

5

3= q

i 162

3= n


a e

34 10+ =4 1+ =4 1 b 6

210+ =+ =x

c d

211 13+ =11+ =11

d 42

5= −= −s e 5 2

3= +5 2= +5 2

y f 6 1

9= −6 1= −6 1

b

g 5 14

3= +5 1= +5 1

y h 7 2

5

9= +7 2= +7 2

r i 2

2

78= −= −v


a x + =3+ =3+ = 1

4 b y + =5+ =5+ = 2

3 c a − =2− =2− = 3

7

d 1

32 1( )2 1( )2 1( )e( )2 1+ =2 1( )+ =( )2 1( )2 1+ =2 1( )2 1 e

1

54 2( )4 2( )4 2( )h( )4 2− =4 2( )− =( )4 2( )4 2− =4 2( )4 2 f

1

23 1( )3 1( )3 1( )x( )3 1+ =3 1( )+ =( )3 1( )3 1+ =3 1( )3 1

g 1

26( )6 2( )6 2+ =( )+ =( )6 2( )6 2+ =6 2( )6 2( )c( )( )+ =( )c( )+ =( ) h

1

45 1( )11( )11 5 1( )5 15 1− =5 1( )− =( )5 1( )5 1− =5 1( )5 15 1−5 15 1( )5 1a5 1( )5 15 1( )5 1− =5 1( )5 1a5 1( )5 1− =5 1( )5 1 i

1

35( )1 7( )1 7− =( )− =( )1 7( )1 7− =1 7( )1 7 −( )u( )( )− =( )u( )− =( )

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SAMPLE


Development


a b − =5

89 b

t − = −4

35 c 3 5

26

3 5x3 53 5−3 5 =

d 1 2

37

1 2+1 2 =x e

4 2

56

4 2+4 2 = −w f

2 3

45

2 3+2 3 =m

g a + =4

6

1

2 h c − =5

8

1

4 i

2 1

2

4

5

e2 1e2 12 1+2 1 =

6 There are thirty-six times as many cars in Australia as trucks. Let C stand for the number of cars and T for the number of trucks.a Write an equation with C as the subject of the equation that correctly describes the

relationship between the number of cars and trucks.b A local community has 120 trucks. How many cars are in the community?


a x x

2 35+ =+ = b

y y

6 212+ =+ =y y+ =y y

c n n

5 103+ =+ =

d c c

2 47− =− = e

s s

5 62− =− = f

m m

7 94− =− =

g 3

2 37

x x− =− = h 5

3 511

a a− =− = i y y

2

2y y2y y

56− =− =

j r r

4 31= −= − k

3

10 28

w w= += + l 3

2

2

32

x x2x x2= += +


a 2 33

2y2 3y2 3

y− =2 3− =2 3+

b x + − −1

21 1=1 1=( )x( )x− −( )− −x− −x( )x− −x 1 1( )1 1 c 2 3

3

2y2 3y2 3

y− =2 3− =2 3+

d x x+x x+x x= +4x x4x x

2

10

3 e d dd d+d d=6d d6d d

3

2 4d d2 4d d +2 4+4

f 3 5

2

2 1

3

x x3 5x x3 5 2 1x x2 13 5x x3 5−3 5x x3 5 = 2 1+2 1

g y y+y y+y y+y y+y y + =1y y1y y

4

1

39 h a a+a a+a a− = −4a a4a a

3

2 3a a2 3a a −2 3−2

1 i 3 5

2

2 1

32

x x3 5x x3 5 2 1x x2 13 5x x3 5−3 5x x3 5 − 2 1+2 1 =

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SAMPLE


2.8 Using formulasA formula is a mathematical relationship between two or more variables. For example:• S D

T= is a formula for relating the speed, distance and time. S, D and T are the variables.

• P = 4L is a formula for fi nding the perimeter of a square, where P is the perimeter and L is the side length of the square. P and L are the variables.

By substituting all the known variables into a formula, we are able to fi nd the value of an unknown variable.

Using a formula

1 Write the formula.2 Replace the variables in the formula with the numbers given in the question.3 Evaluate using the calculator.4 Write the answer to the specifi ed level of accuracy and correct units if

necessary.

Example 24 Using a formula

The cost of hiring a windsurfer is given by the formula

C = 4t + 7

where C is the cost in dollars and t is the time in hours. Kayla wants to sail for 3 hours. How much will it cost her?

Solution1 Write the formula.2 Substitute the value for t into the formula.3 Evaluate. 4 Write your answer in words.

C tC t= +C t= × +=

4 7C t4 7C t= +4 7= +C t= +C t4 7C t= +C t4 3= ×4 3= × 719

It will cost Kayla $19 to hire the windsurfer for 3 hours.

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SAMPLE



The length of a pendulum with a period of oscillation T is

lT=

9802

2

πFind l, correct to two decimal places, if T is 5.

Solution1 Write the formula.

2 Substitute the 5 for T into the formula.

3 Evaluate.

4 Write answer in words correct to two decimal places.

lT=

= ××

=

9802

980= ×980= × 5

2

620 5922498

2

2

π

π.

=== 620 59.

The length of the pendulum is 620.59.


a Write an expression for the perimeter of the trapezium.b If the perimeter is 27 cm, calculate the value of x.

Solution1 Perimeter, P, is the distance around

the outside of a shape. Add all the sides to determine the perimeter.

2 Write the formula.3 Substitute 27 for P into the formula.

4 Evaluate.

a P x x x= +P x= +P x= +

( )x x( )x x+ +( )+ +x x+ +x x( )x x+ +x x( )x x( )x x − +( )− + ( )4 2x x4 2x x+ +4 2+ +x x+ +x x4 2x x+ +x x( )4 2( )+ +( )+ +4 2+ +( )+ +x x+ +x x( )x x+ +x x4 2x x+ +x x( )x x+ +x x( )4 2( )x x( )x x4 2x x( )x x 5 3− +5 3− +( )5 3( )− +( )− +5 3− +( )− + ( )5 3( )x( )x5 3x( )x +( )+5 3+( )+5 2x5 2x= +5 2= +x= +x5 2x= +x

b P x

xx

P x= +P x= +==

5 2P x5 2P x= +5 2= +P x= +P x5 2P x= +P x27 5 2x5 2x= +5 2= +x= +x5 2x= +x25 5

5cm

2x − 5

x + 4x + 3

x

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SAMPLE


Using a formula and a graphics calculator Graphics calculators have an equation mode that may be used to enter a formula. The coloured letters on the keypad are used as the variable.

After the formula is entered, set the unknown variable to zero and enter the values for the known variables. It is not necessary for the known variable to be the subject of the formula.

Select the ‘solver’ key to obtain the answer. The calculator will show the answer and the values of the left-hand side and right-hand side of the equation.

Example 27 Using a formula and a graphics calculator

The circumference, C, of a circle with radius, r, is given by the formula C = 2πr. Find the circumference of a circle with a radius of 5 cm using a graphics calculator.

Solution1 Select the EQUA menu.2 Select SOLVER (F3).3 Enter the formula. To place the variables (C and r)

use the ALPHA key to access the letters in red above the keys.

4 Enter the known variable. The radius is 5 so R = 5.5 Highlight the required variable or C = 0.

6 Press SOLV and the calculator will show the answer. 7 The values of the left (Lft) and right (Rgt) sides of


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SAMPLE


Exercise 2H 1 If A = lb find the value of A when:

a l = 16, b = 2b l = 28, b = 7c l = 30, b = 10

2 If A = s2 find the value of A when:a s = 9b s = 6c s = 32

3 If A bA bhA b=A b1A b1A b2

A b2

A b find the value of A when:

a b = 10, h = 4

b b = 15, h = 2

c b = 2, h = 3

4 If P = 2l + 2b find the value of P when:a l = 8, b = 10b l = 2, b = 11c l = 3, b = 9

5 Find the value of T (correct to one decimal place) in the formula TMv

r=

2

if:

a M = 1.7, v = 4 and r = 3.8

b M = 2.1, v = 1 and r = 2.2

6 Use the formula ab= + −

9= +9= + 18

2 to find the value of a when:

a b = 8 b b = 1 c b = -2

7 If px

x=

+12

4 find the value of p when:

a x = 7 b x = 5 c x = -3

8 The cost of hiring a hall is given by the rule C = 30t + 1000 where C is the total cost in dollars and t is the number of hours for which the hall is hired. Find the cost of hiring the hall for:a 2 hours b 5.5 hours

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SAMPLE


9 The distance, d km, travelled by a truck in t hours at an average speed of s km/h is given by the formula d = st. Find the distance travelled by a truck travelling at a speed of 70 km/h for 5 hours.

10 Given that HE

T= find the value of H when:

a H = 2.6 × 10-11 and T = 100 b E = 7.8 × 10-6 and T = 20

11 The formula used to convert temperature from degrees Fahrenheit to degrees Celsius is C FC F= −C FC F= −C F5C F5C F

9C F

9C F( )C F( )C F= −( )= −C F= −C F( )C F= −C F 32( )32 . Use this formula to convert the following temperatures to degrees

Celsius. Answer correct to the nearest whole number.a 40°F b 110°F

12 The formula v = u + at relates the velocity, acceleration and time. a Make u the subject of the formula.b Make a the subject of the formula.c Make t the subject of the formula.

13 The circumference, C, of a circle with radius, r, is given by the formula C = 2πr. a Make r the subject of the formula.b Find the radii of circles with the following circumferences. (Answer correct to two

decimal places.) i 3 cm ii 6.9 mm

14 The body mass index is Bm

h=

2 where m is the mass in kg and h is the height in m.

a Make m the subject of the formula.b Find the body mass index to the nearest whole number when: i B = 22.78 and h = 1.79 m ii B = 31.8 and h = 1.86 m

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SAMPLE


Development

15 Find the value of v (correct to one decimal place) in the formula v uv u as= += +v u= +v uv u= +v u2= +2= + 2 if: a u = 4, a = 7 and s = 12b u = -2, a = 10 and s = 5

16 Use the formula RV= 3

43

π to find the value of R (correct to two decimal places) when:

a V = 12 b V = 44 c V = 100

17 Find the value of s (correct to one decimal place) in the formula s ut at at= +s u= +s ut a= +t a1t a1t a2

t a2

t a 2 if:

a u = -5, a = 4 and t = 6 b u = 2, a = 5 and t = 15

18 The formula for calculating simple interest is IPRT=100

where P is the principal, R is the

interest rate per annum and T is the time in years. Calculate the interest earned, correct to the nearest cent, on the following investments.a $10 000 at an interest rate of 9% p.a. for 3 yearsb $88 000 at an interest rate of 11.2% p.a. for 2 yearsc $24 000 at an interest rate of 7¾% p.a. for 2 years

19 The volume of a cone is evaluated using V rV r hV r=V r1V r1V r3

V r3

V r2πV rπV r where h is height and r is radius.a Write the formula with h as the subject.b Calculate the height of a cone, correct to two decimal places, if the volume of the

cone is 18 cm3 and the radius is 2 cm.

20 The volume of a sphere is given by the formula V rV rV r=V r4V r4V r3

V r3

V r3πV rπV r where r is the radius.a Write the formula with r as the subject.b What is the radius in metres of a spherical balloon with a volume of 2 m3? Answer

correct to one decimal place.

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SAMPLE

69Chapter 2 — Algebraic manipulationReview

Adding and subtracting like terms 1 Find the like terms.2 Only add or subtract like terms.3 Add or subtract the coefficients of the like terms.

Multiplication and division of

algebraic terms

1 Write in expanded form and cancel any factors in a fraction.

2 Multiply and divide the coefficients and pronumerals. 3 Write the pronumerals in alphabetical order and express

in index notation.

Expanding algebraic expressions 1 Multiply the term outside the grouping symbol by thea First term inside the grouping symbol.b Second term inside the grouping symbol.

2 Simplify and collect like terms if required.

Factorising algebraic expressions 1 Find the largest factor of each term or the HCF.2 Write the HCF outside the grouping symbol.3 Divide the HCF into each term to find the terms inside

the grouping symbols.4 Check the factorisation by expanding the expression.

Linear equations 1 Perform the opposite operation (+ and -, × and ÷).2 Add/subtract the same number to both sides of the equation.3 Multiply/divide both sides of the equation by the same

number.

Equations with fractions • Use the same steps as linear equations.

Substitution 1 Write the algebraic expression.2 Substitute the values and evaluate.

Using formulas 1 Write the formula. 2 Substitute the values and evaluate.

Chapter summary – Algebraic manipulation PowerPoint 1

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SAMPLE

70 Preliminary Mathematics GeneralRe

view 1 Simplify 7xy - 5xy - 4xy.

A -16xy B -2xy C 6xy D 16xy

2 Simplify +x x5

8

2

8.

A x7

8 B

x7

16 C

7

8

2x D

7

16

2x

3 Multiply and simplify the expression 4a × (8a) × 3.A -96a2 B -32a2 + 3 C -4a + 3 D 96a2

4 Simplify 6

53

cc× .

A 9

5

c B 9

5

2c C

18

5

c D

18

5

2c

5 Expand -5(2d + 3e).

A -10d - 15e B -10d + 15e C -10d - 3e D -10d + 3e

6 Factorise 16w2 + 24w by taking out the highest common factor.

A 4(4w2 + 6w) B 4w(4w + 6) C 8w(2w + 3) D 16w(w + 8)

7 Solve the equation 11b - 9 = 26.

A b = 11

35 B b = 11

15 C b = 1

4

11 D b = 3

2

11

8 Find the value of a b2 2a b2 2a ba b+a ba b2 2a b+a b2 2a b if a = 4 and b = 3.

A 7 B 14 C 5 D 25

9 What is the base of a triangle using the formula A bA bhA b=A b1A b1A b2

A b2

A b if A is 14 and h is 4?

A 1.75 B 7 C 28 D 112

Sample HSC – Objective-response questions

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SAMPLE

71Chapter 2 — Algebraic manipulationReview

1 Simplify these expressions.a 11st + 5 - 4st - 4 b d + 4d2 – 2d –3d2

2 Simplify -3s + 5r - 3s + 3r by collecting the like terms.

3 Multiply and simplify:

a 7

25

xx× b

2

12

3

8

3

2

w

y

w

y×

4 Add or subtract the following algebraic fractions.

a c c

6 6+ b 5

3

2

3

y y2y y2− c m m

8

2m m2m m

3+

5 Remove grouping symbols and simplify if required.

a 7(x - 1) b 5(2 + 2r) c 7(5x - 1)

d 3(4a - b) e 7(d + 7e) f -2(8v - 2s)

g -(9b - 2h) h 4(w - 4) + 2(w + 2) i y(y + 2z)

6 Fully factorise these expressions.

a 7b + 35 b 2v - 14 c -3v + 15

d 6y + 9 e 4x - 14y f 12x + 21y

g 21 - 15x h 24b - 16c + 4 i 12s - 15v + 9

7 Solve the following linear equations.

a d - 4 = 7 b 4h = 20 c 5 + r = 8

d 2t + 3 = 11 e 5 = 2x - 5 f 7n + 5 = 33

g r

612= h v + =+ =+ =3+ =3+ =2

31

1

6 i 4 2

1

5m4 2m4 24 2=4 2

Sample HSC – Short-answer questions

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SAMPLE

72 Preliminary Mathematics GeneralRe

view 8 Solve the following linear equations.

a 6x + 4 = 5x - 7 b 3 - 4r = 5r + 12 c 7(n - 2) = n + 4d 8(b - 1) = 2(b + 2) e y - 12 = 2(y - 7) f 6(2v - 7) = -(v + 1)


a u + =3

54 b

z z

4 812+ =+ = c

1

54 2( )4 2( )4 2( )x( )4 2+ =4 2( )+ =( )4 2( )4 2+ =4 2( )4 2

10 The distance d in km that a person can see the horizon from h in metres above the sea level, is given by the formula

dh= ×5= ×5= ×2

.

a Find d when h is 18 metres.b Find h when d is 10 kilometres.

11 Use the formula RV= 2

33

π to fi nd the value of R (correct to two decimal places) when:

a V = 9b V = 24c V = 200

12 Einstein’s equation E = mc2 states that the energy E in joules equals the mass of m kg multiplied by the square of the speed of light c (3 × 108 m/s). Find the amount of energy produced by a:a mass of 500 kgb mass of 200 kg

13 The cost of hiring a hall is given by the formula C = 20t + 2000 where C is the total cost in dollars and t is the number of hours for which the hall is hired. a Make t the subject of the equation.b Find t when C is $2060.

9781107627291c02_p35-72.indd 72 8/29/12 9:27 PM


SAMPLE

algebraic manipulation sample - jschs...

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