What does 'changing the subject of the formula' mean in Cambridge IGCSE Mathematics?

In Cambridge IGCSE Mathematics, 'changing the subject of the formula' refers to rearranging an equation so that a different variable becomes the subject. This involves using algebraic operations to isolate the desired variable on one side of the equation, making it the main focus or 'subject' of the formula.

How do you change the subject of a formula in algebra?

To change the subject of a formula in algebra, you need to perform inverse operations to isolate the desired variable. This involves reversing operations such as addition, subtraction, multiplication, and division, while maintaining the balance of the equation. For example, to make 'x' the subject in the equation y = 2x + 3, you would subtract 3 from both sides and then divide by 2, resulting in x = (y - 3) / 2.

Why is changing the subject of the formula important in mathematics?

Changing the subject of the formula is important in mathematics because it allows you to express relationships between variables in different ways, making it easier to solve problems and understand the underlying concepts. It is a fundamental skill in algebra that is widely used in various mathematical applications, including solving equations and modeling real-world situations.

What are common mistakes to avoid when changing the subject of a formula?

Common mistakes when changing the subject of a formula include not applying inverse operations correctly, forgetting to perform the same operation on both sides of the equation, and mismanaging signs (positive/negative). It's crucial to carefully follow each step and double-check your work to ensure accuracy.

Can you provide an example of changing the subject of a formula in a physics context?

Certainly! Consider the physics formula for speed: v = d/t, where 'v' is speed, 'd' is distance, and 't' is time. If you want to make 't' the subject, you would multiply both sides by 't' to get vt = d, and then divide both sides by 'v' to isolate 't', resulting in t = d/v. This rearrangement is useful for calculating time when distance and speed are known.

Why is "Applying an operation to one side only" a common mistake in Changing the Subject of the Formula?

Why it happens: Students rearrange algebraically without doing the same to both sides. How to avoid it: Whatever you do, do to **both** sides — that's the rule that keeps the equation balanced.

Why is "Forgetting the $\pm$ when taking a square root in a general formula" a common mistake in Changing the Subject of the Formula?

Why it happens: Squaring loses the sign; reversing introduces . How to avoid it: If the variable could physically be negative, write √(). For lengths, +√() only.

Why is "Not factorising when the subject appears in two places" a common mistake in Changing the Subject of the Formula?

Why it happens: Students get to x - yx = c and don't see that they need to factor. How to avoid it: If x appears more than once, your next step is **always** to factor it out. Source: 0580/42 — examiner reports, recurring

Why is "Dividing only one term by the divisor" a common mistake in Changing the Subject of the Formula?

Why it happens: y = mx + c rearranged to x = y/m + c instead of y - c/m. How to avoid it: Subtract first to isolate the term containing x, **then** divide.

AlgebraCambridge IGCSE Maths 0580 Extended

Changing the Subject of the Formula

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Changing the Subject of a Formula — Cambridge IGCSE 0580 Maths Extended (2026)

Rearrange a formula so a different letter sits on its own. The same balancing rules as solving an equation, but you carry letters along instead of numbers.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.4 — Transform simple and more complex formulae.

Basic rearrangement

Move terms across the equals sign by reversing the operation.

Rule. Whatever operation a term goes through, do the inverse to BOTH sides.

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

Subtract $5$ from both sides: $y - 5 = 3x$ .
Divide by $3$ : $\dfrac{y - 5}{3} = x$ .
Final: $x = \dfrac{y - 5}{3}$ .

Worked. Make $r$ the subject of $A = \pi r^{2}$ .

Divide by $\pi$ : $\dfrac{A}{\pi} = r^{2}$ .
Square root both sides: $r = \sqrt{\dfrac{A}{\pi}}$ .

Worked. Make $h$ the subject of $V = \dfrac{1}{3}\pi r^{2} h$ .

Multiply both sides by $3$ : $3V = \pi r^{2} h$ .
Divide by $\pi r^{2}$ : $h = \dfrac{3V}{\pi r^{2}}$ .

Approach. Look at the formula from the OUTSIDE in:

Last operation applied to the subject = first to undo.
Reverse BIDMAS: undo +/- first, then ×/÷, then powers/roots.

Undo the operations in reverse order — the last one done is the first one undone.

Inverse operations: undo +/-/×/÷ in reverse BIDMAS order.
Whatever you do, do to BOTH sides.
Square → square root, cube → cube root.
End with 'subject = …' notation.

When the subject appears more than once

Get all the subject terms on one side, then factorise.

If the new subject appears more than once, you can't isolate by simple steps. Use factorisation to bring it together.

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

Move all $x$ terms to one side: $a x - c x = d - b$ .
Factorise: $x(a - c) = d - b$ .
Divide: $x = \dfrac{d - b}{a - c}$ .

When the subject appears twice: collect, factorise, then divide.

Worked. Make $x$ the subject of $\dfrac{x}{x + 2} = k$ .

Cross-multiply: $x = k(x + 2)$ .
Expand: $x = kx + 2k$ .
Move $kx$ across: $x - kx = 2k$ .
Factorise: $x(1 - k) = 2k$ .
Divide: $x = \dfrac{2k}{1 - k}$ .

Tip. When you spot the subject in two places, the very next move is usually to MULTIPLY out any brackets/fractions, then collect the subject terms.

Subject in two places → factorise.
Cross-multiply to remove fractions, then expand, then collect.
Move all subject terms to one side, all others to the other.
Factorise out the subject, then divide.

Roots and fractions

Square or cube to remove roots; cross-multiply to remove fractions.

Square roots. Square BOTH sides to remove a square root containing the subject.

Worked. Make $x$ the subject of $y = \sqrt{2x + 5}$ .

Square: $y^{2} = 2x + 5$ .
Subtract $5$ : $y^{2} - 5 = 2x$ .
Divide: $x = \dfrac{y^{2} - 5}{2}$ .

Cube roots. Cube BOTH sides similarly.

Fractions. Multiply both sides by the denominator (or cross-multiply if both sides are fractions).

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

Multiply by $b$ : $a + 1 = bc$ .
Subtract $1$ : $a = bc - 1$ .

Combined. Make $r$ the subject of $T = 2\pi\sqrt{\dfrac{r}{g}}$ .

Divide by $2\pi$ : $\dfrac{T}{2\pi} = \sqrt{\dfrac{r}{g}}$ .
Square: $\left(\dfrac{T}{2\pi}\right)^{2} = \dfrac{r}{g}$ → $\dfrac{T^{2}}{4\pi^{2}} = \dfrac{r}{g}$ .
Multiply by $g$ : $r = \dfrac{g T^{2}}{4\pi^{2}}$ .

Square root inside formula → square both sides.
Cube root → cube both sides.
Single fraction → multiply by denominator.
Two fractions equal → cross-multiply.

How it’s examined

Changing the subject is a Paper 4 staple — typically 3-4 marks. Cambridge prefers a real-world formula (volume, area, motion). Paper 2 has it too, simpler form (1-2 marks). Examiner reports flag the 'subject appears twice' case as the recurring difficulty.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.4); 0580/22 May/Jun 2024 — Q13 (rearrange formula); 0580/42 Oct/Nov 2024 — Q12 (subject appears twice); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Changing the Subject of the Formula

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

1Make $x$ the subject of $y = mx + c$
Core• rearranging
Question
Make $x$ the subject of $y = mx + c$ .
Step-by-step solution
1. Step 1
  Subtract $c$ from both sides.
  $y - c = mx$
2. Step 2
  Divide both sides by $m$ .
  $x = \dfrac{y - c}{m}$
Answer
$x = \dfrac{y - c}{m}$
2Subject under a square root
Extended• Adapted from 0580/22 May/Jun 2024 Q12• square root
Question
Make $x$ the subject of $y = \sqrt{x + 3}$ .
Step-by-step solution
1. Step 1
  Square both sides to remove the root.
  $y^{2} = x + 3$
2. Step 2
  Subtract $3$ .
  $x = y^{2} - 3$
Answer
$x = y^{2} - 3$
Examiner tip
Squaring is the inverse of square-rooting. Apply it to both sides simultaneously.
3Rearrange Pythagoras
Extended• square root, Pythagoras
Question
Make $b$ the subject of $a^{2} + b^{2} = c^{2}$ .
Step-by-step solution
1. Step 1
  Subtract $a^{2}$ .
  $b^{2} = c^{2} - a^{2}$
2. Step 2
  Take the positive square root (length is positive).
  $b = \sqrt{c^{2} - a^{2}}$
Answer
$b = \sqrt{c^{2} - a^{2}}$
4Subject appears on both sides
Extended• Adapted from 0580/42 Oct/Nov 2023 Q15• both sides, factorising
Question
Make $x$ the subject of $\dfrac{x + 1}{x - 2} = y$ .
Step-by-step solution
1. Step 1
  Cross-multiply.
  $x + 1 = y(x - 2)$
2. Step 2
  Expand the right-hand side.
  $x + 1 = yx - 2y$
3. Step 3
  Get all $x$ terms on one side.
  $x - yx = -2y - 1$
4. Step 4
  Factorise out $x$ .
  $x(1 - y) = -2y - 1$
5. Step 5
  Divide by the bracket.
  $x = \dfrac{-2y - 1}{1 - y} = \dfrac{2y + 1}{y - 1}$
Answer
$x = \dfrac{2y + 1}{y - 1}$
Examiner tip
When the new subject appears in two places, collect, then factorise to extract a single $x$ . The mark scheme awards method marks for these two steps explicitly.
5Rearrange a formula with a square term
Extended• squaring
Question
Make $r$ the subject of $V = \dfrac{4}{3}\pi r^{3}$ .
Step-by-step solution
1. Step 1
  Multiply both sides by $\dfrac{3}{4\pi}$ .
  $\dfrac{3V}{4\pi} = r^{3}$
2. Step 2
  Take the cube root.
  $r = \sqrt[3]{\dfrac{3V}{4\pi}}$
Answer
$r = \sqrt[3]{\dfrac{3V}{4\pi}}$
6Make $r$ the subject of a circle-area formula
Core• square root, geometry
Question
Make $r$ the subject of $A = \pi r^{2}$ .
Step-by-step solution
1. Step 1
  Divide both sides by $\pi$ .
  $\dfrac{A}{\pi} = r^{2}$
2. Step 2
  Take the positive square root (radius is a length).
  $r = \sqrt{\dfrac{A}{\pi}}$
Answer
$r = \sqrt{\dfrac{A}{\pi}}$
Examiner tip
Take only the positive root because $r$ is a length. Writing $r = \pm\sqrt{A/\pi}$ loses the final mark in geometry contexts.
7Variable inside a square root with subtraction
Extended• Adapted from 0580/42 May/Jun 2024 Q11• square root, rearranging
Question
Make $t$ the subject of $v = \sqrt{2gt - k}$ .
Step-by-step solution
1. Step 1
  Square both sides to remove the root.
  $v^{2} = 2gt - k$
2. Step 2
  Add $k$ to both sides.
  $v^{2} + k = 2gt$
3. Step 3
  Divide both sides by $2g$ .
  $t = \dfrac{v^{2} + k}{2g}$
Answer
$t = \dfrac{v^{2} + k}{2g}$
Examiner tip
The 2024 mark scheme awards method marks for squaring both sides first. Trying to deal with the $-k$ before squaring almost always introduces errors.
8New subject appears twice — collect and factorise
Extended• Adapted from 0580/42 Oct/Nov 2024 Q16• factorising, both sides
Question
Make $x$ the subject of $a x + b = c x + d$ .
Step-by-step solution
1. Step 1
  Get all $x$ terms on one side and constants on the other.
  $a x - c x = d - b$
2. Step 2
  Factorise out $x$ .
  $x(a - c) = d - b$
3. Step 3
  Divide both sides by $(a - c)$ .
  $x = \dfrac{d - b}{a - c}$
Answer
$x = \dfrac{d - b}{a - c}$
Examiner tip
The examiner report flags candidates often divide before factorising and end up with $x = \dfrac{d - b}{a} - cx$ , which is no longer a formula. Always factorise first when the new subject appears in more than one term.
9Change subject when the variable is inside a fraction
Extended• Adapted from 0580/22 Feb/Mar 2024 Q14• fractions, rearranging
Question
Make $x$ the subject of $y = \dfrac{3}{x - 4}$ .
Step-by-step solution
1. Step 1
  Multiply both sides by $(x - 4)$ to clear the denominator.
  $y(x - 4) = 3$
2. Step 2
  Divide both sides by $y$ (assuming $y \neq 0$ ).
  $x - 4 = \dfrac{3}{y}$
3. Step 3
  Add $4$ to both sides.
  $x = \dfrac{3}{y} + 4$
Answer
$x = \dfrac{3}{y} + 4$
Examiner tip
Examiners reward candidates who clear the fraction by multiplying first. Trying to invert both sides directly ( $\frac{1}{y} = \frac{x - 4}{3}$ ) is correct but more error-prone — the method shown is the mark-scheme route.
10Rearrange a formula containing a power
Challenge• Adapted from 0580/42 May/Jun 2023 Q20• index laws, rearranging
Question
The formula for compound interest is $A = P(1 + r)^{n}$ . Make $r$ the subject.
Step-by-step solution
1. Step 1
  Divide both sides by $P$ .
  $\dfrac{A}{P} = (1 + r)^{n}$
2. Step 2
  Take the $n$ -th root of both sides.
  $\sqrt[n]{\dfrac{A}{P}} = 1 + r$
3. Step 3
  Subtract $1$ .
  $r = \sqrt[n]{\dfrac{A}{P}} - 1$
Answer
$r = \sqrt[n]{\dfrac{A}{P}} - 1$
Examiner tip
The mark scheme accepts equivalent forms such as $r = (A/P)^{1/n} - 1$ . The most common slip is forgetting to subtract $1$ at the end — the bracket isolates $1 + r$ , not $r$ itself.

Key Formulae — Changing the Subject of the Formula

The formulae you need to memorise for changing the subject of the formula on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Inverse operations to undo each step
$+ \leftrightarrow -,\ \times \leftrightarrow \div,\ x^{2} \leftrightarrow \sqrt{x},\ x^{3} \leftrightarrow \sqrt[3]{x}$
When to use
Identify what's been done to the desired variable, then apply the inverse to both sides in reverse order.
Collect-and-factor for variable on both sides
$ax + bx = c \implies x(a + b) = c \implies x = \dfrac{c}{a + b}$
When to use
Whenever the new subject appears in more than one term — collect to one side, factorise, divide.

Key Definitions and Keywords — Changing the Subject of the Formula

Definitions to memorise and the exact keywords mark schemes credit for changing the subject of the formula answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Subject of a formula
Examiner keyword
The variable that the formula expresses in terms of the others — usually written as the single term on one side of the equals sign.
Rearrange / make the subject
Examiner keyword
Manipulate a formula using inverse operations until a chosen variable stands alone on one side.
Inverse operation
An operation that undoes another — addition undoes subtraction, multiplication undoes division, squaring undoes square-rooting.

Common Mistakes and Misconceptions — Changing the Subject of the Formula

The traps other students keep falling into on changing the subject of the formula questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Applying an operation to one side only
Why it happens
Students rearrange algebraically without doing the same to both sides.
How to avoid it
Whatever you do, do to both sides — that's the rule that keeps the equation balanced.
✕Forgetting the $\pm$ when taking a square root in a general formula
Why it happens
Squaring loses the sign; reversing introduces $\pm$ .
How to avoid it
If the variable could physically be negative, write $\pm \sqrt{\dots}$ . For lengths, $+\sqrt{\dots}$ only.
✕Not factorising when the subject appears in two places
0580/42 — examiner reports, recurring
Why it happens
Students get to $x - yx = c$ and don't see that they need to factor.
How to avoid it
If $x$ appears more than once, your next step is always to factor it out.
✕Dividing only one term by the divisor
Why it happens
$y = mx + c$ rearranged to $x = \dfrac{y}{m} + c$ instead of $\dfrac{y - c}{m}$ .
How to avoid it
Subtract first to isolate the term containing $x$ , then divide.

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