GCSE Maths · Algebra

Rearranging formulae GCSE Questions and Worked Answers

Rearranging a formula expresses a chosen letter on its own. Apply the same operation to both sides, undoing what is done to that letter, and keep any restrictions on division or roots.

Foundation & Higher6 worked examples10 original questions

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Start with the meaning

What you need to know about rearranging formulae

A rectangle 5 cm long and 3 cm wide has area 15 cm². If you know its area and length but not its width, you calculate 15 ÷ 5 = 3 cm. The same idea works for any rectangle: A = lw means area equals length times width. Dividing by the length gives w = A/l. We have rewritten the relationship so that width is the quantity calculated directly.

See the idea first

Same relationship, different question

The subject is the letter isolated on one side of the equals sign. In A = lw, A is the subject. To make w the subject, undo multiplication by l.

Area ruleA = lwl beside w means l × w

Divide both sides by lA/l = wl must not be zero

Read the width rulew = A/lswapping sides does not change equality

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Undo addition and multiplication

What the problem asks: Make x the subject of y = 3x + 4.

How to solve it: Subtract 4 from both sides, then divide the whole result by 3: x = (y − 4)/3.

Remove a denominator

What the problem asks: Make t the subject of v = d/t, where v and t are non-zero.

How to solve it: Multiply by t to get vt = d, then divide by v: t = d/v.

Undo a power

What the problem asks: Make r the subject of A = πr² for a circle.

How to solve it: Divide by π, then take the positive square root: r = √(A/π). A radius is positive; an unrestricted squared variable can need both signs.

Collect a repeated subject

What the problem asks: Make x the subject of y = ax + bx.

How to solve it: Both right-hand terms contain x. Factorise to y = x(a + b), then divide by a + b, provided it is non-zero.

A reliable routine

When the subject occurs once in a sequence of operations

Undo the operations in reverse order, applying each inverse to both sides. This preserves equality. If the subject occurs more than once, collect and factorise its terms first rather than applying this routine blindly.

Choose the letter that must end up alone.
Identify the last operation applied to the expression containing it.
Undo that operation on both complete sides, using brackets or a fraction bar to preserve grouping.
Repeat, stating non-zero denominators and choosing root signs from the question's context.

Check: Do not replace the other letters with invented numbers. The purpose is to give a rule that works for all allowed values, not just one example.

Fully worked

Rearranging formulae GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Undo a subtracted term

2 marks

Question

Make x the subject of y = 5x − 7.

The formula multiplies x by 5, then subtracts 7. Undo that order: first add 7 to both sides. $y+7=5x$ Divide both sides by 5, keeping y + 7 together.

x=\frac{y+7}{5}

The numerator needs both y and 7.

Example 2

A bracket

2 marks

Question

Make x the subject of p = 4(x + 3).

Divide both sides by 4 before subtracting 3.

\frac p4=x+3

x=\frac p4-3

Expanding first would also work.

Example 3

Subject in the denominator

3 marks

Question

Make t the subject of $s=12/t$ , where s and t are non-zero.

Multiply both sides by t.

st=12

Divide by s.

t=\frac{12}{s}

The denominator cannot be zero.

Example 4

Radius from area

3 marks

Question

A circle has area $A=\pi r^2$ . Make r the subject, for a positive radius.

Divide by π.

r^2=\frac A\pi

A positive radius selects the positive root.

r=\sqrt{\frac A\pi}

The square root covers the whole quotient.

Example 5

A negative coefficient

3 marks

Question

Make t the subject of v = u − at, where a is non-zero.

Subtract u, then divide by −a.

v-u=-at

t=\frac{v-u}{-a}

Equivalently, multiply numerator and denominator by −1.

t=\frac{u-v}{a}

Both forms are correct.

Example 6

Subject on both sides

Higher only4 marks

Question

Make x the subject of y = ax + b − cx, assuming a ≠ c.

Subtract b, then factorise the x terms.

y-b=ax-cx

y-b=x(a-c)

Divide by a − c, which is non-zero.

x=\frac{y-b}{a-c}

If a = c, the original equation instead reduces to y = b and does not determine a unique x.

10 original questions · total 22 marks

Rearranging formulae GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 27 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed

Addition

1 mark

Make a the subject of p = a + b.

Show worked answer

Subtract b from both sides.

a=p-b

Multiplication

1 mark

Make h the subject of V = Ah, where A ≠ 0.

Show worked answer

Divide by A.

h=\frac VA

Two steps

2 marks

Make x the subject of y = 2x + 9.

Show worked answer

y-9=2x

x=\frac{y-9}{2}

Subtract before dividing.

A bracket

2 marks

Make a the subject of P = 3(a − 4).

Show worked answer

\frac P3=a-4

a=\frac P3+4

Divide the whole side by 3.

Triangle height

2 marks

Make h the subject of $A=\frac12bh$ , where b > 0.

Show worked answer

Multiply by 2, then divide by b.

2A=bh

h=\frac{2A}{b}

Denominator

2 marks

Make x the subject of y = 7/x, where x and y are non-zero.

Show worked answer

xy=7

x=\frac7y

Multiply by x before dividing by y.

Square with context

3 marks

A square of positive side length s has area A = s². Make s the subject.

Show worked answer

Take the positive root because a length is positive.

s=\sqrt A

An unrestricted equation s² = A would also allow the negative root when A > 0.

Reverse subtraction

2 marks

Make x the subject of y = 10 − 2x.

Show worked answer

Subtract 10, then divide both sides by −2. $y-10=-2x$ $x=\frac{y-10}{-2}$ Multiplying numerator and denominator by −1 gives $x=\frac{10-y}{2}$

Repeated letter

Higher only3 marks

Make t the subject of p = at + bt, where a + b ≠ 0.

Show worked answer

p=t(a+b)

t=\frac p{a+b}

Factorising gathers the two t terms.

Repeated subject with fraction

Higher only4 marks

Make x the subject of $y=\dfrac{x+2}{x-1}$ . State restrictions.

Show worked answer

The original formula requires x ≠ 1. Multiply by x − 1.

y(x-1)=x+2

yx-y=x+2

yx-x=y+2

x(y-1)=y+2

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

This requires y ≠ 1. The original equation cannot have y = 1, since that would require x + 2 = x − 1.

Examiner-style feedback

Common rearranging formulae mistakes

Dividing only one term

From y − 4 = 3x, divide the whole y − 4 by 3. The fraction bar must cover both terms.

Taking a root too early

From A = πr², first isolate r². A square root does not undo multiplication by π.

Ignoring zero denominators

Dividing by an expression assumes it is non-zero. State or use the restrictions given in the question.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

The subject is the isolated letter.
Preserve equality on both sides.
Use brackets to keep groups intact.
Factorise a repeated subject before division.

Quick answers

Rearranging formulae FAQ

Is changing the subject the same as solving an equation?

It uses the same equality rules, but usually leaves a formula in other letters rather than one numerical answer.

Can different rearranged answers both be correct?

Yes. For example, (u − v)/a and (v − u)/(−a) are equivalent. Check grouping and signs, not just appearance.

Build connected skills

What to revise next

Practise balancing

Linear equations

Apply inverse operations to both sides.

Revise linear equations

Gather a repeated subject

Factorising

Rewrite a sum as a product.

Revise factorising

Content standards

Curriculum and rights review

A5: changing the subject across tiers, with labelled Higher extensions involving a repeated subject and an algebraic denominator. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.

Official specification references

AQA GCSE Maths subject content