Skip to content
Oak National Academy
Oak National Academy
Teachers
Pupils
icon-background-square
New
New
Year 10
Foundation

Checking and securing understanding of changing the subject

I can apply an understanding of inverse operations to a formula in order to make a specific variable the subject.

Download all resources
arrow-right
Share activities with pupils
arrow-right
icon-background-square
New
New
Year 10
Foundation

Checking and securing understanding of changing the subject

I can apply an understanding of inverse operations to a formula in order to make a specific variable the subject.

Download all resources
arrow-right
Share activities with pupils
arrow-right
warning

These resources will be removed by end of Summer Term 2025.

Switch to our new teaching resources now - designed by teachers and leading subject experts, and tested in classrooms.

These resources were created for remote use during the pandemic and are not designed for classroom teaching.

View new resources
arrow-right

Lesson details

Key learning points

  1. Rearranging the formula may be beneficial when what you wish to calculate is not the subject.
  2. You can compare rearranging the formula to solving for one unknown.
  3. Instead of finding a value though, you will find an expression that the variable is equal to.

Keywords

  • Subject of an equation - The subject of an equation/a formula is a variable that is expressed in terms of other variables. It should have an exponent of 1 and a coefficient of 1.

Common misconception

When rearranging multiplicative relationships, the values just 'swap'

Pupils should understand how to manipulate multiplicative relationships properly instead of relying on a 'trick'. This will be useful in other subject areas as well.


To help you plan your year 10 maths lesson on: Checking and securing understanding of changing the subject, download all teaching resources for free and adapt to suit your pupils' needs...

There are opportunities to see the importance of changing the subject in a range of topic areas. For example when rearranging trigonometric ratios, or using geometric formula. There are also examples of linear relationships between x and y which can be linked to identifying the gradient of a line.
speech-bubble
Teacher tip
copyright

Licence

This content is © Oak National Academy Limited (2025), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Lesson video

Skip lesson video

Loading...

Starter quiz

Download starter quiz

6 Questions

Q1.
If the side of an equilateral triangle is $$3n+1$$, write an expression for the perimeter.
Correct Answer: 3(3n+1), 9n + 3
Q2.
If the side of a regular pentagon is $$14y - a$$ write an expression for the perimeter.
Correct Answer: 5(14y - a), 70y - 5a
Q3.
Which of the following is an expression for the area of a square with side length $$3x$$
$$3x^2$$
$$(3x^2)$$
Correct answer: $$(3x)^2$$
Q4.
What is the value of $$x$$ for $$5(x + 7) = 30$$?
Correct Answer: -1
Q5.
If $$x$$ is a positive integer, what is the value of $$x$$ for $$x^2 - 4 = 0$$?
Correct Answer: 2
Q6.
What is the value of $$x$$ for $$2x - 3 = 7$$?
Correct Answer: 5

Exit quiz

Download exit quiz

6 Questions

Q1.
Make $$x$$ the subject of $$3y - x = 15$$
Correct answer: $$x = 3y - 15$$
$$x = 15 - 3y$$
$$x = 3y + 15$$
$$x = -3y - 15$$
Q2.
Make $$y$$ the subject of $$4x + 2y = 8$$
$$y = 2x + 4$$
$$y = -2x - 4$$
Correct answer: $$y = 4 - 2x$$
$$y = 2 - 4x$$
Q3.
Make $$z$$ the subject of $$\frac{z}{2} + 3x = 7$$
$$z = 7 - 3x$$
$$z = 4 - 3x$$
$$z = 14 - 3x$$
Correct answer: $$z = 14 - 6x$$
Q4.
Make $$a$$ the subject of $$b = 3(a - 4) + 2$$
$$a = \frac{b + 2}{3} - 4$$
Correct answer: $$a = \frac{b - 2}{3} + 4$$
$$a = \frac{b - 2}{3} - 4$$
$$a = \frac{b + 6}{3}$$
Q5.
Make $$m$$ the subject of $$\frac{4m + 5}{3} = n$$
$$m = 3n + 5$$
$$m = \frac{3n + 5}{4}$$
Correct answer: $$m = \frac{3n - 5}{4}$$
$$m = \frac{n - 5}{3}$$
Q6.
Make $$p$$ the subject of $$2p + \frac{q}{3} = 4$$
$$p = \frac{8 - q}{6}$$
$$p = 2 - \frac{q}{3}$$
$$p = \frac{4 - q}{3}$$
Correct answer: $$p = 2 - \frac{q}{6}$$