Substitution and rearranging formulas | Year 9

01 / LEARN

The idea, step by step

Substitution must preserve the rule

Replace variables by the supplied values, then follow the order of operations. In $3 x^{2} - 2 y$ , the input $x = - 2$ produces $3 {(- 2)}^{2}$ , not $3 (- 2^{2})$ . Brackets show that the negative number is the base.

Check the meanings and units. The same letter can stand for different quantities in different problems. An area result needs square units, whereas a perimeter needs length units. An unreasonable sign or scale can reveal an input or order error.

Keep an equation balanced

To change the subject, undo operations on both sides. For $v = u + a t$ , subtract $u$ and divide by non-zero $a$ to give $t = \frac{v - u}{a}$ . The whole difference is divided by $a$ .

If the proposed divisor might be zero, state the restriction. When the requested variable occurs in several terms, collect or factor those terms before dividing. This lesson uses linear rearrangements, not solving a quadratic equation.

A common trap

Losing brackets around negative inputs or around a grouped numerator.

Instead: Keep the negative input and the entire difference grouped until the relevant operation has been performed.

02 / WORKED EXAMPLES

See the thinking

Example 1

Substitute a negative value

Evaluate $2 x^{2} - 3 y$ for $x = - 3$ and $y = 4$ .

$2 {(- 3)}^{2} - 3 (4)$ .
$2 (9) - 12 = 18 - 12$ .
The value is 6.

6

Example 2

Rearrange triangle area

For $A = \frac{1}{2} b h$ , make $h$ the subject. Find $h$ if $A = 42$ cm $2$ and $b = 12$ cm.

Multiply by 2: $2 A = b h$ .
Divide by non-zero $b$ : $h = \frac{2 A}{b}$ .
$h = 84 \div 12 = 7$ cm.

h = \frac{2 A}{b}

;

h = 7

Check your understanding

Make $x$ the subject of $y = 5 x - 2$ .

Show answer

$x = \frac{y + 2}{5}$

Add 2 to both sides and divide by 5.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Evaluate $4 a - 3 b$ for $a = 5$ , $b = 2$ .
2 marks
Evaluate $x^{2} + 2 x$ for $x = - 4$ .
2 marks
Make $w$ the subject of $P = 2 l + 2 w$ .
2 marks
Make $r$ the subject of $C = 2 π r$ .
2 marks
The formula $s = u t$ relates distance $s$ , constant speed $u$ and time $t$ , where $u \neq 0$ .
3 marks
1. (a)
  Make $t$ the subject.
  2 marks
2. (b)
  Find $t$ for $s = 84$ m and $u = 7$ m/s.
  1 mark
Make $x$ the subject of $a x + b = c$ . State a restriction.
3 marks

Read the worked solutions

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

Evaluate $4 a - 3 b$ for $a = 5$ , $b = 2$ .
2 marks
$4 (5) - 3 (2) = 20 - 6 = 14$ .
$14$
Evaluate $x^{2} + 2 x$ for $x = - 4$ .
2 marks
${(- 4)}^{2} + 2 (- 4) = 16 - 8 = 8$ .
$8$
Make $w$ the subject of $P = 2 l + 2 w$ .
2 marks
Subtract $2 l$ , then divide by 2.
$w = \frac{P - 2 l}{2}$
Make $r$ the subject of $C = 2 π r$ .
2 marks
Divide by the non-zero constant $2 π$ .
$r = \frac{C}{2 π}$
The formula $s = u t$ relates distance $s$ , constant speed $u$ and time $t$ , where $u \neq 0$ .
3 marks
Divide both sides by $u$ .
$t = 84 \div 7 = 12$ s.
1. (a)
  Make $t$ the subject.
  2 marks
  Divide both sides by $u$ .
  $t = \frac{s}{u}$
2. (b)
  Find $t$ for $s = 84$ m and $u = 7$ m/s.
  1 mark
  $84 \div 7 = 12$ s.
  $12$ s
$t = \frac{s}{u} = 12$ s
Make $x$ the subject of $a x + b = c$ . State a restriction.
3 marks
Subtract $b$ , then divide by $a$ .
The division requires $a \neq 0$ .
$x = \frac{c - b}{a}$ for $a \neq 0$

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum

Year 9 algebraic simplification and equivalent expressions, with consolidation of linear formula skills. Binomial expansion and quadratic factorisation/equations are in the separate Quadratics and Parabolas chapter.

Official curriculum source ↗

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