J560 · 6.03–6.04Equations & InequalitiesFoundation & HigherGrades 4–8

Rearranging Formulae

The formula $v = u + at$ tells you the final velocity. But if you know $v$ , $u$ and $t$ and want the acceleration, you need $a$ on its own: $a = \dfrac{v - u}{t}$ . Making a different letter the $\textbf{subject}$ uses exactly the same inverse operations as solving an equation — the only difference is that the answer is an expression, not a number.

45 min Equations & Inequalities

The big picture

Rearranging is solving an equation where every other quantity is still a letter. Undo the operations around the new subject in reverse order, one step at a time, applying each step to the whole of both sides.

The Higher tier adds two harder cases: formulae with the subject inside a square or root, and formulae where the subject appears in two places. The second needs a new technique — collect the subject’s terms together and factorise it out — and OCR sets it most years.

GCSE Maths - How to Rearrange Formulas (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Before you start

Try a short calculation from the preceding topic without the notes.
Check the tier and grade band shown in this lesson.

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Change the subject of a formula using inverse operations

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Make $x$ the subject of $y = 4x - 7$ .

Show answer and reasoning

$x = \dfrac{y + 7}{4}$ . Add 7, then divide by 4.

After watching 2: Make $r$ the subject of $C = 2\pi r$ .

Show answer and reasoning

$r = \dfrac{C}{2\pi}$ . Divide by $2\pi$ .

What you'll learn

Change the subject of a formula using inverse operations
Rearrange formulae involving fractions, powers and roots
Rearrange formulae where the subject appears more than once (Higher)
Substitute into a rearranged formula

One and two steps

Identify what has been done to the new subject, then undo it in reverse order. In $P = 2l + 2w$ , to make $w$ the subject: subtract $2l$ , then divide by 2.

$w = \dfrac{P - 2l}{2}$ .

Write the new subject on the left at the end.

Example 1Make

u

the subject of

v^{2} = u^{2} + 2as

1Subtract

2as

u^{2} = v^{2} - 2as

2Square root both sides.

Answer

u = \sqrt{v^{2} - 2as}

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

Make $a$ the subject of $v^2 = u^2 + 2as$ .

Your answer

Type it like (v^2-u^2)/(2s).

Tip — When taking a square root in context — a speed, a length — the positive root is normally intended. Write $\pm$ only if negative values make sense.

Fractions, powers and roots

If the subject is in a denominator, multiply both sides by that denominator first to bring it to the top.

Undo a square with a square root, and a square root by squaring. Undo a cube with a cube root.

Square or root the $\textit{whole}$ side, not term by term.

Example 1Make

l

the subject of

T = 2\pi\sqrt{\dfrac{l}{g}}

1Divide by

2\pi

\dfrac{T}{2\pi} = \sqrt{\dfrac{l}{g}}

2Square both sides:

\dfrac{T^{2}}{4\pi^{2}} = \dfrac{l}{g}

3Multiply by

g

Answer

l = \dfrac{gT^{2}}{4\pi^{2}}

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

Make $r$ the subject of $A = 4\pi r^2$ .

Your answer

Type √ as sqrt( ) and π as pi.

Why this works

Squaring $\dfrac{T}{2\pi}$ squares both the $T$ and the $2\pi$ , giving $4\pi^{2}$ on the bottom. Squaring a side means squaring everything on it.

Subject appearing twice

If the subject appears in more than one term: clear any fractions, expand brackets, collect all terms containing the subject on one side, factorise it out, then divide.

This is the only case where factorising is needed, and it is how you recognise the method.

Example 1Make

x

the subject of

y = \dfrac{3x + 2}{x - 4}

1Multiply by

(x - 4)

y(x - 4) = 3x + 2

, so

xy - 4y = 3x + 2

2Collect

x

terms:

xy - 3x = 4y + 2

3Factorise:

x(y - 3) = 4y + 2

4Divide by

(y - 3)

Answer

x = \dfrac{4y + 2}{y - 3}

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

Make $x$ the subject of $y = \dfrac{2x - 1}{x + 3}$ .

Your answer

Type it like (3y+1)/(2-y).

Tip — After collecting, every term with the subject must be on one side and every term without it on the other. If a subject term is left behind, the factorising step will not work.

Using a rearranged formula

Often a question asks you to rearrange and then substitute values. Rearranging first gives a formula you can reuse, and makes each substitution a single calculation.

Alternatively, substitute first and solve the resulting equation — both methods are valid for a single calculation.

Example 1The volume of a cone is

V = \tfrac13\pi r^{2}h

. Find the radius of a cone with volume 300 cm³ and height 12 cm.

1Rearrange:

r = \sqrt{\dfrac{3V}{\pi h}}

2Substitute:

r = \sqrt{\dfrac{900}{12\pi}} = \sqrt{23.87}

Answer

r = 4.89

cm (3 s.f.)

Think like an examiner

Show each inverse operation on a separate line.
Give the final answer with the new subject alone on the left.
When the subject appears twice, show the factorised line clearly.

Rearranging

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

Two-step rearrangement.

A = \pi r^{2} \Rightarrow r = \sqrt{\tfrac{A}{\pi}}

Undoing a square.

ax - bx = c \Rightarrow x = \tfrac{c}{a - b}

Factorising out the subject.

Watch out for these

From

y = 3x + 5

, writing

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

Undo in reverse order: subtract 5 first,

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

Square-rooting term by term:

\sqrt{v^{2} - 2as} = v - \sqrt{2as}

Root the whole side as one expression.

Dividing by

y

while

x

still appears on both sides.

Collect

x

terms and factorise first.

Forgetting to multiply every term when clearing a fraction.

Multiply both entire sides.

Stretch yourself

Make $a$ the subject of $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{c}$ .

Hint — Multiply through by $abc$ to clear all the fractions, then collect the $a$ terms.

Questions students ask

How is rearranging different from solving?

The steps are identical. The difference is that the other quantities stay as letters, so the result is an expression.

When do I need to factorise?

Only when the subject appears in more than one term after clearing fractions and brackets.

Do I need

\pm

when square rooting?

Mathematically both roots satisfy the formula; in context, choose the sign that makes sense and say why if asked.

Key takeaways

Undo operations around the new subject in reverse order.
Clear fractions by multiplying both sides by the denominator.
Square or root whole sides, never term by term.
If the subject appears twice: collect, factorise, divide.
Rearrange once, then substitute as often as needed.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 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 42 marks

Make $t$ the subject of $v = u + at$ .[2]

Question 2Grade 62 marks

Make $x$ the subject of $5(x - a) = 3x + 2a$ .[2]

Question 3Grade 83 marks

Make $p$ the subject of $a = \dfrac{p + 2}{p - 5}$ .[3]

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…

Explore further

OCR J560 official specification (opens in a new tab)
Check the assessed scope and practical or mathematical requirements.
Read or explore this topic (opens in a new tab)
An external explanation or interactive activity to use alongside the lesson. It may include extension material.

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Make $h$ the subject of $A = \dfrac{(a + b)h}{2}$ .
Make $x$ the subject of $y = \sqrt{x} + 3$ .
Make $t$ the subject of $at + b = ct + d$ .
Make $x$ the subject of $y = \dfrac{x}{x + 2}$ .

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 4 lessons

Rearranging Formulae

The formula

v = u + at

tells you the final velocity. But if you know

v

u

and

t

and want the acceleration, you need

a

on its own:

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

. Making a different letter the

\textbf{subject}

uses exactly the same inverse operations as solving an equation — the only difference is that the answer is an expression, not a number.

45 min Equations & Inequalities

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Change the subject of a formula using inverse operations

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: Make $x$ the subject of $y = 4x - 7$ .

Show answer and reasoning

$x = \dfrac{y + 7}{4}$ . Add 7, then divide by 4.

After watching 2: Make $r$ the subject of $C = 2\pi r$ .

Show answer and reasoning

$r = \dfrac{C}{2\pi}$ . Divide by $2\pi$ .

Practise it the way it is examined

Question 1Grade 42 marks

Make $t$ the subject of $v = u + at$ .[2]

Question 2Grade 62 marks

Make $x$ the subject of $5(x - a) = 3x + 2a$ .[2]

Question 3Grade 83 marks

Make $p$ the subject of $a = \dfrac{p + 2}{p - 5}$ .[3]

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Make

h

the subject of

A = \dfrac{(a + b)h}{2}

Make

x

the subject of

y = \sqrt{x} + 3

Make

t

the subject of

at + b = ct + d

Make

x

the subject of

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