Chap 06

Shannon is saving to buy a new computer, which costs $3299. So far he has $449 in the bank and wants to make regular deposits each month until he reaches his target of $3299. If he wants to buy the computer in 8 months’ time, how much does he need to save as a monthly deposit?

One way of solving this problem

is to write and solve an equation for this situation with a pronumeral representing the unknown. This chapter looks at solving different equations and how this skill can be used to solve real-life problems.

6

55eTHINKING

Equations

236 M a t h s Q u e s t 8 f o r V i c t o r i a

READY?areyou

Are you ready?Try the questions below. If you have difficulty with any of them, extra help can be

obtained by completing the matching SkillSHEET. Either click on the SkillSHEET icon

next to the question on the Maths Quest 8 CD-ROM or ask your teacher for a copy.

Flow charts

1 Complete these flow charts to find the output number.

a b

Inverse operations

2 Write the inverse operation for each of the following.

a × 2 b + 8 c − 17 d ÷ −5

Solving equations by backtracking

3 Solve each of the following equations by first completing the flow charts below. Remember to

show the operations needed to backtrack to x.

a 7(x − 4) = 35 b = 3

Combining like terms

4 Simplify each of the following expressions by combining like terms.

a 7v + 3 + 3v + 4 b 6c + 15 − 5c − 8

Expanding expressions containing brackets

5 Expand each of the following expressions containing brackets.

a 2(3x + 5) b −7(m − 1)

Checking solutions by substitution

6 For each equation below there is a solution given. Is the solution correct?

a 5x − 7 = 2x + 2 x = 3 b = 2x − 7 x = 5

Writing equations from worded statements

7 Write an equation for each of the following statements, using x to represent the unknown number.

a When 2 is added to a certain number, the result is 9.

b Eight times a certain number is 40.

c When 11 is subtracted from a certain number, the result is 3.

d Dividing a certain number by 6 gives a result of 2.

Showing inequations on a number line

8 Indicate each of the following on a number line.

a x ≥ 2 b x < 5

c x ≤ 0 d x > −1

6.1

–9×5

4

÷4+3

5

6.2

6.4

x 9+

5------------

x

35

x – 4 7(x – 4) x x + 9 x + 9

3

5

6.5

6.6

6.7

x 9+

2------------

6.8

6.10

C h a p t e r 6 E q u a t i o n s 237

Backtracking — inverse operationsRecall from Year 7 that an equation links two expressions with an equals sign (=).

Equations are very useful for reducing complex problems to simple terms and can be

used to solve many different problems. For example, how much food and water should

x people take on a camping trip that will last for n days?

What we have to remember with equations is to

always keep one side of the equals sign equal to

the other side. A useful skill you learned in Year 7

was to draw flow charts like this one.

The number 7 is called the input number, and by

carrying out the operations in order we find that

the output number is 5.

If we are given the output number, we can find the

input number by working backwards or back-

tracking. For example, look at the flow chart at

right.

Starting with the 3, we can use inverse operations to find the input number.

Remember: Inverse operations undo each other or cancel each other out.

Adding and subtracting are inverse operations.

Multiplying and dividing are inverse operations.

Working backwards,

3 × 7 = 21

21 + 1 = 22

22 ÷ 2 = 11

11 − 5 = 6

7

× 2 – 1 + 2 ÷ 3

7 14 13 15 5

× 2 – 1 + 2 ÷ 3

3

+ 5 × 2 – 1 ÷ 7


So the input number for the third flow chart

on the previous page is 6.

Flow charts can also be followed when

the input number is a pronumeral. Here is an

example.

The first step is easy, add 7.

We must now remember to use a pair of

brackets to show that all of x + 7 is to be

multiplied by 2.

The third step is also easy, subtract 3.

As you can see, the output number is written as an expression.

A line for division must be carefully drawn beneath all the terms that are divided.

For example, 3x + 2y − 3 divided by 7 is written as .

32122116

+ 5 × 2 – 1 ÷ 7

– 5 ÷ 2 + 1 × 7

+ 7 × 2 – 3

x

+ 7 × 2 – 3

x x + 7

+ 7 × 2 – 3

x x + 7 2(x + 7)

+ 7 × 2 – 3

x x + 7 2(x + 7) 2(x + 7) – 3

3x 2 y 3–+

7----------------------------

Find the input number for this flow chart.

THINK WRITE

Copy the flow chart.

Backtrack to find the input number.

The inverse operation of +3 is −3

(7 − 3 = 4).

The inverse operation of ÷ −2 is × −2

(4 × −2 = −8).

The inverse operation of −7 is +7

(−8 + 7 = −1).

Fill in the missing numbers.

1 – 7 ÷ –2 + 3

7

2 – 7 ÷ –2 + 3

+ 7 × –2 – 3

7–1 –8 4

1WORKEDExample

– 7 ÷ –2 + 3

7


Find the output expression for this flow chart.

THINK WRITE

Copy the flow chart and look at the

operations that have been performed.

Multiplying x by 3 gives 3x.

Adding 2 gives 3x + 2.

Now place a line beneath all of 3x + 2

and divide by 4.

1 × 3 + 2 ÷ 4

x

2 × 3 + 2 ÷ 4

x 3x

3 × 3 + 2 ÷ 4

x 3x 3x + 2

4 × 3 + 2 ÷ 4

4x 3x 3x + 2

3x + 2

2WORKEDExample

× 3 + 2 ÷ 4

x

Starting with x, draw the flow chart whose output number is given by the expressions:

a 6 − 2x b −2(x + 6).

THINK WRITE

a Rearrange the expression.

Note: 6 – 2x is the same as –2x + 6.

Multiply x by –2, and then add 6.

a

b The expression x + 6 is grouped in a

pair of brackets, so we must obtain

this part first. Therefore, add 6 to x.

Multiply the whole expression

by –2.

b

1

2

× –2 + 6

x –2x –2x + 6

1

2

+ 6 × – 2

x x + 6 –2(x + 6)

3WORKEDExample

1. To work backwards through a flow chart we use inverse operations.

2. Adding and subtracting are inverse operations.

3. Multiplying and dividing are inverse operations.

remember


Backtracking — inverse operations

1 Find the output number for each of the following flow charts.

a b

c d

e f

g h

i j

2 Find the input number for each of the following flow charts.

a b

c d

e f

g h

i j

k l

6A

SkillSH

EET 6.1

Flow charts– 1 × –5 + 6

3

+ 8 × 3 ÷ 2

–6

– 7 ÷ 2 + 8

5

÷ 3 + 4 – 12

12

× –6 + 8 × –3

0

÷ 10 – 5 × –1

–10

÷ –1 + 8 × –3

–7

– 5 × 12 + 10

4

÷ 8 – 3 × –2

–48

× –8÷ –4 – 2

32

SkillSH

EET 6.2

Inverse operations

WORKED

Example

1

EXCE

L Spreadsheet

Backtracking

+ 6 × 2

28

÷ 5 + 3

7

× –3 + 2

14

– 5 ÷ –4

6

× –2 – 6 × –3

12

÷ –8

–1

+ 5 × 2

+ 11 ÷ –3 – 2

–5

× –3

12

÷ 4 + 7

– 8 ÷ 6 × –5

0

– 5

–11

– 7 × 2

+ 0.5 × 4 – 5.1

1.2

× 5

4

– 2 ÷ 3


3 Find the output expression for each of the following flow charts.

a b

c d

e f

g h

i j

k l

m n

o p

4 Starting with x, draw the flow chart whose output number is:

a 2(x + 7) b −2(x − 8) c 3m − 6 d −3m − 6

e f g −5x + 11 h −x + 11

i −x − 13 j 5 − 2x k l

m n o 3 p

Keeping equations balancedAn equation, as defined earlier, links two expressions with an equals sign. Let’s look at

a very simple equation: x = 3.

This equation can be pictured as a balanced pair of scales

or a seesaw.

WORKED

Example

2

SkillSHEET

6.3

Buildingexpressions

× 2 – 7

x

× 2– 7

w

× –5 + 3

s

× –5+ 3

n

÷ 2 + 7

m

÷ 2+ 7

y

× 6 – 3 ÷ 2

z

+ 5 × –3 ÷ 4

d

× 2 ÷ 5 + 1

e

× –1 + 3 × 4

x

× –2– 5 ÷ 7

w

+ 6 × –3 – 11

z

– 8– 3 ÷ 6

v

– 4× 8 × –7

m

× –5 + 2÷ 6

k

× –5 – 7 ÷ 3

p

WORKED

Example

3

x 5–

8-----------

x

8--- 5–

3x 7–

4---------------

3– x 2–( )4

-----------------------

x 5+8

------------ 3– 7–x

5--- 2–

2x

7------ 4+

1

4---

6x

11------ 3–

x 1 1 1

x = 3


If we double the amount on the left-hand side

(LHS), the scales will balance as long as we

double the amount on the right-hand side (RHS).

Or we can add some mass to both sides:

The scales will remain balanced as long as we do

the same to both sides. By doing the same to both

sides, we can build up very complex equations or

we can solve them by working backwards.

xx1 1 1

1 1 1

2x = 6

xx1 1 1

11

1

1 1

1

1

2x + 2 = 8

Starting with the equation x = 4, write the new equation when we:

a multiply both sides by 4

b take 6 from both sides

c divide both sides by .

THINK WRITE

a Write the equation. a x = 4

Multiply both sides by 4. x × 4 = 4 × 4

Simplify by removing the

multiplication signs. Write numbers

before pronumerals.

4x = 16

b Write the equation. b x = 4

Subtract 6 from both sides. x − 6 = 4 − 6

Simplify. x − 6 = −2

c Write the equation. c x = 4

Dividing by a fraction is the same as

multiplying by its reciprocal.

Multiply both sides by .

x ÷ = 4 ÷

=

Simplify. =

= 10

2

5---

1

2

3

1

2

3

1

2

5

2---

2

5---

2

5---

x5

2---× 4

5

2---×

35x

2------

20

2------

5x

2------

4WORKEDExample

1. An equation links two expressions with an equals sign.

2. An equation is like a pair of balanced scales (or a seesaw). The scales (or

seesaw) will remain balanced as long as we do the same to both sides.

remember


Keeping equations balanced

1 a Write the equation that is represented by the

diagram at right.

b Show what happens when you halve the

amount on both sides. Write the new equation.


diagram at right.

b Show what happens when you take three from

both sides. Write the new equation.


diagram at right.

b Show what happens when you add three to

both sides. Write the new equation.


diagram at right.

b Show what happens when you double the

amount on each side. Write the new equation.

5 Starting with the equation x = 6, write the new equation when we:

a add 5 to both sides b multiply both sides by 7

c take 4 from both sides d divide both sides by 3

e multiply both sides by −4 f multiply both sides by −1

g divide both sides by −1 h take 9 from both sides

i multiply both sides by j divide both sides by

k take from both sides.

6

If we start with x = 5, which of these equations is not true?

A x + 2 = 7 B 3x = 8 C −2x = −10 D E x − 2 = 3

6B

xx 1 1 1 1

x

1 1 1

11

1

11

1 1 1

1 1xx 1

xxx1 1 1

11

1

1

1

WORKED

Example

4

2

3---

2

3---

2

3---

multiple choice

x

5--- 1=


7


8

If we start with x = −6, which of these equations is not true?

9

If we start with 2x = 12, which of these equations is not true?

1 Find the output number for the flow chart at right.

2 By backtracking, find the input number for the

flow chart at right.

3 Find the output expression for the flow chart at right.

4 Draw a flow chart whose output expression is given by .

5 Write the output expression for the flow chart

at right.

6 Write the equation that is represented in the

diagram at right.

Use the following diagram for questions 7 and 8.

7 Write the equation that is represented by the diagram.

8 Write the new equation when you add 3 to both sides.

9 Starting with the equation y = 7, write down the new equation when you multiply both

sides by 5.

10 True or false? If you start with y = −6 and then divide both sides by −1, the new

equation is −y = 6.

A B −2x = −6 C 2x − 6 = 0 D E x − 5 = 2

A −x = 6 B 2x = −12 C x − 6 = 0 D x + 4 = −2 E x − 2 = −8

A B −2x = −12 C 2x − 6 = 2 D 4x = 24 E 2x + 5 = 17

multiple choice

2x

3------ 2=

x

5---

3

5---=

multiple choice

GAM

E time

Equations— 001

WorkS

HEET 6.1

multiple choice

2x

3------ 4=

1

÷ 10 + 2 – 4

– 20

+ 2 ÷ 3 x – 2

– 6

÷ 6 x –1 + 2

m

2z 5+

7---------------

– 9 ÷ 7 – 5

7p p – 9

p – 9

x x x 1 11 1

1 11 1

x 1 111 11 1

1 1


History of mathematicsDIOPHANTUS OF ALEXANDRIA (c . 200–c. 284)

During his life . . .

Gunpowder is

invented in

China.

The Roman

Empire executes

many Christians

to stop the spread

of Christianity.

Like many famous people who lived long

ago, there are few facts known about

Diophantus or his achievements. Even the

dates of his birth and death are approximate.

It is thought that he married at 33, had a son

when he was 42, and died at 84. However,

these ‘facts’ are only known to us through

this riddle that was written about him.

‘. . . his boyhood lasted th of his life; he

married after th more; his beard grew after

th more, and his son was born 5 years later;

the son lived to half his father’s age, and the

father died 4 years after the son.’

Diophantus is considered the creator of

algebra. He published works on algebra and

number systems including a collection of

books known as the Arithmetica. Only six of

the original thirteen volumes of the

Arithmetica still exist, mostly as Arabic

translations. They are mainly about solving

determinate equations numerically.

Diophantus worked mostly on equations

that had positive and rational answers.

Equations without this sort of answer were of

no interest to him because he considered that

their answers were not ‘real’.

As Diophantus had no understanding or

knowledge of the idea of zero, he was

restricted in how he could solve equations.

He partially got around this by examining

them in the formats ax + bx = c, ax = bx + c

and ax + c = bx. This approach meant that he

did not have to use negative values.

Diophantus is credited with introducing a

form of algebraic symbolism. He introduced

symbols for subtraction and for powers of the

unknown.

The letters of the Greek alphabet were also

used as digits. A line over the top of the

letters distinguished them as numbers. For

example, to represent ‘take a number,

multiply it by itself and then multiply this

result by 3 before adding 15’, Diophantus

would write . Nowadays, we would

write this as 3x2 + 15. With the introduction

of symbols, mathematics became more of a

language, allowing problems to be written in

a shorter form.

Work done by Diophantus in the third

century inspired the seventeenth century

French mathematician Pierre de Fermat to

further investigate number theory. The

symbol used today as the equal sign (=) can

be traced back to Diophantus.

Questions

1. What multi-volume work did

Diophantus publish?

2. How did Diophantus solve equations

without knowing about zero?

3. Which seventeenth century

mathematician was inspired by his

work?

Research

Find out more about the symbols used by

Diophantus to write problems in shorter

algebraic form.

LIBYA

A

UNITEDARABREPUBLIC (EGYPT)

Y

TURKEYGREECE

CRETE

I

SYRIA

Red S

ea

Mediterranean Sea

Alexandria

Cairo

Istanbul

E

1

6---

1

7---

1

12------

∆Y

ς

∆Y∆

KY

KYK

∆KY

M unity (x°)

subtraction symbol

number (x)

square (x2)

cube (x3)

square-square (x4)

square-cube (x5)

cube-cube (x6)

°

∆Y–γ

–ι εM

°


Using algebra to solve equationsA linear equation is an equation in which the variable has an index (power) of 1.

For example, it never contains terms like x2 or .

Let us take another look at a simple linear equation and see

what happens when we change both sides.

Remember: To keep an equation true, balance it by doing

the same to both sides.

x = 7

Multiply both sides by 3. 3x = 21

Subtract 1 from both sides. 3x − 1 = 20

We can show this process on a flow chart.

The flow chart at right shows us that

to backtrack from 3x − 1 we must add

1 and then divide by 3. Let us solve the

equation 3x − 1 = 20 by backtracking

and doing the same to both sides.

3x − 1 = 20

Add 1 to both sides. 3x − 1 + 1 = 20 + 1

3x = 21

Divide both sides by 3. =

x = 7

To solve a linear equation, perform the same operations

on both sides of the equation until the pronumeral is left by itself.

A flow chart is useful to show you the order of operations applied to x, so that the

reverse order and inverse operations can be used to solve the equation. Once you

become confident with solving equations algebraically (doing the same to both sides),

you can leave out the flow chart steps.

x

3x

3------

21

3------

Solve these one-step equations by doing the same to both sides.

a p − 5 = 11 b = -2

THINK WRITE

a Write the equation. a p − 5 = 11

Draw a flow chart and fill in the arrow

to show what has been done to p.

Backtrack from 11.

x

16------

1

2

p

–5

p –5

3p

–5

16 11

p –5

+ 5

5WORKEDExample

× 3 – 1

÷ 3 + 1

7

x

21 20

3x 3x – 1


The equations in worked example 5 are called one-step equations because we need to

perform one operation to obtain the value of the pronumeral.

THINK WRITE

Add 5 to both sides. p − 5 + 5 = 11 + 5

Give the solution. p = 16

b Write the equation. b = –2

Draw a flow chart and fill in the arrow

to show what has been done to x.

Backtrack from −2.

Multiply both sides by 16. × 16 = −2 × 16

Give the solution. x = −32

4

5

1x

16------

2

x

÷ 16

x

16

3

x

÷ 16

16

x

16

–32 –2

4x

16------

5

Solve these two-step equations by doing the same to both sides.

a 2(x + 5) = 18 b

THINK WRITE

a Write the equation. a 2(x + 5) = 18

Draw a flow chart and fill in the arrows to

show what has been done to x.

Backtrack from 18.

Continued over page

x

3--- 1+ 7=

1

2

× 2+ 5

x x + 5 2(x + 5)

3

× 2

4 9 18

+ 5

– 5 ÷ 2

x x + 5 2(x + 5)

6WORKEDExample


The equations in worked example 6 are called two-step equations because we need to

perform two operations to obtain the value of x.

THINK WRITE


x + 5 = 9

Subtract 5 from both sides. x + 5 − 5 = 9 − 5

x = 4Give the solution.

b Write the equation. b



Backtrack from 7.

Subtract 1 from both sides. + 1 − 1 = 7 − 1

= 6

Multiply both sides by 3. × 3 = 6 × 3

Give the solution. x = 18

42 x 5+( )

2--------------------

18

2------

5

6

1x

3--- 1+ 7=

2

÷ 3 + 1

+ 1x3

x–

3

x–

3

× 3

÷ 3 + 1

+ 1

– 1

18

x

6 7

3

x–

3

x–

4x

3---

x

3---

5x

3---

6

Solve the following equations by doing the same to both sides.

a3(m − 4) + 8 = 5 b 6 = –18

THINK WRITE

a Write the equation. a 3(m − 4) + 8 = 5


show what has been done to m.

x

2--- 5+

1

2

m m – 4 3(m –4) 3(m – 4)+8

– 4 + 8

7WORKEDExample


The equations in worked example 7 are called three-step equations. Why?

THINK WRITE

Backtrack from 5.

Subtract 8 from both sides. 3(m − 4) + 8 − 8 = 5 − 8

3(m − 4) = −3


m − 4 = −1

Add 4 to both sides. m − 4 + 4 = −1 + 4

Give the solution. m = 3

b Write the equation. b 6 = –18



Backtrack from −18.


+ 5 = −3

Subtract 5 from both sides. + 5 − 5 = −3 − 5

= −8

Multiply both sides by 2. × 2 = −8 × 2

Give the solution. x = −16

3

m m – 4 3(m – 4) 3(m – 4)+8

– 4 + 8

+ 4 ÷3 – 8

53 – 1 – 3

4

53 m 4–( )

3---------------------

3–

3------

6

7

1x

2--- 5+

2

x + 5 6( + 5)

÷ 2 + 5

x

2x

2x

2

3

x + 5 6( + 5)

2 + 5

– 5 ÷ 6

–18–16 – 8 –3

x

2

x

2

x

2

÷

4

6x

2--- 5+

6---------------------

18–

6---------

x

2---

5x

2---

x

2---

6x

2---

7

1. A linear equation is an equation where the variable has an index (power) of 1.

2. When solving linear equations, perform the same operations on both sides of

the equation until the pronumeral is left by itself.

3. You can draw a flow chart to help you to decide what to do next.

remember


Using algebra to solve equations

1 Solve these one-step equations by doing the same to both sides.

a x + 8 = 7 b 12 + r = 7 c 31 = t + 7

d w + 4.2 = 6.9 e = m + f = j + 3

g q − 8 = 11 h −16 + r = −7 i 21 = t − 11

j y − 5.7 = 8.8 k – = z − l − = f − 1

2 Solve these one-step equations by doing the same to both sides.

a 11d = 88 b 7p = −98 c 5u = 4

d 2.5g = 12.5 e 8m = f – = 9j

g = 3 h = −12 i −5.3 =

j = k = − l − =

3 Solve these two-step equations by doing the same to both sides.

a 3m + 5 = 14 b −2w + 6 = 16 c −5k − 12 = 8

d 4t − 3 = −15 e 2(m − 4) = −6 f −3(n + 12) = 18

g 5(k + 6) = −15 h −6(s + 11) = −24 i 2m + 3 = 10

j 40 = −5( p + 6) k 5 − 3g = 14 l 11 − 4f = –9

m 2q − 4.9 = 13.2 n 7.6 + 5r = −8.4 o 13.6 = 4t − 0.8

p −6k + 7.3 = 8.5 q −4g − = r − = 2f −

4 Below is Alex’s working to solve

the equation 2x + 3 = 14.

2x + 3 = 14

+ 3 =

x + 3 = 7

x + 3 − 3 = 7 − 3

x = 4

a Is the solution correct?

b If not, can you find where the

error is and correct it?

5 Solve these two-step equations by doing the same to both sides.

a b c

d e f

g h i

j k = –4.5 l – 3.2 = –5.8

6C

SkillSH

EET 6.4

Solvingequations bybacktracking

WORKED

Example

5a5

8---

1

8---

2

7---

11

7------

2

3---

9

13------

WORKED

Example

5b

Mat

hcad

Doing thesame toboth sides 1

4---

3

5---

Mat

hcad

Solvingequations

t

8---

k

5---

l

4---

v

6--- 2

3---

c

9--- 5

27------

7

12------

h

5---

WORKED

Example

6a

EXCE

L Spreadsheet

2-stepequations

1

5---

4

5---

3

8---

18

8------

2x

2------

14

2------

GC pr

ogram– TI

Solvingequations

GC pr

ogram– Casio

Solvingequations

WORKED

Example

6b x

3--- 2+ 9=

x 5–

4----------- 1=

m 3+

2------------- 7–=

h

3–------ 1+ 5=

m–

5------- 3– 1=

2w

5------- 4–=

3m–

7---------- 1–=

c 7–

3----------- 2–=

5m–

4---------- 10=

t 2+

7----------- 5–=

c 21–

9--------------

x

8---


6 Solve these equations by doing the same to both sides. They will need more than

2 steps.

a 2(m + 3) + 7 = 3 b c

d e f

g – 2 = 1 h – 3 = 5 i + 2 = –5

j 6 – = –2 k 8 – = 2 l –9 – + –4

m n −7(5w + 3) = 35 o

p q r

7 Simplify the left-hand side of the following equations by collecting like terms, and

then solve.

a 3x + 5 + 2x + 4 = 19 b 13v − 4v + 2v = −22

c −3m + 6 − 5m + 1 = 15 d −3y + 7 + 4y − 2 = 9

e −3y − 7y + 4 = 64 f 5t + 4 − 8t = 19

g 5w + 3w − 7 + w = 13 h w + 7 + w − 15 + w + 1 = −5

i 7 − 3u + 4 + 2u = 15 j 7c − 4 − 11 + 3c − 7c + 5 = 8

8 A repair person calculates his service fee using the equation F = 40t + 55, where F is

the service fee in dollars and t is the number of hours spent on the job.

a How long did a particular job take if the service fee was $155?

b Explain what the numbers 40 and 55 could represent as costs in the service fee

equation.

9 Lyn and Peta together raised $517 from their cake stalls at the school fete. If Lyn

raised l dollars and Peta raised $286, write an equation that represents the situation

and determine the amount Lyn raised.

10 a Write an equation that represents the perimeter of

the figure at right and then solve for x.

b Write an equation that represents the perimeter of

the figure at right and then solve for x.

EXCEL Spreadsheet

3-stepequations

WORKED

Example

7

2 x 5+( )–

5----------------------- 6= 5m 6+

3----------------- 4=

3x 2–

7--------------- 1= 4 2x–

3--------------- 6= x– 3+

2--------------- 4–=

3x

7------

4b

5------

7 f

9------

4z

3-----

6m

5-------

5u

11------

3m 5–

2–---------------- 7= 5

x

2--- 6–

10–=

d 7–

2------------ 10+ 8= 3n 1+

4--------------- 5– 2= 3 t 5–( )

7------------------- 9+ 6=

SkillSHEET

6.5

Combininglike terms

2x + 19

5x + 18

3x + 17

Perimeter = 184 cm

3x – 21

2x + 16

2x + 30

4x – 13

Perimeter = 287 cm


Equations with the pronumeral on both sides

Some equations have the pronumeral on both sides of

the equation. Take, for example, the equation:

4x + 1 = 2x + 5.

If we draw this equation as a pair of scales, it looks

like this:

The scales will remain balanced as long as we do

the same to both sides. Let’s remove 2x from each side

and look again at the diagram.

We can easily see that 2x + 1 = 5.

Starting from 4x + 1 = 2x + 5

we have taken 2x from both sides.

4x + 1 − 2x = 2x + 5 − 2x

2x + 1 = 5

If we continue:

2x = 4

x = 2

Equations like 4x + 1 = 2x + 5 are easy to solve if we first remove the x term from

one side.

4x + 1 = 2x + 5

xx

xx1

1 1 1

1 1x

x

2x + 1 = 5

1x

x 1 1 1

1 1

Solve the equation 5t − 8 = 3t + 12 and check your solution.

THINK WRITE

Write the equation. 5t − 8 = 3t + 12

Subtract the smaller pronumeral (that

is, 3t) from both sides and simplify.

5t − 8 − 3t = 3t + 12 − 3t

2t − 8 = 12

Add 8 to both sides and simplify. 2t − 8 + 8 = 12 + 8

2t = 20

Divide both sides by 2 and simplify. =

t = 10

Check the solution by substituting

t = 10 into the left-hand side and then

the right-hand side of the equation.

If t = 10,

LHS = 5t – 8

= 50 − 8

= 42

If t = 10,

RHS = 3t + 12

= 30 + 12

= 42

Comment on the answers obtained. Since the LHS and RHS are equal, the equation

is true when t = 10.

1

2

3

42t

2-----

20

2------

5

6

8WORKEDExample


Solve the equation 3n + 11 = 6 − 2n and check your solution.

THINK WRITE

Write the equation. 3n + 11 = 6 − 2n

The inverse of − 2n is + 2n.

Therefore, add 2n to both sides and

simplify.

3n + 11 + 2n = 6 − 2n + 2n

5n + 11 = 6

Subtract 11 from both sides and

simplify.

5n + 11 − 11 = 6 − 11

5n = −5


n = −1


n = −1 into the left-hand side and then

the right-hand side of the equation.

If n = −1,

LHS = 3n + 11

= −3 + 11

= 8

If n = −1,

RHS = 6 – 2n

= 6 − −2

= 6 + 2

= 8

Comment on the answers obtained. Since the LHS and RHS are equal, the equation

is true when n = –1.

1

2

3

45n

5------

5–

5------

5

6

9WORKEDExample

Expand the brackets and then solve the following equation, checking your solution.

a 3(s + 2) = 2(s + 7) + 4 b 4(d + 3) – 2(d + 7) + 4 = 5(d + 2) + 7

THINK WRITE

a Write the equation. a 3(s + 2) = 2(s + 7) + 4

Expand the brackets on each side of

the equation first and then simplify.

3s + 6 = 2s + 14 + 4

3s + 6 = 2s + 18

Subtract the smaller pronumeral

(that is, 2s) from both sides and

simplify.

3s + 6 − 2s = 2s + 18 − 2s

s + 6 = 18


simplify.

s + 6 − 6 = 18 − 6

s = 12


s = 12 into the left-hand side and

then the right-hand side of the

equation.

If s = 12,

LHS = 3(s + 2)

= 3(12 + 2)

= 3(14)

= 42

Continued over page

1

2

3

4

5

10WORKEDExample


Note: When solving equations with the pronumeral on both sides, it is good practice to

remove the smaller pronumeral from the relevant side.

THINK WRITE

If s = 12,

RHS = 2(s + 7) + 4

= 2(12 + 7) + 4

= 2(19) + 4

= 38 + 4

= 42

Comment on the answers obtained. Since the LHS and RHS are equal, the

equation is true when s = 12.

b Write the equation. b 4(d + 3) − 2(d + 7) + 4 = 5(d + 2) + 7

Expand the brackets on each side of

the equation first, and then simplify.

4d + 12 − 2d − 14 + 4 = 5d + 10 + 7

2d + 2 = 5d + 17

Subtract the smaller pronumeral

(that is, 2d) from both sides and

simplify.

2d − 2d + 2 = 5d − 2d + 17

2 = 3d + 17

Rearrange the equation so that the

pronumeral is on the left-hand side

of the equation.

3d + 17 = 2


simplify.

3d + 17 − 17 = 2 − 17

3d = −15

Divide both sides by 3 and simplify. = –

d = −5


d = −5 into the left-hand side and

then the right-hand side of the

equation.

If d = −5,

LHS = 4(d + 3) − 2(d + 7) + 4

= 4(−5 + 3) − 2(−5 + 7) + 4

= 4(−2) − 2(2) + 4

= −8 − 4 + 4

= −8

If d = −5,

RHS = 5(−5 + 2) + 7

= 5(−3) + 7

= −15 + 7

= −8

Comment on the answers obtained. Since the LHS and RHS are equal, the

equation is true when d = −5.

6

1

2

3

4

5

63d

3------

15

3------

7

8


Equations with the pronumeral on both sides

1 Solve the following equations and check your solutions.

a 8x + 5 = 6x + 11 b 5y − 5 = 2y + 7 c 11n − 1 = 6n + 19

d 6t + 5 = 3t + 17 e 2w + 6 = w + 11 f 4y − 2 = y + 9

g 3z − 15 = 2z − 11 h 5a + 2 = 2a − 10 i 2s + 9 = 5s + 3

j k + 5 = 7k − 19 k 4w + 9 = 2w + 3 l 7v + 5 = 3v − 11

2 Solve the following equations and check your solutions.

a 3w + 1 = 11 − 2w b 2b + 7 = 13 − b c 4n − 3 = 17 − 6n

d 3s + 1 = 16 − 2s e 5a + 12 = −6 − a f 7m + 2 = −3m + 22

g p + 7 = −p + 15 h 3 + 2d = 15 − 2d i 5 + m = 5 − m

j 7s + 3 = 15 − 5s k 3t − 7 = −17 − 2t l 16 − 2x = x + 4

3 Expand the brackets and then solve the following equations, checking your solutions.

a 3(2x + 1) + 3x = 30 b 2(4m − 7) + m = 76

c 3(2n − 1) = 4(n + 5) + 1 d t + 4 = 3(2t − 7)

e 3d − 5 = 3(4 − d) f 4(3 − w) = 5w + 1

g 2(k + 5) − 3(k − 1) = k − 7 h 4(2 − s) = −2(3s − 1)

i −3(z + 3) = 2(4 − z) j 5(v + 2) = 7(v + 1)

k 2m + 3(2m − 7) = 4 + 5(m + 2) l 3d + 2(d + 1) = 5(3d − 7)

m 4(d + 3) – 2(d + 7) + 5 = 5(d + 12) n 5(k + 11) + 2(k – 3) – 7 = 2(k – 4)

o 7(v – 3) – 2(5 – v) + 25 = 4(v + 3) – 8 p 3(l – 7) + 4(8 – 2l) – 7 = –4(l + 2) – 6

1. When a pronumeral appears on both sides of an equation, remove the

pronumeral term from one side. It is good practice to remove the smaller

pronumeral from the relevant side.

2. For a positive term we can remove by subtraction. For example,

7x + 7 = 5x − 3 (subtract 5x from both sides).

3. For a negative term we can remove by addition. For example,

3x + 11 = 7 − 2x (add 2x to both sides).

remember

6DSkillSHEET

6.6

Expandingexpressionscontainingbrackets

Mathcad

Pronumeralson both

sides

GC

program–

Casio

Solvingequations

WORKED

Example

8

GC program

– TI

Solvingequations

EXCEL Spreadsheet

Pronumeralson both

sidesWORKED

Example

9

WORKED

Example

10

MA

TH

S Q

UEST

CHALL

EN

GE

CHALL

EN

GE

MA

TH

S Q

UEST

1 The cost of a cricket ball and a glove is $25 and the cost of two cricketballs and three gloves is $65.

a What is the cost of a cricket ball and two gloves? Explain youranswer.

b What is the cost of one glove?

c What is the cost of one cricket ball?

2 Of 3 coins, one is known to be a fake, which is slightly lighter than theother two. But which one? Explain how, with just one weighing on apair of scales, you can determine which is the fake coin.

Kidneys . . .

4 9 –8 –6 –3 3 13 13

–3 15 13 4 4 –8 –16 9 2

–1 5 –5 2 6 5 –7 –3

–3 15 13 4

–8 1 –6 3

2 5 1 –8

–8 9 –6 3 –16 3 1 –5 –3

2A – 5 = A 3D = 10 – 2D 7+5E = 2E – 2 15 – F = F + 7

Solve the equations todiscover the puzzle answer code.

6I + 11 = 74 – I 3K + 8 = K – 6 L = 4L + 24 M + 6 = 3M – 6

3S – 6 – S = S + 9 2(T – 6) = 3(2T +4) 9 – 2Y = 5Y + 16U = 48 + 4 U

6(N + 3) = 2N – 2 2O + 4 = 3O – 9 12 – 4I = 8 I 3(2R – 1) = 5R



1 Find the output expression for the flow chart:

2 Draw a flow chart whose output expression is given by .

3 Starting with the equation b = 9 write the new equation when you divide both

sides by .

4 What operation must have occurred to get d − 2 = 4 from d = 6?

5 Solve 10 − g = 8.

6 Solve + 5 = 9.

7 True or false? The solution to the equation is h = 12.

8 Solve 6v + 6 = 2v − 6.

9 Solve 3m + 10 = −6 − m.

10 Expand the brackets first and then solve the equation 5(3 − t) = −2(2t − 1).

Checking solutions

Lucy is not able to solve the equation ,

but Jo thinks that the solution is m = 3. Is she correct?

Lucy decides to check Jo’s solution by substituting

m = 3 into both sides of the equation.

Left-hand side: =

= 1

Right-hand side: =

=

= 1

Both sides equal 1, so is a true statement

when m = 3. This means that m = 3 is the solution to the equation, so Jo is correct!

2

– 4 x 5 + 3

f

4t

9--- 10+

–

1

3---

y

4---

2 h 5+( )3

-------------------- 8– 4=

m 5+8

-------------4m 7–

5----------------=

m 5+8

-------------3 5+

8------------

4m 7–

5----------------

4 3 7–×5

---------------------

12 7–

5---------------

m 5+8

-------------4m 7–

5----------------=


a For each of the following, determine whether x = 7 is the solution to the equation.

i = 2x – 12 ii 3x − 7 = 2x + 3

b For each of the following, determine whether x = 2 is a possible solution to the equation.

i x5 × x2 = x7 ii (x3)3 = x6

THINK WRITE

a ii Write the equation. a ii = 2x − 12

Write the left-hand side of the

equation and substitute x = 7.

If x = 7,

LHS =

Perform the calculation. =

= 2

Write the right-hand side of the


If x = 7,

RHS = 2x − 12

Perform the calculation. = 14 − 12

= 2

Comment on the result. x = 7 is the solution because the LHS of

the equation equals the RHS.

ii Write the equation. ii 3x − 7 = 2x + 3



If x = 7,

LHS = 3x − 7

Perform the calculation. = 21 − 7

= 14



If x = 7,

RHS = 2x + 3

Perform the calculation. = 14 + 3

= 17

Comment on the result. x = 7 is not the solution because the LHS

of the equation does not equal the RHS.

b ii Write the equation. b ii x5 × x2 = x7



If x = 2,

LHS = x5 × x2

Perform the calculation. = 25 × 22

= 32 × 4

= 128



If x = 2,

RHS = x7

Perform the calculation. = 27

= 128

x 1–

3------------

1x 1–

3-----------

2

x 1–

3-----------

37 1–

3------------

4

5

6

1

2

3

4

5

6

1

2

3

4

5

11WORKEDExample


Checking solutions

1 For each of the following, determine whether:

a x = 5 is the solution to the equation

b x = 10 is the solution to the equation

c x = 3 is a solution to the equation x2 + 2 = 2x + 5

d x = 4 is a solution to the equation x2 − 3 = 3x + 3

e x = −2 is the solution to the equation 4 − x = 5x + 16

f x = −1 is the solution to the equation

g x = 7 is a solution to the equation x2 = 7x

h x = 29 is a solution to the equation x2 + 9 = 30x + 20

i x = 87 is the solution to the equation

j x = 2.7 is a solution to the equation x2 + x = 8.29

k x = 3.6 is the solution to the equation 2.1x + 0.44 =

THINK WRITE

Comment on the result. x = 2 is a possible solution because the

LHS of the equation equals the RHS.

However, it is not the only possible

solution.

ii Write the equation. ii (x3)3 = x6



If x = 2,

LHS = (23)3

Perform the calculation. = (8)3

= 512



If x = 2,

RHS = x6

Perform the calculation. = 26

= 64

Comment on the result. x = 2 is not a possible solution because

the LHS of the equation does not equal

the RHS.

6

1

2

3

4

5

6

We can check our solution to an equation by substituting a guess into the left- and

right-hand sides of that equation.

remember

6ESkillSHEET

6.7

Checkingsolutions bysubstitution

WORKED

Example

11 x 3+

2------------ 3x 10–=

2x 4+

6--------------- x 6–=

Mathcad

Checkingsolutions

5x 3+

2---------------

4x 2–

6---------------=

5x 15+

9------------------

x 13+

2---------------=

x

0.5------- 0.36–


l x = is the solution to the equation

m x = 2 is a possible solution to x3 × x2 = x5

n x = 5 is a possible solution to x4 × x2 = x7

o x = 4 is a possible solution to (x2)2 = x4

p x = 3 is a possible solution to (x4)4 = x8

q x = 6, y = 3 is a possible solution to (x × y)2 = x2y2

r x = 3, y = 4 is a possible solution to (x × y)3 = xy3

s x = 12, y = 6 is a possible solution to (x ÷ y)2 =

t x = 15, y = 15 is a possible solution to x4 ÷ y4 = 1.

2 a Copy and complete the table at right to

find the values of and .

b Find the solution to the equation

.

3 a Copy and complete the table at right to

find the values of m2 and 2m + 8.

b What is a solution to m2 = 2m + 8?

4 a Determine whether x = 2 is the solution to the

equation 6x2 − 15 = 9.

b Determine whether x = −2 is the solution to the

equation 6x2 − 15 = 9.

c Solve the equation 6x2 − 15 = 9 using the given

flow chart and backtracking, and prove that

parts 4a and 4b are both correct.

Note: Recall that the square root of a number

produces both a positive and a negative result.

5 Solve the following equations using the flow chart and backtracking method outlined in

question 4c.

a 2x2 + 6 = 24

b 5x2 − 7 = −2

c 3x2 − 60 = 48

2

7---

3

5---x

1

35------+

7

10------x=

x2

y2

-----

Mat

hcad

Guess

and

check

x

1 3

2 3.25

3

4

5

x 5+

2------------

3x 7+

4---------------x 5+

2------------

3x 7+

4---------------

x 5+

2------------

3x 7+

4---------------=

EXCE

L Spreadsheet

Guess

and

check

m m2

2m + 8

0 0 8

1

2

3

4

5 18

x x2

( )2

÷ 6 + 15

× 6

9

– 15


Solving word problemsEquations can be used to solve many everyday puzzles and problems. We must write

the problem as an equation and then solve it by backtracking or doing the same to both

sides. Here is an example.

Sarah is 3 years older than her brother, Jason. If the sum of their ages is 35 years,

then how old is Sarah?

Step 1 Read the question carefully.

The important facts are these:

Sarah is 3 years older than Jason.

Sarah’s age plus Jason’s age equals 35.

Step 2 Turn the information into an equation.

We are asked to find Sarah’s age, so we

call it x. We usually say, ‘Let Sarah’s age

equal x.’

Jason is younger by 3 years, so his age

must be x − 3, and since their ages add up

to 35 we know that x + x − 3 = 35. This is

our equation.

Step 3 Simplify and then solve the equation.

x + x − 3 = 35

2x − 3 = 35

2x = 38

x = 19

Step 4 Check your solution.

If Sarah is 19, then Jason is 16, and their

total age is 19 + 16 = 35. This fits the

information we were given.

Step 5 Write your answer in words.

Sarah is 19 years old.

If t is a whole number, then write the value of:

a the number that is 15 less than t

b the number that is 15 times as large as t

c the number of weeks in t days.

THINK WRITE

a To make it less than t, subtract 15 from t. a t − 15

b To find 15 times t, multiply t by 15. b 15t

c To change days to weeks, divide by 7. ct

7---

12WORKEDExample


a The total distance around the rectangular field at right

is 580 m. Find the value of x and use this to find the

length and width of the field.

b The area of the triangle at right is 148 cm2. Find the

value of x and use this to find the height of the triangle.

THINK WRITE

a Write the rule for the area of a

rectangle.

a P = 2(l + w)

Substitute the known values into the

rule.

580 = 2(3x + 20 + 2x + 10)

Simplify the terms inside the pair of

brackets.

580 = 2(5x + 30)

Divide both sides of the equation by

2.

=

290 = 5x + 30



of the equation.

5x + 30 = 290

Subtract 30 from both sides of the

equation and simplify.

5x + 30 − 30 = 290 − 30

5x = 260


x = 52

Evaluate the length and width of the

rectangle.

l = 3x + 20 w = 2x + 10

= 3 × 52 + 20 = 2 × 52 + 10

= 156 + 20 = 104 + 10

= 176 m = 114 m

The length and the width of the rectangle are

176 m and 114 m respectively.

Check solutions by placing values

into the perimeter formula.

Check: P = 2(l + w)

= 2(176 + 114)

= 2(290)

= 580 m

This corresponds to the initial information

given.

b Write the rule for the area of a

triangle.

b A =

Substitute the known values into the

rule.

148 =

(3x + 20) m

(2x + 10) m

160 cm

(x – 10) cm

1

2

3

4580

2---------

2 5x 30+( )

2--------------------------

5

6

75x

5------

260

5---------

8

9

1bh

2------

2160 x 10–( )

2-----------------------------

13WORKEDExample


THINK WRITE

Simplify the right-hand side of the

equation.

148 = 80(x − 10)

Divide both sides of the equation by

80.

=

1.85 = x − 10



of the equation.

x − 10 = 1.85

Add 10 to both sides of the equation

and simplify.

x − 10 + 10 = 1.85 + 10

x = 11.85

Evaluate the height of the triangle. h = x − 10

= 11.85 − 10

= 1.85 cm

The height of the triangle is 1.85 cm.

Check solutions by placing values

into the area formula.

Check: A =

=

=

= 148 cm2

This corresponds to the initial information

given.

3

4148

80---------

80 x 10–( )

80--------------------------

5

6

7

8bh

2------

160 1.85×

2-------------------------

296

2---------

Annette and Patrick each have part-time jobs and worked a total of 28 hours over a week. Annette earned $14 per hour, while Patrick earned $16 per hour. Together they earned $416. How many hours did Patrick work?

Continued over page

THINK WRITE

Write the important information. Annette’s hours + Patrick’s hours = 28

Annette’s earnings at $14 per hour + Patrick’s

earnings at $16 per hour = $416

State the unknown as a pronumeral. Let x be the number of hours Patrick worked.

So Patrick earned $16 × x or 16x.

Write the number of hours

Annette worked in terms of this

pronumeral.

Annette has worked (28 − x) hours. So she has

earned $14 × (28 − x) or 14(28 − x).

Write an equation for the total earnings. 14(28 − x) + 16x = 416

Simplify the equation by expanding the

brackets.

392 − 14x + 16x = 416

392 + 2x = 416

1

2

3

4

5

14WORKEDExample


Solving word problems

1 If m is a whole number, then write the value of:

a the number that is 5 less than m

b the number that is 10 times as large as m

c the number that is half of m

d the number that is 4 more than m

e the next largest number after m

f the number of weeks in m days

g the number of months in m years

h the (mean) average of 6, 15, and m

i the cost of m theatre tickets at $65 per ticket

j the number of minutes in m seconds.

2

For each of the following, choose the correct response.

a Heidi is y years old. In 4 years’ time her age will be:

A 4y B 4 C y − 4 D y + 4 E 8

b Jake is twice as old as Kyna. If Jake is w years old, then Kyna’s age is:

A w + 2 B 2w C D 2 E w − 2

THINK WRITE

Subtract 392 from both sides. 392 + 2x − 392 = 416 − 392

2x = 24


x = 12

Check the solution. If Patrick worked 12 hours, then Annette

worked 16 hours.

Total earnings = 14 × 16 + 16 × 12

= $416

The solution is correct.

Write the answer in words. Patrick worked 12 hours.

6

72x

2------

24

2------

8

9

Solving word problems involves 5 steps.

Step 1 Read the question carefully.

Step 2 Turn the information into an equation.

Step 3 Simplify and then solve the equation.

Step 4 Check the solution.

Step 5 Write the answer in words.

remember

6FWORKED

Example

12

multiple choice

w

2----


c If t is an even number, the next (greater) even number is:

A 2t B 2 C 4 D t + 2 E t + 1

d If t is an odd number, the next (greater) odd number is:

A t + 1 B 3t C t + 3 D 3 E t + 2

e Sonja is 7 cm taller than Jodie. If Sonja is h cm tall, then Jodie’s height is:

A h + 7 B 7h C h − 7 D 107 E

f Mikhail is twice as old as Natasha. If Natasha is w years old, then Mikhail’s age is:

A w + 2 B 2w C D 2 E w − 2

g Nhan has h dollars, but Dyson has 7 dollars more. In total they have:

A h + 7 B h + 14 C 2h + 7 D 2h + 14 E 2h − 7

h Natalie is t years old. Compared with Natalie, Lauren is 4 years older, but Fiona is

3 years younger. The total of their ages is:

A 3t + 7 B 3t + 1 C 3t − 1 D 2t + 1 E t + 1

i Prashant and Anapam together earned $98 last week. If Prashant earned k dollars,

then Anapam earned:

A 98 + k B k + 98 C k − 98 D 98 − k E k

Use the following information for parts j and k.

Tom walks k metres to school. Jared walks 80 m further than Tom, and Katrina walks

twice as far as Jared.

j Katrina walks:

A 160 B 2k C 2k + 80 D 2(k + 80) E 2k − 160

k The total distance that the three students walk to school is:

A 4k + 240 B k + 240 C 3k + 80 D 2k + 240 E 3k + 240

l A tennis court can be hired for $15

per hour. How much does it cost to

hire the court for m hours?

A m + 15 B 15 C

D E 15m

3 Terry is 7 cm shorter than Jacinta. If

their total height is 291 cm, then how

tall is Jacinta?

4 Yolanda and Janelle play netball

together and between them they have

scored 103 goals, but Janelle has scored

15 more goals than Yolanda. How many

goals has Yolanda scored?

5 In a Mathematics test, Duy scored twice

as many marks as Paul but three marks

fewer than Kate. The three students

scored a total of 178 marks. How much

did each one score?

6 At a certain store, a football costs $15 more than a basketball. If 3 basketballs and

2 footballs cost a total of $115, find the cost of each ball.

h

100--------- 7–

w

2----

SkillSHEET

6.8

Writingequations

from wordedstatements

m

15------

15

m------


7 Three consecutive odd numbers add up to 159. What are the three numbers?

8 Four consecutive even numbers add up to 100. What is the largest of the four numbers?

9 The total distance around this rectangular field is

480 metres. Find the value of x and use this

result to find the length and width of the field.

10 The area of the triangle at right is 131.75 cm2.

Find the value of x and use this to find the height

of the triangle.

11 These two rectangles have the same area. Write an

equation and solve it to find x. What is the area of

each rectangle? (All lengths are given in metres.)

12 Amanda, at tenpin bowling, has bowled games of

118 and 131. How much must she score in her next

game to attain an average of 125?

13 The volume of blood, in litres, in your body is

approximately equal to your mass in kilograms

multiplied by 0.08.

a Write an equation to represent this situation

and state what each pronumeral represents.

b According to this rule, how many litres of

blood does a 60-kg person have?

c Estimate the mass of a person with 3.8 litres of

blood.

14 Write a word problem that can be answered by

solving the equation 3x − 10 = 26. Then solve the

problem.

15 The formula to convert degrees Fahrenheit (F) to

degrees Celsius (C) is C = (F – 32). Find the

temperature in degrees Fahrenheit when the

thermometer reads 40 degrees Celsius.

16 Sandra and Merv worked a total of 20 hours.

Sandra earned $12 per hour while Merv earned $10

per hour. Together they earned $218 over the week.

How many hours did Sandra work?

17 Oscar earns the same amount for babysitting every week. Each week he saves all but

$10 of his earnings. If he has $160 after 8 weeks, how much does he earn each week?

(Write an equation to solve for this situation first.)

18 Alex, Stuart and Ali receive an average of $12 per week in pocket money. Alex receives

$2 more than Stuart but $5 less than Ali. How much does each one receive?

Mat

hcad

Solvingequations

EXCE

L Spreadsheet

Solvingequations

GC pr

ogram– TI

Solvingequations

GC pr

ogram– Casio

Solvingequations

WORKED

Example

13a

(x +

50) m

(3x + 10) m

(x – 8) cm

170 cm

WORKED

Example

13b

x + 22

3040

x – 2

5

9---

WORKED

Example

14

WorkS

HEET 6.2


Inequalities and inequationsAn inequation is an expression that contains an inequality sign. We learned about

these signs back in chapter 2, but let us go over them again here.

There are four different inequality signs:

< means ‘is less than’

≤ means ‘is less than or equal to’

> means ‘is greater than’

≥ means ‘is greater than or equal to’.

Just like equations, inequations can be true or false. For example:

8 > 6 This is true because 8 is greater than 6.

5 < 3 This is false because 5 is not less than 3.

What about the inequation x > 2? This sentence could be either true or false, depending

on the value of x. There are many values of x that would make the sentence true, for

example, x = 3 or x = 200. Can you think of some others? Of course we can include

fractions and decimals like 2 , 2 , 2.95 or 2.000 01. We cannot possibly list all of the

solutions to the inequation x > 2, but we can show them using a number line.

You should now be in a position to assist Shannon in working out how much he

needs to save each month to reach his target of $3299 in 8 months. He has $449 in

the bank already.

1 Write an equation for this situation, stating what the pronumeral represents.

2 Solve the equation and write a sentence advising Shannon on how much he

should save each month so he can buy the computer.

Shannon’s friend Kate is

saving for a new sound

system that costs $629.

She has $29 in the bank.

3 How much does Kate

need to save each

month to be able to

buy the new sound

system at the same

time as Shannon buys

a computer?

4 If Kate was able to save the same amount as Shannon each month, how much

sooner would Kate be able to buy her sound system?

COMMUNICATION Save, save, save!

1

2---

7

8---


Note the careful spacing. The arrows tell us that the line continues in both directions.

On a number line, the numbers get larger as you move to the right.

We show the inequation on a number line by placing a dot over the numbers that are

greater than 2.

There are so many dots that they will form a continuous line, but there is no dot above

2 (because 2 is not greater than 2). We can show the inequation simply by placing an

open circle above 2, and ruling a line above the numbers that we want. The arrow tells

us that the line keeps going forever in one direction. Adding a label completes the

solution.

The inequation x ≥ 2 (x is greater than or equal to 2) is slightly different because we

need to include the value x = 2. We do this by shading the circle.

The inequation x > 2 has an infinite number of solutions. In other words, the list of

solutions is never ending.

–3 –2 –1 0 1 2 3 4 5 6 7 x

–3 –2 –1 0 1 2 3 4 5 6 7 x

–3 –2 –1 0 1 2

x > 2

3 4 5 6 7 x

–3 –2 –1 0 1 2 3 4 5 6 7

x ≥ 2

x

Referring, if necessary, to the number line at

right, place either < or > between the

following pairs of numbers to make a true

statement.

a 1 4 b 1 −4

THINK WRITE

a On a number line, the numbers get larger as

you move to the right, so 1 is less than 4.

1 < 4

b On a number line, the numbers get larger as

you move to the right, so 1 is greater than −4.

1 > −4

15WORKEDExample

–5 –4 –3 –2 –1 0 1 2 3 4 5


State whether each of the ordered pairs satisfies the given inequality.

a x2 > 2y (3.5, 6) b x

3 < 10y (5.2, 8)

THINK WRITE

a Write the inequation. a x2 > 2y


inequation and substitute x = 3.5.

If x = 3.5,

LHS = x2

Perform the calculation. LHS = 3.52

LHS = 12.25


inequation and substitute y = 6.

If y = 6,

RHS = 2y

Perform the calculation. RHS = 2 × 6

RHS = 12

Comment on the result. x2 > 2y

12.25 >12

The inequation is correct in this case; there-

fore, the point (3.5, 6) satisfies x2 > 2y.

b Write the inequation. b x3 < 10y


inequation and substitute x = 5.2.

If x = 5.2,

LHS = x3

Perform the calculation. LHS = 5.23

LHS = 140.608


inequation and substitute y = 8.

If y = 8,

RHS = 10y

Perform the calculation. RHS = 10 × 8

RHS = 80

Comment on the result. x3 < 10y

140.608 < 80

The inequation is incorrect in this case;

therefore, the point (5.2, 8) does not satisfy

x3 < 10y.

1

2

3

4

5

6

1

2

3

4

5

6

16WORKEDExample

Draw a neatly ruled and clearly labelled number line to show the inequation x < −1.

THINK WRITE

Rule a number line and mark several

numbers either side of and including −1.

Less than means ‘do not include the value’,

so draw an open circle above x = −1.

Draw a line to the left, because numbers get

smaller as we move to the left.

1 –3 –2 –1 0 1 2 3 x

2

–3 –2 –1 0 1 2 3 x

3

–3 –2 –1 0 1 2 3 x

17WORKEDExample


Like an equation, an inequation can be solved by doing the same to both sides.

However, as we saw in the investigation above, if multiplying or dividing by a

negative number is required, we reverse the inequality sign to keep the inequation

true.

Consider the statement 2 < 6. Is this statement true?

1 After applying each of the operations described below, state whether the

statement is still true. If it is not, explain what needs to be done to make it a true

statement.

a Add 4 to both sides of 2 < 6.

b Subtract 7 from both sides of 2 < 6.

c Multiply both sides of 2 < 6 by 3.

d Multiply both sides of 2 < 6 by −3.

e Divide both sides of 2 < 6 by 2.

f Divide both sides of 2 < 6 by −2.

2 Write a summary of what you have found.

3 Investigate whether this occurs with other inequalities.

COMMUNICATION Operations on inequalities

Solve the following inequations by doing the same to both sides.

a 2x < 4 b −3x < 12 c 4x − 3 ≥ 17 d > −1

THINK WRITE

a Write the inequation. a 2x < 4

To make x the subject, divide both sides

by 2.

<

Simplify. x < 2

b Write the inequation. b −3x < 12

Divide both sides by −3. Remember to

reverse the inequality sign, since we are

dividing by a negative number.

>

x > −4

Simplify.

c Write the inequation. c 4x − 3 ≥ 17

Add 3 to both sides. 4x − 3 + 3 ≥ 17 + 3

Divide both sides by 4. 4x ≥ 20

≥

Simplify. x ≥ 5

1 2x–

7---------------

1

22x

2------

4

2---

3

1

23x–

3–---------

12

3–------

3

1

2

3

4x

4------

20

4------

4

18WORKEDExample


Inequalities and inequations

1 Copy and complete the following statements about number lines.

a As you move to the right the numbers become .

b As you move to the left the numbers become .

2 Referring, if necessary, to the number line below, place either < or > between the

following pairs of numbers to make a true statement.

a 1 13 b 1 −3 c 3 −3 d 3 1

e −1 −3 f −1 0 g −3 3 h −3 −1

i −4 −3 j −3 −4

3 This number line has both numbers and pronumerals. Write each statement below,

placing either > or < between the two terms in order to make a true statement.

a h x b x h c −5 h d m x

e y −2 f −2 y g h y h y h

i 0 m j m 0

THINK WRITE

d Write the inequation. d > −1

Multiply both sides by 7. × 7 > −1 × 7

1 − 2x > −7

Subtract 1 from both sides. 1 − 2x − 1 > −7 − 1

−2x > −8

Divide both sides by −2. Remember to

reverse the inequality sign, since we are

dividing by a negative number.

<

x < 4

Simplify.

11 2x–

7---------------

21 2x–( )

7--------------------

3

42x–

2–---------

8–

2–------

5

1. An inequation contains an inequality sign.

2. There are 4 different inequality signs: < means ‘is less than’, ≤ means ‘is less

than or equal to’, > means ‘is greater than’, ≥ means ‘is greater than or equal

to’.

3. On a number line, the numbers get larger as you move to the right.

4. An inequation can be solved by doing the same to both sides.

5. Multiplying or dividing both sides by a negative number reverses the inequality

sign.

6. An inequation has an infinite number of solutions.

remember

6G

WORKED

Example

15

–5 –4 –3 –2 –1 0 1 2 3 4 5

–5 h –2 0 mx y 4 5


4 State whether each inequation below is true (T) or false (F).

a 5 > 3 b 6 > 6 c 6 ≤ 6 d 2.1 < 2

e −1 > 0 f −8 < 5 g 0 > −5 h −3 < −3

i 1.99 < 2 j 5.0001 > 5 k −5 > −6 l −3 > 2

m −4 ≥ −4 n 0 > −10

5 State whether each of the ordered pairs satisfies the given inequation.

a x2 > 2y (4.8, 7) b x

3 < 10y (3.5, 12)

c 3x3 < y2 (2.8, 6.8) d 4x

4 ≥ y2 (1, 2)

e 3x3 − y2 < 16 (5.1, 12) f 34 ≤ 4x

3 − y3 (3.9, 6)

6

For each of these number lines, select the correct label.

a A x < 1 B x > 1 C x < 5

D x ≤ 1 E x > 5

b A x < 1 B x > 1 C x < 5

D x ≥ 1 E x ≥ 5

c A x > 6 B x > 10 C x < 6

D x < 10 E x > 8

d A x > 6 B x > 10 C x < 6

D x < 10 E x > 8

e A x > 300 B x ≥ 100 C x < 300

D x > 500 E x < 500

f A x ≤ 300 B x > 100 C x < 300

D x ≤ 500 E x < 500

g A x < 0 B x < −4 C x > −4

D x < −8 E x ≥ −4

h A x < 0 B x < −4 C x > −4

D x < −8 E x ≥ −4

7 Draw a neatly ruled and clearly labelled number line to show the following

inequations.

a x ≥ 5 b x < 6 c x ≥ 0 d x ≤ −2

e x > −5 f x < 0 g x > 2.3 h x ≤ 3

i x > −50 j x ≤ 120

8 Solve the following inequations by doing the same to both sides.

a x + 1 < 7 b x − 4 > 3 c x − 5 ≤ 1 d x + 20 < −10

e 2x > 8 f 3x ≤ 9 g −5x ≥ 20 h −2x < 50

i −4x < −16 j 8x > − 4


a 5x − 1 > 4 b 2x + 7 ≤ 3 c 3x + 4 < −2 d 4x − 3 > 13

e −2x + 1 ≥ −5 f −6x − 3 < 3 g −4x − 7 < 5 h −3x + 9 ≥ −9

i > 1 j < 3 k ≤ −3 l ≥ −8

m > 5 n < 3 o + 2 ≤ 7 p 1 − ≥ −5

q 10 − > 4 r −4 + > 11

WORKED

Example

16SkillSH

EET 6.9

Checking solutions to inequations

multiple choice

–3 –2 –1 0 1 2 3 4 5 6 7 x

–3 –2 –1 0 1 2 3 4 5 6 7 x

0 1 2 3 4 5 6 7 8 9 10 x

0 1 2 3 4 5 6 7 8 9 10 x

0 100 200 300 400 500 x

0 100 200 300 400 500 x

–3 –2–5 –4–7 –6–8 –1 0 1 2 x

–3 –2–5 –4–7 –6–8 –1 0 1 2 x

WORKED

Example

17

1

2---

SkillSH

EET 6.10

Showing inequations on a number line

EXCE

L Spreadsheet

Inequations

WORKED

Example

18a, b

WORKED

Example

18c, d

x 2+

7------------

x 4–

5-----------

1 x–

2-----------

5 x–

3-----------

4 3x–

2---------------

7 2x–

5---------------

5x

6------

3x

4------

2x

5------

5x

3------


For questions 10, 11 and 12, write and solve an inequation for each situation.

10 To obtain a driver’s licence, a person needs to be 18 or older. Selena has stated that

she will try for her licence in 5 years’ time. What is Selena’s age?

11 A pair of rollerblades can be hired for an initial fee of $12 plus $4 per hour. For how

many hours can you hire the rollerblades and still spend less than $50?

12 A land owner has 100 hectares of land and wants to keep at least 10 hectares. The rest

is to be divided and sold in 6-hectare lots. How many of these lots can the land owner

offer for sale?

Rows in a theatre are often elevated so each person can look over the head of the

person in front. Theatre floors must either be sloped or have steps. Whether the

floor can be sloped depends on the depth of the row, d, and the amount of rise, r.

A floor can be sloped only if r ≤ .

1 How deep can the rows be if the floor is sloped and has a rise of 12.5 cm?

2 Many theatres have 80-cm-deep rows. What is the maximum rise for a sloped

floor?

3 An Imax cinema claims to give everyone an unobstructed view of the screen.

Suppose the rise is 27 cm. What is the minimum row depth for a sloped floor?

Why do you think these theatres have stepped floors?

4 Investigate different situations with varying amounts of rise and row depth for

theatre seating. What are the optimum conditions? Justify your answer, showing

all working. Include a list of advantages and disadvantages with stepped and

sloped floors.

Kim had taken his son, Ethan, to the circus. Ethan was fascinated by the tight-rope

walker who used a long pole to help him maintain his balance. Kim explained the

principle of balancing weights attached to rods. As a practical example, he decided

to construct a mobile using rods and string with balls of plasticine attached. A

diagram of the mobile is shown below.

GAME time

Equations— 002

WorkS

HEET 6.3

COMMUNICATION Theatre design

d

8---

COMMUNICATION Mobiles

1g

Kim carefully weighed a 1-g ball of plasticine

and attached it via string to the rod in the

position shown.

Write instructions for Kim to assist him to

determine the masses of all balls so that the

mobile will balance evenly. Provide reasoning

for all your answers.


Copy the sentences below. Fill in the gaps by choosing the correct word or

expression from the word list that follows.

1 operations are operations that undo one another.

2 Addition and are inverse operations.

3 and division are inverse operations.

4 When solving we must do the same to both sides.

5 A is useful to determine what to do next.

6 Replacing a pronumeral with a number is called .

7 The to an equation is the value that makes the equation

.

8 After guessing a solution and checking, you should before

you .

9 An inequation is a statement containing an sign.

10 There are different inequality signs.

11 An inequation has an number of solutions, which are easily

shown on a .

12 An circle indicates the value is not .

13 A circle indicates the value is included.

14 An inequation can be solved by doing the same to sides.

15 Multiplying or dividing both sides of an inequation by a negative number

the inequality sign.

summary

W O R D L I S T

equations

flow chart

four

number line

inequality

infinite

inverse

open

comment

multiplication

solution

substitution

both

closed

subtraction

think

true

reverses

included


1 Find the output number for each of these flow charts.

a b

c d

2 Draw the flow chart whose output number is given by the following expressions.

a −3(m + 4) b c d 7 − 15w

3 a Write an equation that is represented by

the diagram at right.

b Show what happens when you take

2 from both sides, and write the new

equation.

4

If we start with x = 4, which of these equations

is not true?

A x + 2 = 7 B 3x = 12 C −2x = −10 D E x − 2 = 3

5


A B −2x = −6 C 2x − 6 = 0 D E x − 5 = 2

6 Solve these equations by doing the same to both sides.

a z + 7 = 18 b –25 + b = –18 c – = z –

d 9t = e –8.7 = f – =

7 Solve these equations by doing the same to both sides.

a 5v + 3 = 18 b 5(s + 11) = 35 c

d −2(r + 5) − 3 = 5 e f − 3 = 2

8 Solve the following equations and check each solution.

a 5k + 7 = k + 19 b 4s − 8 = 2s − 12

c 3t − 11 = 5 − t d 5x + 2 = −2x + 16

6A

CHAPTERreview

– 1 × 5

x

÷ 8 × 3

x

+ 8 ÷ 5

x

+ 3 × 3 − 7

x

÷ 2

6An

3--- 5+

m 7–

5------------- 4–

6B

represents an unknown amount

represents 1

Keymultiple choice 6B

x

5--- 1=

multiple choice

6B2x

3−−− 2=

x

5---

3

5−=

6C8

9---

4

3---

1

3---

l

5--- 6

13------

h

8---

6Cd 7–

4------------ 10=

2y 3–

7--------------- 9=

x

5---

6D


9 Expand the brackets first and then solve the following equations.

a 5(2v + 3) − 7v = 21 b 3(m − 4) + 2m = m + 8

10 Determine whether m = 4 is the solution to the equation = 2m − 5.

11 For each of the following, determine whether:

a x = 2 is a possible solution to (x3)2 = x5

b x = 3 is a possible solution to (x5)3 = x15

c x = 5, y = 2 is a possible solution to (x × y)2 = x2y

2

d x = 6, y = 2 is a possible solution to (x × y)3 = xy3.

12 Nhan is 7 cm taller than Dyson and the sum of their heights is 333 cm. How tall is Nhan?

13 A bicycle rental store charges a flat fee of $15 plus $5 per day to rent a mountain bike.

a If x is the number of days a mountain bike is rented, write an equation to show the cost of

renting a mountain bike.

b If a person was charged $85 for renting a mountain bike, for how many days did the

person rent the bike?

14 The total distance around the rectangular field at right is 840 m.

Find the value of x and use this to find the length and width of

the field.

15 The area of the triangle at right is 438 cm2. Find the value of x and use

this to find the base of the triangle.

16 Write true or false for each of these statements.

a 3 < 3 b −3 < 3

c −5 > −2 d ≥

17 State whether each of the ordered pairs satisfies the given inequality.

a 2x2 > 5y (3.7, 6.5) b 3x

3 < 13y (2.5, 7.1)

c 6x3 < y2 (0.8, 2.8) d 4x

4 ≥ y2 (1.3, 3.5)

18 Draw neat number lines to show the following sets of numbers.

a x ≤ 3 b x > −7

c x < −100 d x ≥ 0


a x + 8 > −5 b 3x − 7 ≤ 11

c 4 − 5x > 9 d > 5

e ≤ −6 f 2 − > 10

20 A camera shop charges $15 to develop a roll of film plus $0.80 for each extra print. How

many extra prints can Neale get if he cannot spend more than $22 on photographs? (First

write and solve an inequation for this situation.)

6D

6Em 5+

3-------------

6E

6F6F

6F(2x – 25) m

(3x + 10) m

24 cm

(x – 12) cm

6F

6G1

4---

1

3---

6G

6G

6G

3x 2+

4---------------

8 2x–

3---------------

4x

7------

testtest

CHAPTER

yourselfyourself

6

6G

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Chap 06

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