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 $A = \pi r^2$ or $v = u + at$ . 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. $3x^2$ 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.

Open on YouTube

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 $\textbf{in brackets}$ , 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 $2a = 2 \times (-3) = -6$ .

Example 1Find the value of

4p - 3q

when

p = 5

and

q = -2

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 $5m + 2n$ when $m = -3$ and $n = 4$ .

Your answer

Show worked solution

Solution

1 $5(-3) + 2(4)$ .
2 $= -15 + 8 = -7$ .

Answer: $-7$

Example 2Find the value of

2x^2 - 5x

when

x = -3

1Substitute:

2(-3)^2 - 5(-3)

2Indices first:

(-3)^2 = 9

, so

2 \times 9 = 18

-5 \times -3 = +15

, so the total is

18 + 15

Answer

33

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

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

Your answer

Show hint

Square first: $(-2)^2 = 4$ .

Show worked solution

Solution

1 $3(-2)^2 + 4(-2)$ .
2 $= 3 \times 4 - 8 = 12 - 8 = 4$ .

Answer: $4$

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 $C = 3 + 1.8m$ .

Example 1Use

s = ut + \frac{1}{2}at^2

to find

s

when

u = 4

a = 10

and

t = 3

ut = 4 \times 3 = 12

\frac{1}{2}at^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 $v^2 = u^2 + 2as$ to find $v^2$ when $u = 6$ , $a = 3$ and $s = 10$ .

Your answer

Show worked solution

Solution

1 $u^2 = 36$ and $2as = 2 \times 3 \times 10 = 60$ .
2 $v^2 = 36 + 60 = 96$ .

Answer: $v^2 = 96$

Example 2A plumber charges a £45 call-out fee plus £38 per hour. Write a formula for the total cost

C

pounds for

h

hours, and find the cost of a 3.5-hour job.

1Cost

= 45 + 38 \times

hours, so

C = 45 + 38h

C = 45 + 38 \times 3.5 = 45 + 133

Answer

C = 45 + 38h

; the job costs £178

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.

Example 1Make

u

the subject of

v = u + at

u

has

at

added to it. Subtract

at

from both sides.

Answer

u = v - at

Example 2Make

x

the subject of

y = \dfrac{3x - 4}{5}

1Multiply by 5:

5y = 3x - 4

2Add 4:

5y + 4 = 3x

3Divide by 3.

Answer

x = \dfrac{5y + 4}{3}

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

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

Your answer

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

Show worked solution

Solution

1Subtract 9: $y - 9 = 4x$ .
2Divide by 4: $x = \dfrac{y - 9}{4}$ .

Answer: $x = \dfrac{y - 9}{4}$

Example 3Make

r

the subject of

A = \pi r^2

1Divide by

\pi

\dfrac{A}{\pi} = r^2

2Square root both sides.

Answer

r = \sqrt{\dfrac{A}{\pi}}

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

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

Your answer

Type √ as sqrt( ), e.g. sqrt(d/5).

Show worked solution

Solution

1Divide by 5: $t^2 = \dfrac{d}{5}$ .
2Square root: $t = \sqrt{\dfrac{d}{5}}$ .

Answer: $t = \sqrt{\dfrac{d}{5}}$

Check with numbers

Test a rearrangement with numbers. In $y = 4x + 9$ , $x = 2$ 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.

Example 1Make

x

the subject of

ax + 3 = bx - 7

1Collect

x

terms on one side:

ax - bx = -7 - 3

2Factorise:

x(a - b) = -10

3Divide by

(a - b)

Answer

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

Example 2Make

x

the subject of

y = \dfrac{x + 2}{x - 1}

1Multiply by

(x - 1)

y(x - 1) = x + 2

, so

xy - y = x + 2

2Collect

x

terms:

xy - x = y + 2

3Factorise:

x(y - 1) = y + 2

Answer

x = \dfrac{y + 2}{y - 1}

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

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

Your answer

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

Show hint

Expand, then get both $x$ terms on the left.

Show worked solution

Solution

1Expand: $px + 3p = 2x + 5$ .
2Collect: $px - 2x = 5 - 3p$ .
3Factorise: $x(p - 2) = 5 - 3p$ , so $x = \dfrac{5 - 3p}{p - 2}$ .

Answer: $x = \dfrac{5 - 3p}{p - 2}$

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, \quad -3^2 = -9

Brackets round negatives when substituting.

y = mx + c \Rightarrow x = \dfrac{y - c}{m}

Undo in reverse order.

x(a - b) = c \Rightarrow x = \dfrac{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

2x^2 = 36

Square first:

2 \times 9 = 18

Making

x

the subject of

y = 3x + 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 $T = 2\pi\sqrt{\dfrac{L}{g}}$ 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 $2\pi$ first, then square both sides to remove the root.

Reveal solution

Solution

\frac{T}{2\pi} = \sqrt{\frac{L}{g}}

, so

\frac{T^2}{4\pi^2} = \frac{L}{g}

and

L = \frac{gT^2}{4\pi^2}

. With

T = 2

and

g = 9.8

L = \frac{9.8 \times 4}{4\pi^2} = \frac{9.8}{\pi^2} = 0.993

(3 s.f.).

\blacksquare

Questions students ask

What does "the subject" mean?

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

A = lw

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

$P = 3a^2 - 2b$ . Work out the value of $P$ when $a = -4$ and $b = 7$ .[2]
Show mark scheme
- M1Correct substitution, e.g. $3 \times (-4)^2 - 2 \times 7$ or $3 \times 16 - 14$
- A1 $34$
Answer: $34$

Question 2Grade 54 marks

(a)Make $h$ the subject of $V = \frac{1}{3}\pi r^2 h$ .[2]
Show mark scheme (a)
- M1A correct first step, e.g. $3V = \pi r^2 h$
- A1 $h = \dfrac{3V}{\pi r^2}$
Answer: $h = \dfrac{3V}{\pi r^2}$
(b)A cone has volume 300 cm³ and radius 5 cm. Work out its height, to 1 decimal place.[2]
Show mark scheme (b)
- M1Substitutes into their rearrangement: $\dfrac{900}{25\pi}$
- A1 $11.5$ cm (11.45…)
Answer: $11.5$ cm

Question 3Grade 73 marks

Make $a$ the subject of $\dfrac{a + 4}{3} = \dfrac{2a - b}{5}$ .[3]
Show mark scheme
- M1Removes fractions: $5(a + 4) = 3(2a - b)$
- M1Expands and collects $a$ terms: e.g. $5a - 6a = -3b - 20$
- A1 $a = 3b + 20$
Answer: $a = 3b + 20$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

Find $3a + b$ when $a = 4$ and $b = -5$ .
- $7$
- $17$
- $-7$
- $2$
Show answer and explanation
Answer: $7$
$12 + (-5) = 7$ .
Find $x^2$ when $x = -6$ .
- $36$
- $-36$
- $-12$
- $12$
Show answer and explanation
Answer: $36$
$(-6) \times (-6) = 36$ .
Find $4x^2$ when $x = 3$ .
- $36$
- $144$
- $24$
- $48$
Show answer and explanation
Answer: $36$
Square first: $4 \times 9$ .
Use $F = \frac{9}{5}C + 32$ to convert $C = 25$ .
- $77$
- $45$
- $57$
- $89$
Show answer and explanation
Answer: $77$
$\frac{9}{5} \times 25 = 45$ , then add 32.
Make $a$ the subject of $P = 2a + 2b$ .
- $a = \frac{P - 2b}{2}$
- $a = \frac{P}{2} - 2b$
- $a = P - b$
- $a = 2P - 2b$
Show answer and explanation
Answer: $a = \frac{P - 2b}{2}$
Subtract $2b$ , then divide by 2.
Make $r$ the subject of $C = 2\pi r$ .
- $r = \frac{C}{2\pi}$
- $r = C - 2\pi$
- $r = 2\pi C$
- $r = \frac{2\pi}{C}$
Show answer and explanation
Answer: $r = \frac{C}{2\pi}$
Divide both sides by $2\pi$ .
Make $x$ the subject of $y = \sqrt{x + 3}$ .
- $x = y^2 - 3$
- $x = \sqrt{y} - 3$
- $x = (y - 3)^2$
- $x = y^2 + 3$
Show answer and explanation
Answer: $x = y^2 - 3$
Square both sides, then subtract 3.
Make $x$ the subject of $kx + 1 = 3x + m$ .
- $x = \frac{m - 1}{k - 3}$
- $x = \frac{m + 1}{k + 3}$
- $x = \frac{m - 1}{k}$
- $x = m - 1 - k + 3$
Show answer and explanation
Answer: $x = \frac{m - 1}{k - 3}$
$kx - 3x = m - 1$ , so $x(k - 3) = m - 1$ .

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