IGCSEMathsCambridge (CIE)ExtendedRevision NotesAlgebra & SequencesRearranging FormulasFormulas where Subject Appears Once

Formulas where Subject Appears Once (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

Exam code: 0580 & 0980

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 27 January 2026

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Understand this topic with deeper explanations and understanding checks.

Simple rearranging

What are formulas?

A formula is a rule, definition or relationship between different quantities, written in shorthand using letters (variables)
- Formulas include an equals sign
Some examples you should be familiar with are:
- The equation of a straight line
  - $y = m x + c$
- The area of a trapezium
  - $Area = \frac{(a + b) h}{2}$
- Pythagoras' theorem
  - $a^{2} + b^{2} = c^{2}$

How do I rearrange formulas?

The letter (variable) that is on its own on one side is called the subject
- y is the subject of y = mx + c
To make a different letter the subject, we need to rearrange the formula
- This is also called changing the subject
The method is as follows:
- First, remove any fractions
  - Multiply both sides by the lowest common denominator
- Then use inverse (opposite) operations to get the variable on its own
  - This is similar to solving equations
For example, make $x$ the subject of $\frac{5 x + 6}{2} = y$
- First remove fractions
  - Multiply both sides by 2
    $5 x + 6 = 2 y$
- Then get $x$ on its own
  - Subtract 6 from both sides
    $5 x = 2 y - 6$
  - Divide both sides by 5
    $x = \frac{2 y - 6}{5}$
- There may be more than one correct way to write an answer
  - The following are acceptable alternative forms
    $x = \frac{2 y}{5} - \frac{6}{5}$
    $x = \frac{2 (y - 3)}{5}$
    $x = 0.4 (y - 3)$
    $x = 0.4 y - 1.2$

Should I expand brackets?

If the variable is inside the brackets, you can either:
- expand the brackets
- or divide both by the coefficient if it is not zero
For example, make $x$ the subject of $3 (1 + x) = y$
- You can expand and then rearrange
  - $3 + 3 x = y$
  - $\begin{array}{rcl} 3 x & = & y - 3 \end{array}$
  - $\begin{array}{rcl} x & = & \frac{y - 3}{3} \end{array}$
- You can divide and then rearrange
  - $1 + x = \frac{y}{3}$
  - $x = \frac{y}{3} - 1$
- Both answers are equivalent
  - $\frac{y - 3}{3}$ is the same as $\frac{y}{3} - 1$

Examiner Tips and Tricks

Stop and think when you see a bracket. If the variable you are trying to make the subject is not inside the bracket, then you do not need to expand.

For example, to make $x$ the subject of $(1 + k) x = y$ , you can simply divide both sides by $(1 + k)$ to get $x = \frac{y}{1 + k}$ .

What if I get fractions in fractions?

Some rearrangements can lead to fractions in fractions
- $x = \frac{\frac{3}{t}}{2}$
Either rewrite with a divide sign, $\div$ , then use the method of dividing two fractions
- $x = \frac{3}{t} \div 2$
  $\begin{array}{l} x = \frac{3}{t} \div \frac{2}{1} \\ x = \frac{3}{t} \times \frac{1}{2} \\ x = \frac{3}{2 t} \end{array}$
Or multiply top and bottom by the lowest common denominator of the two fractions and cancel
- $x = \frac{\frac{5}{y}}{\frac{t}{8}}$ becomes $x = \frac{\frac{5}{y} \times 8 y}{\frac{t}{8} \times 8 y} = \frac{40}{t y}$

What if I end up dividing by a negative?

Remember that $\frac{a}{- b}$ (minus below) is the same as $\frac{- a}{b}$ (minus above) and the same as $- \frac{a}{b}$ (minus outside)
- Though be careful, as $\frac{- a}{- b}$ is $\frac{a}{b}$
$- 2 x = y - 3$ becomes $x = \frac{y - 3}{- 2}$ (minus below)
- This is the same as $x = \frac{- (y - 3)}{2}$ (minus above) or $- \frac{y - 3}{2}$ (minus outside)
  - brackets are required for minus above
  - brackets are assumed for minus outside
- You can also expand the brackets
  $\frac{- (y - 3)}{2} = \frac{- y + 3}{2} = \frac{3 - y}{2}$

Examiner Tips and Tricks

Mark schemes will accept different forms of the same answer, as long as they are correct and fully simplified.

Worked Example

Make $x$ the subject of the following.

(a) $4 m + 5 x = 3$

Answer:

Get 5x on its own by subtracting 4m from both sides

$5 x = 3 - 4 m$

Get x on its own by dividing both sides by 5

$x = \frac{3 - 4 m}{5}$

(b) $3 t = \frac{2}{x}$

Answer:

Remove fractions by multiplying both sides by the denominator, x

$3 t x = 2$

Get x on its own by dividing both sides by 3t

$x = \frac{2}{3 t}$

Answer:

Remove fractions by multiplying both sides by the denominator, 2g

$2 g A = 9 (1 - 4 x)$

x is inside the brackets
Expand the brackets to release the x term

$2 g A = 9 - 36 x$

One way to get x on its own is by subtracting 9 then dividing by -36
Or you can first add 36x to both sides, to create positive 36x on the left

$2 g A + 36 x = 9$

Now get x on its own by subtracting 2gA then dividing by 36

$\begin{array}{rcl} 36 x & = & 9 - 2 g A \\ x & = & \frac{9 - 2 g A}{36} \end{array}$

$x = \frac{9 - 2 g A}{36}$

Other accepted forms of the answer are $\frac{2 g A - 9}{- 36}, \frac{- (2 g A - 9)}{36}, - \frac{2 g A - 9}{36}, \frac{1}{4} - \frac{g A}{18}$

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Questions students ask about Formulas where Subject Appears Once

Real questions from students revising this topic on Save My Exams

How do you rearrange a formula when the subject is squared?

Get the squared term on its own first, then square root both sides.

Remove any fractions by multiplying both sides by the denominator
Use inverse operations to get the squared term alone on one side
Square root the whole of each side
Write ± in front if the subject could be positive or negative; if it is a length, the positive root is enough

Example: make a the subject of Pythagoras' theorem, a² + b² = c²

Subtract b² from both sides: a² = c² − b²
Square root both sides: a = √(c² − b²)
a is a length, so you only need the positive root

Watch out: √(c² − b²) is not c − b. The root applies to the whole side, so keep it under one root sign.

Squaring and square rooting are inverse operations, just like multiplying and dividing, so you use them the same way when changing the subject.

How do you divide both sides of an equation when rearranging a formula?

Divide the whole of each side by the same number or expression — every term on a side gets divided, not just one of them.

The simplest way is to write the whole side as a single fraction over the divisor.

Page example: 5x = 2y − 6

Divide both sides by 5: x = (2y − 6)/5
The 5 divides both the 2y and the 6, so this can also be written 2y/5 − 6/5
Writing x = 2y − 6/5 is wrong, because that only divides the 6

Dividing by a letter or a bracket works the same way:

3tx = 2 → divide by 3t → x = 2/(3t)
(1 + k)x = y → divide by (1 + k) → x = y/(1 + k)
3(1 + x) = y → divide by 3 → 1 + x = y/3, so x = y/3 − 1

Tips:

Remove fractions first; dividing is usually the last step
You can only divide by a coefficient that is not zero

Formulas where Subject Appears Once (Cambridge (CIE) IGCSE Maths: Extended): Revision Note

Simple rearranging

What are formulas?

How do I rearrange formulas?

Should I expand brackets?

Examiner Tips and Tricks

What if I get fractions in fractions?

What if I end up dividing by a negative?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Questions students ask about Formulas where Subject Appears Once

How do you rearrange a formula when the subject is squared?

How do you divide both sides of an equation when rearranging a formula?

Author: Mark Curtis

Reviewer: Dan Finlay