Rearranging simple formulas | Year 8

01 / LEARN

The idea, step by step

Change the subject

The subject is the variable alone on one side. In $C = 4 n + 7$ , $C$ is the subject. To make $n$ the subject, undo adding $7$ and then undo multiplying by $4$ . Keep equality by applying each operation to both sides.

The result is $n = \frac{C - 7}{4}$ . The whole difference $C - 7$ is divided by $4$ , so brackets or a complete fraction bar are essential. The relationship itself has not changed.

Use and check the rearranged rule

From $A = l w$ , divide both sides by $l$ to get $w = \frac{A}{l}$ , provided $l \neq 0$ . A rectangle has positive length, so that restriction agrees with the situation.

Check a rearrangement using a known set of values. Measurements must also give sensible units: area divided by length is a length. Do not divide a sum by only one of its terms; place the complete required numerator over the divisor.

A common trap

Writing $n = C - \frac{7}{4}$ from $C = 4 n + 7$ .

Instead: Subtract $7$ first and divide the whole difference: $n = \frac{C - 7}{4}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Undo addition then multiplication

Make $n$ the subject of $C = 4 n + 7$ .

Subtract $7$ : $C - 7 = 4 n$ .
Divide by $4$ .
Write the required subject on the left.

n = \frac{C - 7}{4}

Example 2

Use a rearranged area formula

From $A = l w$ , find $w$ if $A = 42$ square centimetres and $l = 7$ cm.

Rearrange: $w = \frac{A}{l}$ .
Substitute: $w = 42 / 7 = 6$ .
Check $7 \times 6 = 42$ .

6

Check your understanding

Make $x$ the subject of $y = 3 x - 2$ .

Show answer

$x = \frac{y + 2}{3}$

Add $2$ to both sides, then divide the whole sum by $3$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Make $x$ the subject of $y = x + 8$ .
2 marks
Make $p$ the subject of $C = 5 p$ .
2 marks
Make $t$ the subject of $d = v t$ , where $v \neq 0$ .
2 marks
Make $n$ the subject of $T = 3 n + 6$ .
2 marks
Make $a$ the subject of $P = 2 a + 2 b$ .
2 marks
A repair costs $C = 6 h + 18$ dollars. Make $h$ the subject and find $h$ when $C = 54$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Make $x$ the subject of $y = x + 8$ .
2 marks
Subtract $8$ from both sides.
$x = y - 8$
Make $p$ the subject of $C = 5 p$ .
2 marks
Divide both sides by $5$ .
$p = \frac{C}{5}$
Make $t$ the subject of $d = v t$ , where $v \neq 0$ .
2 marks
Divide both sides by $v$ .
$t = \frac{d}{v}$
Make $n$ the subject of $T = 3 n + 6$ .
2 marks
Subtract $6$ , then divide by $3$ .
$n = \frac{T - 6}{3}$
Make $a$ the subject of $P = 2 a + 2 b$ .
2 marks
Subtract $2 b$ , then divide the full difference by $2$ .
$a = \frac{P - 2 b}{2}$
A repair costs $C = 6 h + 18$ dollars. Make $h$ the subject and find $h$ when $C = 54$ .
3 marks
Subtract $18$ and divide by $6$ .
$(54 - 18) / 6 = 6$ .
$h = \frac{C - 18}{6}$ ; $6$ hours.

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Curriculum reference & lesson scope

Australian Curriculum v9 Mathematics with SA curriculum context

Combining like terms; substitution with signed/rational values; multiplying/dividing algebraic terms; expanding a single bracket; factorising common factors; equivalent linear expressions; rearranging simple formulas.

Official curriculum source ↗

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