A2, A5AlgebraFoundation & HigherGrades 2–7

Substitution & Formulae

Q: What does "the subject" mean?

The letter on its own on one side of the formula. In A = lw, A is the subject.

A formula is a rule connecting quantities, such as $A = π r^{2}$ or $v = u + a t$ . Substituting means replacing the letters with numbers to find a value. Rearranging means changing which letter the formula is "for" — the subject — so you can find a different quantity.

40 min Video by Maths Genie AQA GCSE Maths

The big picture

Substitution is where arithmetic and algebra meet, so the order of operations decides everything. $3 x^{2}$ with $x = - 2$ is $3 \times 4 = 12$ , not $(- 6)^{2} = 36$ : the power acts before the multiplication.

Rearranging a formula uses exactly the same balancing as solving an equation. The only difference is that the answer is an expression in other letters rather than a single number.

Changing the Subject of a FormulaCurated by Kepler, not produced by us. Watch it, then read the notes below.

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What you'll learn

Substitute positive and negative numbers into expressions and formulae
Write a formula from a description in words
Rearrange formulae to change the subject
Rearrange formulae where the subject appears twice (Higher)

Substituting values

Replace each letter with its value $in brackets$ , then follow the order of operations: brackets, indices, division and multiplication, addition and subtraction.

Brackets around negative numbers stop sign errors: with $a = - 3$ , write $a^{2} = (- 3)^{2} = 9$ and $2 a = 2 \times (- 3) = - 6$ .

1Substitute:

4 (5) - 3 (- 2)

= 20 - (- 6) = 20 + 6

Answer

26

Your turnSame method, new numbers — try it before moving on.

Find the value of $5 m + 2 n$ when $m = - 3$ and $n = 4$ .

Your answer

Your turnSame method, new numbers — try it before moving on.

Find the value of $3 y^{2} + 4 y$ when $y = - 2$ .

Your answer

Tip — On a calculator, type negative numbers in brackets: $(- 3)^{2}$ gives 9, but $- 3^{2}$ gives $- 9$ .

Using and writing formulae

Real formulae work the same way. Substitute, calculate, and give units where the question uses them.

To write a formula from words, choose letters for the quantities and translate each phrase into an operation. "A taxi charges £3 plus £1.80 per mile" becomes $C = 3 + 1.8 m$ .

u t = 4 \times 3 = 12

\frac{1}{2} a t^{2} = \frac{1}{2} \times 10 \times 9 = 45

s = 12 + 45

Answer

s = 57

Your turnSame method, new numbers — try it before moving on.

Use $v^{2} = u^{2} + 2 a s$ to find $v^{2}$ when $u = 6$ , $a = 3$ and $s = 10$ .

Your answer

Changing the subject

The subject is the letter on its own on one side. To change it, undo the operations around the new subject in reverse order, doing the same to both sides — just like solving an equation.

Square roots undo squares, and squares undo square roots.

u

has

a t

added to it. Subtract

a t

from both sides.

Answer

u = v - a t

Your turnSame method, new numbers — try it before moving on.

Make $x$ the subject of $y = 4 x + 9$ .

Your answer

Use brackets for fractions, e.g. (y - 9)/4.

Your turnSame method, new numbers — try it before moving on.

Make $t$ the subject of $d = 5 t^{2}$ .

Your answer

Type √ as sqrt( ), e.g. sqrt(d/5).

Check with numbers

Test a rearrangement with numbers. In $y = 4 x + 9$ , $x = 2$ gives $y = 17$ . Put $y = 17$ into your answer: $\frac{17 - 9}{4} = 2$ ✓. It takes ten seconds and catches most errors.

Subject appearing twice (Higher)

If the new subject appears in two places, collect every term containing it on one side and everything else on the other. Then factorise so the subject appears once, and divide.

1Collect

x

terms on one side:

a x - b x = - 7 - 3

2Factorise:

x (a - b) = - 10

3Divide by

(a - b)

Answer

x = \frac{- 10}{a - b} = \frac{10}{b - a}

Your turnSame method, new numbers — try it before moving on.

Make $x$ the subject of $p (x + 3) = 2 x + 5$ .

Your answer

Use brackets, e.g. (5 - 3p)/(p - 2).

Think like an examiner

Write the substitution with brackets before calculating — it earns the method mark and makes sign errors visible.
When rearranging, show one operation per line. The final answer must have the subject alone on one side, with no copies of it on the other.
$x = \frac{y - 9}{4}$ and $x = \frac{y}{4} - \frac{9}{4}$ are both correct; $x = y - \frac{9}{4}$ is not.

Methods

(- 3)^{2} = 9, - 3^{2} = - 9

Brackets round negatives when substituting.

y = m x + c \Rightarrow x = \frac{y - c}{m}

Undo in reverse order.

x (a - b) = c \Rightarrow x = \frac{c}{a - b}

Subject twice: collect, factorise, divide.

Watch out for these

With

x = - 4

x^{2} = - 16

(- 4)^{2} = 16

— a square is never negative.

With

x = 3

2 x^{2} = 36

Square first:

2 \times 9 = 18

Making

x

the subject of

y = 3 x + 2

x = \frac{y}{3} - 2

Subtract 2 before dividing:

x = \frac{y - 2}{3}

Leaving

x

on both sides:

x = \frac{5 + x}{3}

Collect all

x

terms on one side and factorise.

Stretch yourself

The formula $T = 2 π \frac{L}{g}$ gives the time for a pendulum swing. Make $L$ the subject, then find $L$ when $T = 2$ and $g = 9.8$ , to 3 significant figures.

Hint — Divide by $2 π$ first, then square both sides to remove the root.

Questions students ask

What does "the subject" mean?

The letter on its own on one side of the formula. In

A = l w

A

is the subject.

Is rearranging the same as solving?

It uses the same balancing steps. Solving gives a number; rearranging gives an expression in the other letters.

Do I need a

\pm

when I square root?

In a formula about a real quantity, like a length or time, take the positive root. In a pure equation such as

x^{2} = 9

, both

3

and

- 3

are solutions.

Key takeaways

Substitute with brackets and follow the order of operations.
A formula from words: choose letters, then translate each phrase.
To change the subject, undo operations in reverse order on both sides.
Square roots undo squares.
If the subject appears twice, collect its terms, factorise, then divide.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 32 marks

$P = 3 a^{2} - 2 b$ . Work out the value of $P$ when $a = - 4$ and $b = 7$ .[2]

Question 2Grade 54 marks

(a)Make $h$ the subject of $V = \frac{1}{3} π r^{2} h$ .[2]
(b)A cone has volume 300 cm³ and radius 5 cm. Work out its height, to 1 decimal place.[2]

Question 3Grade 73 marks

Make $a$ the subject of $\frac{a + 4}{3} = \frac{2 a - b}{5}$ .[3]

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Substitution & Formulae

Substituting values

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Subject appearing twice (Higher)

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Substitution & Formulae

Substituting values

Using and writing formulae

Changing the subject

Subject appearing twice (Higher)

Practise it the way it is examined

Skill drill

Practise this topic

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