Make $h$ the subject of $A=2\pi r^2+2\pi rh$.

Subtract $2\pi r^2$: $A-2\pi r^2=2\pi rh$. Divide by $2\pi r$: $h=\dfrac{A-2\pi r^2}{2\pi r}$. Answer: $h=\dfrac{A-2\pi r^2}{2\pi r}$

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Edexcel IGCSE Maths (4MA1) · Unit 2: Algebra Basics

Edexcel IGCSE 4MA1 Rearranging Formulae: exam questions and answers, notes and worked examples

Updated October 2026 · By Pete Bromfield

To change the subject of a formula, undo what has been done to the new subject in reverse order, doing the same to both sides each time. If the new subject appears twice, collect those terms on one side and factorise it out: from $p x + 4 = q x$ you get $x = \frac{4}{q - p}$ . Substituting into a formula is the reverse job: replace each letter with its value, in brackets if it is negative.

Unit 2: Algebra Basics
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Notes
Worked examples
Mistakes
FAQ

Rearranging Formulae notes

Substituting into a formula

Replace each letter with its value and keep negative numbers in brackets. With $v = u + a t$ , $u = 12$ , $a = - 3$ and $t = 5$ : $v = 12 + (- 3) (5) = - 3$ . Work out powers before multiplying: in $E = \frac{1}{2} m v^{2}$ with $m = 4$ and $v = 3$ , $E = \frac{1}{2} \times 4 \times 9 = 18$ .

One-step and two-step rearranging

Undo the operations in reverse order. For $y = 5 x + 2$ , subtract 2 and then divide by 5: $x = \frac{y - 2}{5}$ . For $P = \frac{a + b}{2}$ , multiply by 2 and then subtract $b$ : $a = 2 P - b$ .

Powers and roots

Undo a square with a square root, and keep $\pm$ unless the context rules out a negative value, as it does for a radius: $A = π r^{2}$ gives $r = \sqrt{\frac{A}{π}}$ . Undo a square root by squaring both sides: $T = 2 \sqrt{L}$ gives $L = \frac{T^{2}}{4}$ .

When the subject appears twice

Clear fractions and expand brackets, collect every term containing the new subject on one side, then factorise it out and divide. For $t = \frac{3 s + 1}{s - 2}$ : $t s - 2 t = 3 s + 1$ , so $t s - 3 s = 2 t + 1$ , $s (t - 3) = 2 t + 1$ and $s = \frac{2 t + 1}{t - 3}$ .

Rearranging Formulae worked examples

Worked example 1: Subject inside a term

Make $h$ the subject of $A = 2 π r^{2} + 2 π r h$ .

Solution

Subtract $2 π r^{2}$ : $A - 2 π r^{2} = 2 π r h$ .
Divide by $2 π r$ : $h = \frac{A - 2 π r^{2}}{2 π r}$ .

3 steps

Tip: Isolate the term containing h first.

Worked example 2: Subject appears twice

Make $x$ the subject of $a x - b = c x + d$ .

Solution

Collect $x$ terms: $a x - c x = b + d$ .
Factorise: $x (a - c) = b + d$ .
$x = \frac{b + d}{a - c}$ .

4 steps

Tip: When the subject appears twice, collect and factorise.

Worked example 3: Subject on the top and bottom of a fraction · 3 marks

Make $m$ the subject of $k = \frac{2 m - 3}{m + 4}$ .

Mark scheme

M1 km + 4k = 2m − 3
M1 km − 2m = −3 − 4k and factorises m(k − 2)
A1 m = (4k + 3)/(2 − k) oe

Answer: $m = \frac{4 k + 3}{2 - k}$

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Rearranging Formulae exam-style questions

Try each question before you open the answer (one has its full worked solution open). The mark schemes use Edexcel-style marks: M for method, A for accuracy (after the method mark), B for an independent result.

Question 1 · easy · 2 marks

Make $x$ the subject of $a x + b = c x$ .

Show the answer and mark scheme

M1 collects x terms on one side and factorises, e.g. x(c − a) = b
A1 x = b/(c − a) oe

Answer: $x = \frac{b}{c - a}$

Question 2 · easy · 2 marks

Make $a$ the subject of $P = 2 a + a b$ .

Show the answer and mark scheme

M1 P = a(2 + b)
A1 a = P/(2 + b)

Answer: $a = \frac{P}{2 + b}$

Question 3 · medium · 3 marks

Make $p$ the subject of $3 p + q = p r - 2$ .

Full worked solution Free, no sign-in

M1 collects p terms, e.g. 3p − pr = −2 − q
M1 factorises p(3 − r)
A1 p = (q + 2)/(r − 3) oe

Answer: $p = \frac{q + 2}{r - 3}$

Every question here has a full worked solution. See them all →

Question 4 · medium · 3 marks

Make $x$ the subject of $y = \frac{x + 2}{x - 5}$ .

Show the answer and mark scheme

M1 y(x − 5) = x + 2
M1 xy − x = 5y + 2 and factorises x(y − 1)
A1 x = (5y + 2)/(y − 1) oe

Answer: $x = \frac{5 y + 2}{y - 1}$

Question 5 · hard · 3 marks

Make $m$ the subject of $E = \frac{m (v + u)}{m + 2}$ .

Show the answer and mark scheme

M1 Em + 2E = mv + mu
M1 factorises m(E − v − u) = −2E oe
A1 m = 2E/(v + u − E) oe

Answer: $m = \frac{2 E}{v + u - E}$

Question 6 · hard · 4 marks

Make $t$ the subject of $v = \sqrt{\frac{t + a}{t - b}}$ .

Show the answer and mark scheme

M1 squares both sides: v² = (t + a)/(t − b)
M1 v²t − v²b = t + a
M1 t(v² − 1) = a + bv²
A1 t = (a + bv²)/(v² − 1) oe

Answer: $t = \frac{a + b v^{2}}{v^{2} - 1}$

Question 7 · very hard · 3 marks

Make $x$ the subject of $\frac{1}{x} + \frac{1}{y} = \frac{1}{f}$ .

Show the answer and mark scheme

M1 multiplies through by xyf: fy + fx = xy
M1 fy = x(y − f)
A1 x = fy/(y − f) oe

Answer: $x = \frac{f y}{y - f}$

Common rearranging formulae mistakes

Square rooting term by term. $\sqrt{v^{2} - 2 a s}$ is not $v - \sqrt{2 a s}$ . Square root the whole side as one.
Dividing only one term. Dividing $5 x + 2 = y$ by 5 gives $x + \frac{2}{5} = \frac{y}{5}$ : every term is divided. Subtract the 2 first and the division is simpler.
Leaving the subject on both sides. $x = \frac{b + c x}{a}$ is not finished, because $x$ is still on the right. Collect the $x$ terms and factorise.
Losing a minus sign. From $2 y - a y = - 5$ , $y (2 - a) = - 5$ , so $y = \frac{- 5}{2 - a}$ , which is the same as $y = \frac{5}{a - 2}$ . Both are right; check the sign by substituting a number.

Where students lose marks across 4MA1, from the examiner reports →

Rearranging Formulae FAQ

How do I rearrange when the subject appears twice?

Get rid of fractions and brackets, move every term containing the new subject to one side and everything else to the other, then factorise the subject out and divide by the bracket.

Do I need the ± sign when I take a square root?

Keep ± unless the context makes a negative value impossible, such as a length, a radius or a time. If in doubt, keep it.

Can I use a calculator for rearranging formulae in 4MA1?

Yes. A calculator may be used in all four 4MA1 papers, so both Higher papers (1H and 2H) allow one. Most marks are for the method you write down, so show your working.

How long are the 4MA1 Higher papers?

Papers 1H and 2H are 2 hours each and 100 marks each, worth 50% each. Each tier has its own formulae sheet in the exam.

Next steps

Rearranging Formulae revision notes · 4MA1 formula sheet · Taking Cambridge 0580 instead? Cambridge 0580 rearranging formulae questions

Every question and worked example here is original, written for IGCSE Math Revision rather than copied from Pearson papers, and every result was re-solved independently before publishing.

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Edexcel IGCSE 4MA1 Rearranging Formulae: exam questions and answers, notes and worked examples

Try it first

Rearranging Formulae notes

Substituting into a formula

One-step and two-step rearranging

Powers and roots

When the subject appears twice

Rearranging Formulae worked examples

Worked example 1: Subject inside a term

Solution

Worked example 2: Subject appears twice

Solution

Worked example 3: Subject on the top and bottom of a fraction · 3 marks

Mark scheme

Rearranging Formulae exam-style questions

Question 1 · easy · 2 marks

Question 2 · easy · 2 marks

Question 3 · medium · 3 marks

Full worked solution Free, no sign-in

Question 4 · medium · 3 marks

Question 5 · hard · 3 marks

Question 6 · hard · 4 marks

Question 7 · very hard · 3 marks

Common rearranging formulae mistakes

Rearranging Formulae FAQ

Next steps

Practise rearranging formulae with your progress saved

More Unit 2 topics: Algebra Basics