Substitution and rearranging formulas | Year 9

01 / LEARN

Understand the idea.

Substitution must preserve the rule

Substitution replaces each variable by its stated value without changing the structure of the formula. Put negative inputs in brackets: when $x = - 2$ , $3 x^{2}$ becomes $3 {(- 2)}^{2}$ . The exponent acts on the complete input.

Apply the order of operations after substituting: powers first, then multiplication and division, then addition and subtraction. Keep track of the quantities and their units; an area has square units, while a side length has length units.

Keep an equation balanced

To make a variable the subject, isolate it by applying the same operation to both sides. Undo surrounding addition or subtraction before undoing multiplication or division. A fraction bar groups the entire numerator, so $\frac{v - u}{a}$ divides the whole difference.

Division requires a nonzero divisor. State that condition if a letter is used as the divisor. For a geometric triangle the base is positive, so dividing its area equation by the base is valid. Check a numerical result by substitution into the original formula.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Substitute a negative value

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

Worked solution

1
Replace both variables with bracketed inputs.
$2 x^{2} - 3 y = 2 {(- 3)}^{2} - 3 (4)$
The square applies to the whole negative input -3.
2
Evaluate the power before multiplying.
$2 (9) - 3 (4)$
The square of -3 is positive 9.
3
Multiply, then subtract.
$18 - 12 = 6$
The multiplication operations precede the subtraction.

Answer

$6$

EXAMPLE 02

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.

Worked solution

1
Multiply the area equation by 2.
$A = \frac{1}{2} b h ⟹ 2 A = b h$
Applying the same operation to both sides preserves equality.
2
Divide by the nonzero base.
$h = \frac{2 A}{b}, b \neq 0$
This isolates the height; a geometric triangle has a positive base.
3
Substitute the given area and base.
$h = \frac{2 (42)}{12} cm = 7 cm$
Square centimetres divided by centimetres gives centimetres.
4
Check the original area.
$\frac{1}{2} (12) (7) {cm}^{2} = 42 {cm}^{2}$
The calculated height reproduces the stated area.

Answer

$h = \frac{2 A}{b}$ ; $h = 7$ cm

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

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

Add 2 to both sides to obtain $y + 2 = 5 x$ , then divide both sides by 5. Thus $x = \frac{y + 2}{5}$ ; the whole sum is divided by 5.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Substitution must preserve the rule

Keep an equation balanced

See the method in action.

Substitute a negative value

Replace both variables with bracketed inputs.

Evaluate the power before multiplying.

Multiply, then subtract.

Rearrange triangle area

Multiply the area equation by 2.

Divide by the nonzero base.

Substitute the given area and base.

Check the original area.

Try it yourself.

Check your understanding

Keep practising