What does rearranging formulae mean in the context of Edexcel IGCSE Mathematics?

Rearranging formulae in Edexcel IGCSE Mathematics involves changing the subject of a formula. This means manipulating the equation to solve for a different variable, which is essential for solving problems where you need to find the value of a specific variable given other known values.

How do I rearrange a formula to make a different variable the subject?

To rearrange a formula to make a different variable the subject, follow these steps: 1) Identify the variable you want to isolate. 2) Use inverse operations to move other terms to the opposite side of the equation. 3) Simplify the equation if necessary. This process often involves operations like addition, subtraction, multiplication, division, and sometimes factoring or expanding.

What are some common mistakes to avoid when rearranging formulae?

Common mistakes include not applying operations to all terms on both sides of the equation, forgetting to reverse the operation when moving terms across the equals sign, and losing track of negative signs. It's crucial to perform each step carefully and check your work to ensure accuracy.

Can you provide an example of rearranging a formula in Mathematics?

Certainly! Consider the formula for the area of a rectangle, A = l × w, where A is the area, l is the length, and w is the width. To make l the subject, divide both sides by w to get l = A/w. This rearrangement allows you to calculate the length if you know the area and the width.

Why is it important to learn rearranging formulae for the Edexcel IGCSE exam?

Rearranging formulae is crucial for the Edexcel IGCSE exam because it enhances problem-solving skills and understanding of mathematical relationships. It is widely applicable in various topics, such as physics and chemistry, where you need to manipulate equations to find unknown values. Mastery of this skill is essential for success in both exams and real-world applications.

Why is "Applying an operation to only one side" a common mistake in Rearranging Formulae?

Why it happens: Hurried. How to avoid it: ALWAYS apply the same operation to BOTH sides of the equation. This preserves equality.

Why is "Squaring individual terms, e.g. $(y - 2)^{2} \to y^{2} - 4$" a common mistake in Rearranging Formulae?

Why it happens: Distributing the square. How to avoid it: (y - 2)^2 = y^2 - 4y + 4 (using square of binomial). Three terms, not two.

Why is "Trying to solve when subject is on both sides without factoring" a common mistake in Rearranging Formulae?

Why it happens: Confusion about what to do next. How to avoid it: Gather all x terms on ONE side. Factor out x. Divide. Mark scheme awards each step. Source: 4MA1/1H — examiner reports 2022-2024

Equations, Formulae and IdentitiesEdexcel IGCSE Mathematics (4MA1)

Rearranging Formulae

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Rearranging Formulae Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

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

What you’ll learn

Mapped to the Cambridge 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 $x = \dfrac{y - 5}{3}$ .

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: $x = \dfrac{1 + 2y}{y - 1}$ .

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.

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.

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Step-by-step worked examples — Rearranging Formulae

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.

Key Definitions and Keywords — Rearranging Formulae

Definitions to memorise and the exact keywords mark schemes credit for rearranging formulae answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Subject (of a formula)
Examiner keyword
The variable on its own on one side of the formula. To 'make $x$ the subject' is to rearrange the formula so $x$ alone appears on one side.
Example
In $y = 3x + 5$ , $y$ is the subject. After rearranging, $x = (y-5)/3$ has $x$ as subject.
Inverse operation
The opposite of an operation. Inverse of $+$ is $-$ . Inverse of $\times$ is $\div$ . Inverse of squaring is square root.

Common Mistakes and Misconceptions — Rearranging Formulae

The traps other students keep falling into on rearranging formulae questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Applying an operation to only one side
Why it happens
Hurried.
How to avoid it
ALWAYS apply the same operation to BOTH sides of the equation. This preserves equality.
✕Squaring individual terms, e.g. $(y - 2)^{2} \to y^{2} - 4$
Why it happens
Distributing the square.
How to avoid it
$(y - 2)^{2} = y^{2} - 4y + 4$ (using square of binomial). Three terms, not two.
✕Trying to solve when subject is on both sides without factoring
4MA1/1H — examiner reports 2022-2024
Why it happens
Confusion about what to do next.
How to avoid it
Gather all $x$ terms on ONE side. Factor out $x$ . Divide. Mark scheme awards each step.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Rearranging Formulae — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Rearranging Formulae

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Make xxx the subject

2Rearrange with a power

3Subject appears on both sides

4Rearrange with a square root

Subject (of a formula)

Inverse operation

✕Applying an operation to only one side

✕Squaring individual terms, e.g. (y−2)2→y2−4(y - 2)^{2} \to y^{2} - 4(y−2)2→y2−4

✕Trying to solve when subject is on both sides without factoring

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Rearranging Formulae — frequently asked questions

1Make $x$ the subject

✕Squaring individual terms, e.g. $(y - 2)^{2} \to y^{2} - 4$