Rearranging Formulae · Study Notes · Edexcel IGCSE 4MA1 Higher

Make a different variable the subject. Apply inverse operations to both sides. Standard A* technique on every Higher Tier paper.

At a glance

Apply inverse operations to BOTH sides.
$+$ and $-$ are inverses; $\times$ and $\div$ are inverses; square and square root.
Subject on both sides: gather, factor, divide.
When squaring or square-rooting: do the WHOLE expression.
Show every step — Edexcel awards method marks for each.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.6 — Change the subject of a formula.
2.6 — Apply inverse operations consistently.
2.6 — Handle formulae where the subject appears on both sides.

The standard method

▼

Apply inverse operations to both sides until the desired variable is alone.

Goal: Get the desired variable on its own.

Standard procedure:

Identify all operations on the desired variable.
Apply them in REVERSE order, using INVERSE operations.
ALWAYS to both sides.

Inverse operations:

Operation	Inverse
Add $a$	Subtract $a$
Subtract $a$	Add $a$
Multiply by $a$	Divide by $a$
Divide by $a$	Multiply by $a$
Square ( $x^{2}$ )	Square root
Cube ( $x^{3}$ )	Cube root
Square root	Square
Reciprocal ( $1/x$ )	Reciprocal

Example. Make $x$ the subject of $y = 3x + 5$ .

The $x$ has: ' $\times 3$ ', then ' $+ 5$ '. Reverse order: subtract 5, then divide by 3.

Step 1: $y - 5 = 3x$ .
Step 2: $\dfrac{y - 5}{3} = x$ .

So $x = \dfrac{y - 5}{3}$ .

To change the subject, undo each operation in the reverse order with its inverse.

Example with power. Make $r$ the subject of $V = \dfrac{4}{3}\pi r^{3}$ .

$r^{3} = \dfrac{3V}{4\pi}$ .
$r = \sqrt[3]{\dfrac{3V}{4\pi}}$ .

Worked qualitative. Why does the order of inverses matter?

Original: $y = 3x + 5$ . To $x$ , we did ' $\times 3$ ' then ' $+ 5$ '.
To reverse, we 'unwind' in reverse: first undo $+ 5$ (subtract), then undo $\times 3$ (divide).
BIDMAS in reverse.

Edexcel tip. Show every step on its own line. Mark schemes credit each transformation.

Inverse operations to both sides.
Reverse order of original operations.
Show every step.
BIDMAS in reverse.

Powers and roots in rearranging

▼

Square root undoes squaring. Apply to the WHOLE side.

Squaring as inverse of square root.

If you have $\sqrt{...} =$ something, square BOTH sides to remove the root:

$\sqrt{x + 4} = y$ → $x + 4 = y^{2}$ .

If you have $x^{2} =$ something, take square root:

$x^{2} = 16$ → $x = \pm 4$ (technically two solutions, but in formula context usually positive).

Example. Make $x$ the subject of $y = \sqrt{x + 4} - 2$ .

Add $2$ : $y + 2 = \sqrt{x + 4}$ .
Square: $(y + 2)^{2} = x + 4$ .
Subtract $4$ : $x = (y + 2)^{2} - 4$ .

KEY POINT. When squaring or rooting, apply to the WHOLE expression — not term-by-term.

$(y + 2)^{2} \ne y^{2} + 4$ . $(y + 2)^{2} = y^{2} + 4y + 4$ .

Worked qualitative. Why doesn't $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ ?

Test: $\sqrt{9 + 16} = \sqrt{25} = 5$ , but $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$ .
Square root doesn't distribute over addition.

Edexcel tip. When you have $\sqrt{...}$ on one side, ISOLATE the root first, THEN square. Squaring an unisolated side won't help.

Square root: square both sides.
Square: take square root.
Apply to WHOLE expression.
Don't distribute roots over addition.

Subject on both sides

▼

Gather all subject terms on one side. Factor out. Divide.

The challenge: The variable you want as subject appears on BOTH sides.

Method:

Multiply out any fractions or brackets.
Move all terms WITH the desired variable to ONE side.
Move everything else to the OTHER side.
Factor out the desired variable from its side.
Divide.

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

Multiply by $(x - 2)$ : $y(x - 2) = x + 1$ .
Expand: $yx - 2y = x + 1$ .
Move $x$ terms to LHS, others to RHS: $yx - x = 1 + 2y$ .
Factor out $x$ : $x(y - 1) = 1 + 2y$ .
Divide: $x = \dfrac{1 + 2y}{y - 1}$ .

When the subject is on both sides: expand, gather its terms, factor it out, then divide.

Why this works. Once $x$ has been factored, you can isolate it cleanly with division.

Worked qualitative. Make $x$ the subject of $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$ .

Multiply through by $xyz$ : $yz + xz = xy$ .
Move $x$ terms: $xz - xy = -yz$ .
Factor: $x(z - y) = -yz$ .
Divide: $x = \dfrac{-yz}{z - y} = \dfrac{yz}{y - z}$ .

Edexcel tip. When you see the subject on both sides, MULTIPLY OUT FIRST. Mark schemes deduct for missed terms.

Multiply out brackets/fractions.
Gather subject terms.
Factor out subject.
Divide.

Quick recap

Apply inverse operations to both sides.
Reverse BIDMAS order.
Powers/roots: apply to whole expression.
Subject on both sides: gather, factor, divide.
Show every step.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Inverse operations to both sides.
Square root undoes square.
Square undoes square root.
Subject on both sides: gather, factor, divide.

How it’s examined

Rearranging formulae appears every Paper 1H + 2H (3-5 marks). Often combined with substitution. Examiner reports flag (1) one-sided operations, (2) wrong-sided distribution of squares, (3) failure to factor when subject is on both sides.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2023. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on rearranging formulae, written exactly the way a tutor would explain them at the board.

1Make $x$ the subject
Foundation• rearrange
▼
Question
Make $x$ the subject of $y = 3x + 5$ .
Step-by-step solution
1. Step 1
  Subtract $5$ from both sides: $y - 5 = 3x$ .
2. Step 2
  Divide by $3$ :
  $x = \frac{y - 5}{3}$
Answer
$x = \dfrac{y - 5}{3}$
2Rearrange with a power
Higher• Adapted from 4MA1/1H May/Jun 2023 Q12• rearrange, power
▼
Question
Make $r$ the subject of $V = \dfrac{4}{3}\pi r^{3}$ .
Step-by-step solution
1. Step 1
  Multiply both sides by $\dfrac{3}{4\pi}$ :
  $\frac{3V}{4\pi} = r^{3}$
2. Step 2
  Take cube root:
  $r = \sqrt[3]{\frac{3V}{4\pi}}$
Answer
$r = \sqrt[3]{\dfrac{3V}{4\pi}}$
Examiner tip
Reverse the operations one at a time, applying the inverse to both sides. Cube root undoes cubing.
3Subject appears on both sides
Higher• rearrange, factorise
▼
Question
Make $x$ the subject of $y = \dfrac{x + 1}{x - 2}$ .
Step-by-step solution
1. Step 1
  Multiply both sides by $(x - 2)$ :
  $y(x - 2) = x + 1$
2. Step 2
  Expand: $yx - 2y = x + 1$ .
3. Step 3
  Get $x$ on one side: $yx - x = 1 + 2y$ .
4. Step 4
  Factor out $x$ : $x(y - 1) = 1 + 2y$ .
5. Step 5
  Divide:
  $x = \frac{1 + 2y}{y - 1}$
Answer
$x = \dfrac{1 + 2y}{y - 1}$
Examiner tip
When the subject appears on both sides, gather it on one side, factor out, then divide. Always show this collection step.
4Rearrange with a square root
Higher• rearrange, square root
▼
Question
Make $x$ the subject of $y = \sqrt{x + 4} - 2$ .
Step-by-step solution
1. Step 1
  Add $2$ :
  $y + 2 = \sqrt{x + 4}$
2. Step 2
  Square BOTH sides:
  $(y + 2)^{2} = x + 4$
3. Step 3
  Subtract $4$ :
  $x = (y + 2)^{2} - 4$
Answer
$x = (y + 2)^{2} - 4$
Examiner tip
When squaring to remove a square root, square the WHOLE expression on both sides — including any constants.
5Undoing a cube [1 mark]
Getting started• Paper 1H/2H style — 1 mark• rearranging formulae, roots
▼
Question
Make $x$ the subject of $y = x^{3}$ . [1]
Step-by-step solution
1. Step 1
  The inverse of cubing is taking the cube root.
2. Step 2
  B1 Take the cube root of both sides.
  $x = \sqrt[3]{y}$
Answer
$x = \sqrt[3]{y}$ or $x = y^{\frac{1}{3}}$
Examiner tip
Examiner tip: A cube root has only one real value, so there is no $\pm$ here, unlike a square root.

Common mistake: Writing $x = \dfrac{y}{3}$ . Dividing by $3$ undoes multiplying by $3$ , not cubing.
6Perimeter of a rectangle: make w the subject [2 marks]
Getting started• Paper 1H/2H style — 2 marks• rearranging formulae, perimeter
▼
Question
The perimeter of a rectangle is $P = 2l + 2w$ .

Make $w$ the subject of the formula. [2]
Step-by-step solution
1. Step 1
  M1 Subtract $2l$ from both sides.
  $2w = P - 2l$
2. Step 2
  A1 Divide the whole of the right-hand side by $2$ .
  $w = \frac{P - 2l}{2} = \frac{P}{2} - l$
Answer
$w = \dfrac{P - 2l}{2}$ or $w = \dfrac{P}{2} - l$
Examiner tip
Examiner tip: Both forms are correct. If you split the fraction, divide every term by $2$ : $\dfrac{2l}{2} = l$ .

Common mistake: Writing $w = \dfrac{P}{2} - 2l$ , dividing only the first term. Examiners report splitting a fraction and dividing only one term as a frequent error.
7Motion formula: make t the subject [2 marks]
Getting started• Paper 1H/2H style — 2 marks• rearranging formulae, kinematics
▼
Question
Make $t$ the subject of $v = u + at$ . [2]
Step-by-step solution
1. Step 1
  M1 Subtract $u$ from both sides.
  $v - u = at$
2. Step 2
  A1 Divide both sides by $a$ .
  $t = \frac{v - u}{a}$
Answer
$t = \dfrac{v - u}{a}$
Examiner tip
Examiner tip: Undo the operations in the reverse order to how they were applied: $t$ was multiplied by $a$ , then $u$ was added, so subtract $u$ first, then divide by $a$ .

Common mistake: Dividing first and writing $t = \dfrac{v}{a} - u$ . Dividing both sides by $a$ divides every term, so the correct line is $\dfrac{v}{a} = \dfrac{u}{a} + t$ , giving $t = \dfrac{v}{a} - \dfrac{u}{a}$ .
8Two fractions equal to each other [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• rearranging formulae, algebraic fractions
▼
Question
Make $x$ the subject of $\dfrac{x + 3}{4} = \dfrac{2x - y}{5}$ . [3]
Step-by-step solution
1. Step 1
  M1 Multiply both sides by $20$ , keeping each numerator in a bracket.
  $5(x + 3) = 4(2x - y) \;\Rightarrow\; 5x + 15 = 8x - 4y$
2. Step 2
  M1 Collect the $x$ terms on the side with the larger coefficient, so they stay positive.
  $15 + 4y = 8x - 5x = 3x$
3. Step 3
  A1 Divide by $3$ .
  $x = \frac{15 + 4y}{3}$
Answer
$x = \dfrac{15 + 4y}{3}$ oe
Examiner tip
Examiner tip: Check with $y = 0$ : the formula gives $x = 5$ , and then both sides of the original are $2$ .

Common mistake: Writing $5x + 3 = 8x - y$ , multiplying only the first term of each numerator. The brackets are essential when you clear fractions.
9Area of a ring: make r the subject [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• rearranging formulae, square roots
▼
Question
The area of a ring between two circles of radii $R$ and $r$ is $A = \pi(R^{2} - r^{2})$ .

Make $r$ the subject of the formula, where $r > 0$ . [3]
Step-by-step solution
1. Step 1
  M1 Divide both sides by $\pi$ .
  $\frac{A}{\pi} = R^{2} - r^{2}$
2. Step 2
  M1 Add $r^{2}$ and subtract $\dfrac{A}{\pi}$ , so that $r^{2}$ is positive and on its own.
  $r^{2} = R^{2} - \frac{A}{\pi}$
3. Step 3
  A1 Take the positive square root of the whole right-hand side.
  $r = \sqrt{R^{2} - \frac{A}{\pi}}$
Answer
$r = \sqrt{R^{2} - \dfrac{A}{\pi}}$ oe
Examiner tip
Examiner tip: An equivalent form is $r = \sqrt{\dfrac{\pi R^{2} - A}{\pi}}$ . Either scores full marks.

Common mistake: Writing $r = R - \sqrt{\dfrac{A}{\pi}}$ , taking the root of each term separately. The square root must cover the whole difference.
10Subject twice, then evaluate [4 marks]
Stretch• Paper 1H/2H style — 4 marks• rearranging formulae, factorising, substitution
▼
Question
(a) Make $g$ the subject of $\dfrac{g + h}{g} = 5h$ . [3]

(b) Work out the value of $g$ when $h = 0.6$ [1]
Step-by-step solution
1. Step 1
  M1 (a) Multiply both sides by $g$ to clear the fraction.
  $g + h = 5gh$
2. Step 2
  M1 $g$ now appears twice. Collect the $g$ terms on one side and factorise.
  $h = 5gh - g = g(5h - 1)$
3. Step 3
  A1 Divide by the bracket.
  $g = \frac{h}{5h - 1}$
4. Step 4
  B1 (b) Substitute $h = 0.6$ .
  $g = \frac{0.6}{5 \times 0.6 - 1} = \frac{0.6}{2} = 0.3$
Answer
(a) $g = \dfrac{h}{5h - 1}$ oe (b) $0.3$
Examiner tip
Examiner tip: Check (b) in the original: $\dfrac{0.3 + 0.6}{0.3} = 3$ and $5 \times 0.6 = 3$ .

Common mistake: Stopping at $g = 5gh - h$ , or any line with $g$ still on both sides. Examiners report that not factorising is the usual reason candidates cannot finish when the subject appears twice.

Question 1
Paper 1H/2H style — 1 mark1 mark
The volume of a cuboid is $V = lwh$ .

Make $h$ the subject of the formula. [1]
Model answer
$h = \dfrac{V}{lw}$
Why this scores
Mark scheme:
- B1 for $h = \dfrac{V}{lw}$ oe
Why this scores full marks: $h$ is multiplied by both $l$ and $w$ , so both go into the denominator in one step. $h = \dfrac{V}{l} \times w$ is wrong: it multiplies by $w$ instead of dividing.
Question 2
Paper 1H/2H style — 2 marks2 marks
Make $x$ the subject of $y = 8 - \dfrac{x}{2}$ . [2]
Model answer
$\dfrac{x}{2} = 8 - y$

$x = 2(8 - y)$

Answer: $x = 16 - 2y$
Why this scores
Mark scheme:
- M1 for $\dfrac{x}{2} = 8 - y$ or $2y = 16 - x$ oe
- A1 for $x = 16 - 2y$ oe
Why this scores full marks: Adding $\dfrac{x}{2}$ to both sides first makes the $x$ term positive, so there are no negative signs to divide through. The bracket in $2(8 - y)$ makes sure both terms are doubled.
Question 3
Paper 1H/2H style — 2 marks2 marks
The volume of a cylinder is $V = \pi r^{2} h$ .

Make $r$ the subject of the formula, where $r > 0$ . [2]
Model answer
$r^{2} = \dfrac{V}{\pi h}$

Answer: $r = \sqrt{\dfrac{V}{\pi h}}$
Why this scores
Mark scheme:
- M1 for $r^{2} = \dfrac{V}{\pi h}$
- A1 for $r = \sqrt{\dfrac{V}{\pi h}}$ oe
Why this scores full marks: Everything multiplying $r^{2}$ is divided out before the square root is taken, and the root covers the whole fraction. A root that covers only $V$ loses the A mark.
Question 4
Paper 1H/2H style — 2 marks2 marks
Make $x$ the subject of $y = 4x^{3} - 1$ . [2]
Model answer
$y + 1 = 4x^{3}$

$x^{3} = \dfrac{y + 1}{4}$

Answer: $x = \sqrt[3]{\dfrac{y + 1}{4}}$
Why this scores
Mark scheme:
- M1 for $x^{3} = \dfrac{y + 1}{4}$ oe
- A1 for $x = \sqrt[3]{\dfrac{y + 1}{4}}$ oe
Why this scores full marks: The script undoes the $-1$ and the $\times 4$ before the cube root, so the root is applied to $x^{3}$ alone. Writing $\sqrt[3]{y + 1} \div 4$ takes the root too early and is wrong.
Question 5
Paper 1H/2H style — 3 marks3 marks
Make $c$ the subject of $a = \dfrac{5b}{c - 2}$ . [3]
Model answer
$a(c - 2) = 5b$

$c - 2 = \dfrac{5b}{a}$

Answer: $c = \dfrac{5b}{a} + 2$
Why this scores
Mark scheme:
- M1 for $a(c - 2) = 5b$ oe
- M1 for $c - 2 = \dfrac{5b}{a}$ or $ac = 5b + 2a$
- A1 for $c = \dfrac{5b}{a} + 2$ or $c = \dfrac{5b + 2a}{a}$ oe
Why this scores full marks: Multiplying by the whole denominator $(c - 2)$ , in a bracket, brings the subject to the top. After that it is two ordinary inverse operations.
Question 6
Paper 1H/2H style — 3 marks3 marks
Make $y$ the subject of $4(y - x) = x(y + 3)$ . [3]
Model answer
$4y - 4x = xy + 3x$

$4y - xy = 7x$

$y(4 - x) = 7x$

Answer: $y = \dfrac{7x}{4 - x}$
Why this scores
Mark scheme:
- M1 for expanding both brackets correctly, $4y - 4x = xy + 3x$
- M1 for collecting the $y$ terms and factorising, $y(4 - x) = 7x$ oe
- A1 for $y = \dfrac{7x}{4 - x}$ oe
Why this scores full marks: $y$ appears on both sides, so the script collects and factorises instead of trying to divide. The $x$ terms combine correctly: $3x + 4x = 7x$ , with no sign slip. Check: $x = 1$ gives $y = \dfrac{7}{3}$ , and both sides then equal $\dfrac{16}{3}$ .
Question 7
Paper 1H/2H style — 3 marks3 marks
The formula $F = \dfrac{9C}{5} + 32$ converts a temperature of $C$ degrees Celsius to $F$ degrees Fahrenheit.

(a) Make $C$ the subject of the formula. [2]

(b) Work out the value of $C$ when $F = 95$ [1]
Model answer
(a) $F - 32 = \dfrac{9C}{5}$

$5(F - 32) = 9C$

Answer: $C = \dfrac{5(F - 32)}{9}$

(b) $C = \dfrac{5(95 - 32)}{9} = \dfrac{5 \times 63}{9}$

Answer: $C = 35$
Why this scores
Mark scheme:
- M1 for $F - 32 = \dfrac{9C}{5}$ or $5F = 9C + 160$ oe
- A1 for $C = \dfrac{5(F - 32)}{9}$ oe
- B1 for $35$ (ft their formula from (a))
Why this scores full marks: Subtracting $32$ before multiplying by $5$ keeps $F - 32$ together, and the bracket is kept when multiplying. $C = \dfrac{5F - 32}{9}$ is the usual wrong answer.
Question 8
Paper 1H/2H style — 4 marks4 marks
Make $x$ the subject of $\dfrac{x}{2} - \dfrac{a}{3} = \dfrac{ax}{6} + 1$ . [4]
Model answer
$3x - 2a = ax + 6$

$3x - ax = 6 + 2a$

$x(3 - a) = 6 + 2a$

Answer: $x = \dfrac{6 + 2a}{3 - a}$
Why this scores
Mark scheme:
- M1 for multiplying every term by $6$ , $3x - 2a = ax + 6$
- M1 for collecting the $x$ terms on one side, $3x - ax = 6 + 2a$ oe
- M1 for factorising, $x(3 - a)$ oe
- A1 for $x = \dfrac{6 + 2a}{3 - a}$ oe
Why this scores full marks: The lowest common multiple $6$ clears all three fractions at once, and every term, including the $1$ , is multiplied. Check: $a = 1$ gives $x = 4$ , and both sides of the original equal $\dfrac{5}{3}$ .
Question 9
Paper 1H/2H style — 4 marks4 marks
Make $t$ the subject of $y = \dfrac{t^{2} + 2}{t^{2} - 5}$ . [4]
Model answer
$y(t^{2} - 5) = t^{2} + 2$

$yt^{2} - 5y = t^{2} + 2$

$yt^{2} - t^{2} = 5y + 2$

$t^{2}(y - 1) = 5y + 2$

Answer: $t = \pm\sqrt{\dfrac{5y + 2}{y - 1}}$
Why this scores
Mark scheme:
- M1 for $y(t^{2} - 5) = t^{2} + 2$
- M1 for collecting the $t^{2}$ terms, $yt^{2} - t^{2} = 5y + 2$ oe
- M1 for factorising, $t^{2}(y - 1) = 5y + 2$ oe
- A1 for $t = \pm\sqrt{\dfrac{5y + 2}{y - 1}}$ oe
Why this scores full marks: The script treats $t^{2}$ as a single quantity, collects and factorises it, and takes the square root only at the end. Nothing restricts $t$ to be positive, so both roots are given. Check: $t^{2} = 6$ gives $y = 8$ , and $\dfrac{42}{7} = 6$ .
Question 10
Paper 1H/2H style — 5 marks5 marks
A solid is made from a cylinder of radius $r$ cm and height $h$ cm with a hemisphere of radius $r$ cm on top. Its volume, $V$ cm³, is

$V = \pi r^{2} h + \frac{2}{3}\pi r^{3}$

(a) Make $h$ the subject of the formula. [3]

(b) The solid has volume $800$ cm³ and radius $5$ cm. Work out the height of the cylinder. Give your answer correct to 3 significant figures. [2]
Model answer
(a) $V - \dfrac{2}{3}\pi r^{3} = \pi r^{2} h$

$h = \dfrac{V - \frac{2}{3}\pi r^{3}}{\pi r^{2}}$

Answer: $h = \dfrac{V}{\pi r^{2}} - \dfrac{2r}{3}$

(b) $h = \dfrac{800}{25\pi} - \dfrac{10}{3} = 10.1859\ldots - 3.3333\ldots = 6.8526\ldots$

Answer: $6.85$ cm
Why this scores
Mark scheme:
- M1 for $V - \dfrac{2}{3}\pi r^{3} = \pi r^{2} h$
- M1 for dividing the whole of $V - \dfrac{2}{3}\pi r^{3}$ by $\pi r^{2}$
- A1 for $h = \dfrac{V}{\pi r^{2}} - \dfrac{2r}{3}$ oe
- M1 for substituting $V = 800$ and $r = 5$ into their formula, or into the original
- A1 for awrt $6.85$
Why this scores full marks: Dividing both terms by $\pi r^{2}$ simplifies the second one to $\dfrac{2r}{3}$ , which makes (b) quick. The calculator values are kept unrounded until the final answer; rounding $10.19 - 3.33$ first gives $6.86$ and loses the A mark.

4MA1 Rearranging Formulae

At a glance

What you’ll learn

The standard method

Powers and roots in rearranging

Subject on both sides

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Make $x$ the subject

2Rearrange with a power

3Subject appears on both sides

4Rearrange with a square root

5Undoing a cube [1 mark]

6Perimeter of a rectangle: make w the subject [2 marks]

7Motion formula: make t the subject [2 marks]

8Two fractions equal to each other [3 marks]

9Area of a ring: make r the subject [3 marks]

10Subject twice, then evaluate [4 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Rearranging Formulae

At a glance

What you’ll learn

The standard method

Powers and roots in rearranging

Subject on both sides

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Make xxx the subject

2Rearrange with a power

3Subject appears on both sides

4Rearrange with a square root

5Undoing a cube [1 mark]

6Perimeter of a rectangle: make w the subject [2 marks]

7Motion formula: make t the subject [2 marks]

8Two fractions equal to each other [3 marks]

9Area of a ring: make r the subject [3 marks]

10Subject twice, then evaluate [4 marks]

Model AnswersFull mark-scheme-style answers

1Make $x$ the subject