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Year 9

Changing the subject with simple formula

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

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New
New
Year 9

Changing the subject with simple formula

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

Download all resources
Share activities with pupils

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

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Lesson details

Key learning points

  1. Changing the subject can be thought of as solving an equation in one unknown.
  2. Throughout the rearrangement, the expressions on each side of the equals sign remain equal.
  3. Changing the subject can offer different perspectives on the relationship.
  4. It can make it easier to calculate the unknown value if that is the subject.

Keywords

  • Formula - A formula is a rule linking sets of physical variables in context. The plural of formula is formulae.

  • Subject - 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

The subject of a formula is just the first term in the formula.

Draw attention to the variety of equations and formulae used in the lesson.


To help you plan your year 9 maths lesson on: Changing the subject with simple formula, download all teaching resources for free and adapt to suit your pupils' needs...

It will help pupils if they can link this skill back to solving equations. Pupils will need to be confident in forming expressions correctly so this can be an important skill to practise before this lesson.
Teacher tip

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

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Starter quiz

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6 Questions

Q1.
Division in the __________ operation of multiplication.
improper
inequal
invariant
Correct answer: inverse
Q2.
Select the rearrangements of the additive relationship 389.16 + 417.4 = 806.56.
Correct answer: 417.4 + 389.16 = 806.56
806.56 + 417.4 = 389.16
417.4 − 806.56 = 389.16
Correct answer: 806.56 − 417.4 = 389.16
417.4 − 389.16 = 806.56
Q3.
Select the rearrangements of the multiplicative relationship $$y=5x$$.
$$5y=x$$
$${5\over{y}}=x$$
Correct answer: $${y\over{5}}=x$$
Correct answer: $${y\over{x}}=5$$
$${x\over{y}}=5$$
Q4.
The solution to $$x-13=63$$ is $$x=$$ .
Correct Answer: 76, seventy six, seventy-six
Q5.
Laura begins to solve the equation $$x-12=13$$ by writing $$x-12+12=13$$. What error has Laura made?
Laura should add $$13$$, not $$12$$.
Correct answer: Laura has only performed the operation $$+12$$ to one side of the equation.
The answer is obviously just $$1$$. Laura doesn't need a method here.
Laura should do $$-12$$, not $$+12$$
Q6.
Which of the below are rearrangements of the equation $$17-x=2y$$ ?
$$17+2y=x$$
Correct answer: $$17=2y+x$$
$$x-17=2y$$
Correct answer: $$17-2y=x$$
$$17-2x=y$$

Exit quiz

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6 Questions

Q1.
The __________ of a formula is a variable expressed in terms of other variables.
solution
Correct answer: subject
value
variable
Q2.
$$y^2$$ is not the subject of the equation $$y^2=25-x^2$$. Why not?
Because $$y^2$$ is not the solution.
Because $$x$$ is the subject.
Correct answer: Because the exponent of $$y$$ is not 1.
Because this is not an equation.
Q3.
Match each equation to its subject.
Correct Answer:$$c + d + e = f$$,$$f$$

$$f$$

Correct Answer:$$c = f-(d+e)$$,$$c$$

$$c$$

Correct Answer:$$c-d-f=e$$,$$e$$

$$e$$

Correct Answer:$$d=c-ef$$,$$d$$

$$d$$

Q4.
Laura wants to make $$x$$ the subject of the equation $$x+3y=17$$. Which operation should she apply to both sides of the equation?
$$-17$$
$$\div3$$
$$-x$$
Correct answer: $$-3y$$
$$+3y$$
Q5.
Make $$n$$ the subject of $$T=25-n$$.
$$n=25$$
$$n+T=25$$
$$n=T-25$$
Correct answer: $$n=25-T$$
$$n=T-25$$
Q6.
The formula for the volume of a cylinder is $$V=\pi{r^2}h$$. Select the correct arrangement of the formula to make $$h$$ the subject.
$$V\pi{r^2}=h$$
$${V\over{\pi}r}=h^2$$
Correct answer: $${V\over{\pi}r^2}=h$$
$$h={\pi{r^2}\over{V}}$$
$${V\over{h}}={\pi}{r^2}$$