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In order to access this I need to be confident with:

Expanding brackets Factorising Solving equations

This topic is relevant for:

GCSE Maths Algebra Equations Rearranging Equations

Make X The Subject

Make x The Subject

Here we will learn about making x the subject of an equation or a formula.

There are also rearranging equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What does it mean to make x the subject?

Making x the subject of a formula or equation means rearranging the equation or formula so that we have a single x variable equal to the rest of it.

For example,

Make x the subject.

Step-by-step guide: Rearranging equations

What does it mean to make x the subject?

How to make x the subject

In order to make x the subject:

Isolate the variable by:
– Removing any fractions by multiplying by the denominator(s).
– Dividing by the coefficient of the variable.
– Adding or subtracting terms near to the variable.
– Taking a root or power of both sides of the equation.
Rearrange the equation so each term containing x is on the same side of the =.
Factorisation may be needed if there are multiple terms containing x.

E.g. factorise 2x + 3xy to x(2+3y)
*not always required*
Perform an operation to ensure only a single x variable is left as the subject.

Explain how to make x the subject in 4 steps

Make x the subject worksheet

Get your free make x the subject worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Make x the subject worksheet

Get your free make x the subject worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

Make x the subject examples

Example 1: multiple step but with single variable

Make x the subject.

Example 2: questions involving x²

Example 3: questions involving √x

Example 4: factorisation of the variable is required

Example 5: factorisation of the variable is required

Common misconceptions

Perform the same operation to both sides

When we perform an operation to the left hand side of the equation we have to perform the same operation to the right hand side.

E.g

\[\frac{x}{2}=3 y+5\]

To isolate the variable x we need to multiply both sides by 2.

\[x=6y+5\]

This is wrong because we have only multiplied the 3y by 2.
The correct answer should be:

\[x=2(3y+5)=6y+10\]

This is correct because we have multiplied everything by 2 using brackets.

Inverse operations

E.g.
To isolate the variable x from 5x we need to perform the inverse operation.

5x = 5 × x, so the inverse operation is divide by 5.

E.g.

\[\frac{x}{5}=5 \div x\]

so the inverse operation is × 5.

E.g.
To isolate the variable x from x+5 we need to perform the inverse operation:

The inverse operation of +5 is −5.

E.g.
To isolate the variable x from x−5 we need to perform the inverse operation:

The inverse operation of −5 is +5.

Factorising

To make x the subject there can only be one x visible at the end. We need to get all x’s on one side of the equal sign and then factorise.

E.g.

\[x = ax+y \]

\[\begin{aligned} x-a x &=y \\ x(1-a) &=y \\ x &=\frac{y}{1-a} \end{aligned}\]

Square rooting a term

When we square rooting a number/variable as an inverse operation the answer can be positive or negative.

E.g.

\begin{align} & {{x}^{2}}=4 \\ & x=\pm \sqrt{4}=\pm 2 \\ \end{align}

\[\sqrt{x} \text { should be written as } \pm \sqrt{x}\]

Practice make x the subject questions

$x = y – \frac{3}{4}$

$x = \frac{y}{6} + 8$

$x = \frac{y}{6} – 8$

$x = y - 8$

y = 6(x + 8)

Divide both sides by $6$

\frac{y}{6} = x + 8

Then subtract $8$ from both sides

x = \frac{y}{6} – 8

$x=\pm\sqrt{3b+4p}$

$x=\pm\sqrt{4b-3p}$

$x=4b+3p$

$x=\pm\sqrt{4b+3p}$

$3p={x}^2-4b$

Add $4b$ to both sides

$3p+4b=x^{2}$

Square root both sides

$x=\pm\sqrt{4b+3p}$

$x=\frac{6{g}^2+8}{7}$

$x=\frac{36{g}^2-8}{7}$

$x=\frac{36{g}^2+8}{7}$

$x=\frac{6g+8}{7}$

$6g=\sqrt{7x-8}$

Square both sides

$(6g)^{2}=(\sqrt{7x-8})^{2}$

$36g^{2}=7x-8$

Add $8$ to both sides

$36g^{2}+8=7x$

Divide both sides by $7$

$x=\frac{36{g}^2+8}{7}$

$x=\frac{f}{5y-4}$

$x=\frac{-f}{5y+4}$

$x=\frac{f}{5y+4}$

$x=\frac{-f}{5y-4}$

$y=\frac{4x-f}{5x}$

Multiply both sides by $5x$

$5xy=4x-f$

Subtract $4x$ from both sides

$5xy-4x=-f$

Factorise the left hand side

$x(5y-4)=-f$

Divide both sides by the quantity in the bracket

$x=\frac{-f}{5y-4}$

$x=\frac{18+3y}{y+6}$

$x=\frac{18-3y}{y-6}$

$x=\frac{18-3y}{y+6}$

$x=\frac{18+3y}{y-6}$

$\frac{y}{3}=\frac{6-2x}{x+3}$

Multiply each side of the equation by the denominator of the other side.

$xy+3y=18-6x$

To both sides, add $6x$ and subtract $3y$

$xy+6x=18-3y$

Factorise the left hand side

$x(y+6)=18-3y$

Divide by the quantity in the bracket

$x=\frac{18-3y}{y+6}$

Make x the subject GCSE Questions

1. Make $x$ the subject of the formula

$y=5x-7$

(2 marks)

Show answer

$y+7=5x$

(1)

$\frac{y+7}{5} = x$

(1)

2. Make $x$ the subject of the formula

$z^{2}=x^{2}-5 a y$

(2 marks)

Show answer

$z^{2}+5 a y=x^{2}$

(1)

$\pm\sqrt{z^{2}+5 a y}=x$

(1)

3. Make $x$ the subject of the formula

$y=\frac{3(t+5 x)}{x}$

(4 marks)

Show answer

$y x=3 t+15 x$

(1)

$y x-15 x=3 t$

(1)

$x(y-15)=3 t$

(1)

$x=\frac{3t}{y-15} \quad$

(1)

Learning checklist

You have now learned how to:

Understand and use standard mathematical formulae

Rearrange formulae to change the subject

The next lessons are

Still stuck?

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