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Year 11
Higher

Changing the subject where the variable appears in multiple terms

I can change the subject where the desired variable appears in multiple terms.

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Year 11
Higher

Changing the subject where the variable appears in multiple terms

I can change the subject where the desired variable appears in multiple terms.

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Lesson details

Key learning points

  1. The variable to become the subject may appear in more than one term.
  2. It is possible the terms cannot be additively combined.
  3. In this situation, it is possible to factorise with one of the factors being the desired variable.

Keywords

  • Subject of an equation/formula - 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

Not fully isolating the required variable by leaving other terms containing the variable in the expression.

To make a variable the subject, it must be expressed in terms of other variables only. Factorising helps isolate a variable if it is in multiple terms.


To help you plan your year 11 maths lesson on: Changing the subject where the variable appears in multiple terms, download all teaching resources for free and adapt to suit your pupils' needs...

Task A contains a question where pupils are rearranging the formula for the perimeter of a sector. You may wish to recap perimeter of a sector here and allow pupils to apply the formula to different questions. They could think about making different variables the subject as required.
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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.
In which of these is $$x$$ the subject?
$$x = \frac{xy+4}{2}$$
$${1\over x} = {1\over y+2}$$
Correct answer: $$\frac{2y+2z}{3y+3z}=x$$
Correct answer: $$x=2(3z+y) + \frac{z}{y}$$
$$\frac{3w(2+y)}{x(2 + y)}=x$$
Q2.
Make $$x$$ the subject of $$3x+4=5x-b$$.
$$x = 2 + b$$
Correct answer: $$x=\frac{4+b}{2}$$
$$x=\frac{4-b}{2}$$
$$x=\frac{3x+4+b}{5}$$
$$x=\frac{5x-4-b}{3}$$
Q3.
Factorise $$x^2 - 6x + 8$$.
$$(x-8)(x-1)$$
$$(x-8)(x+1)$$
$$(x-6)(x+8)$$
Correct answer: $$(x-4)(x-2)$$
$$(x-4)(x+2)$$
Q4.
Shown is the formula for the perimeter of a rectangle. Make $$b$$ the subject.
An image in a quiz
$$b = P - a$$
$$b=2P-a$$
$$2b=P-2a$$
$$b=\frac{P+2a}{2}$$
Correct answer: $$b=\frac{P}{2}-a$$
Q5.
Make $$z$$ the subject of $$\frac{4}{x} +\frac{2}{y} = \frac{3}{z}$$.
$$z=\frac{x+y}{2}$$
$$z=\frac{2y+x}{3}$$
$$z=\frac{3x+6y}{4}$$
Correct answer: $$z=\frac{3xy}{4y+2x}$$
$$z= 3-\frac{4y+2x}{xy}$$
Q6.
Simplify $$\frac{3x-12}{2x-6} + \frac{x}{3x-9}$$
$$\frac{7x^2-7x+9}{6x-18}$$
Correct answer: $$\frac{11x-36}{6x-18}$$
$$\frac{11x-4}{6x-3}$$
$$\frac{2x+9}{6}$$
$$\frac{4}{5}$$

Exit quiz

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6 Questions

Q1.
What is the most efficient first step to make $$x$$ the subject of $$ax + b = 2x + c$$?
Divide both sides by 2.
Factorise both sides of the equation.
Correct answer: Subtract $$2x$$ from both sides.
Q2.
What is the most efficient next step to make $$x$$ the subject of $$ax-2x = c-b$$ ?
Add $$b$$ to both sides.
Divide both sides by 2.
Divide both sides by $$a$$.
Correct answer: Factorise the expressions.
Q3.
Make $$a$$ the subject of $$a(b+c)=3+4(a+2b)$$.
Correct answer: $$a=\frac{3+8b}{b+c-4}$$
$$a=\frac{3+8b}{b+c+4}$$
$$a=\frac{14b}{b+c-7}$$
$$a=\frac{3+2b-c}{b+4}$$
$$a=\frac{3+4a+8b}{b+c}$$
Q4.
Make $$r$$ the subject of $$p = \frac{r+5}{r+q}$$.
$$r=\frac{pq}{5}$$
$$r=\frac{r+5}{p}-q$$
$$r=\frac{5+pq}{p}$$
$$r=\frac{5-q}{p-1}$$
Correct answer: $$r=\frac{5-pq}{p-1}$$
Q5.
What would be the most efficient first step to make $$a$$ the subject of $$\frac{8a+4b}{6a+3b}+\frac{a+1}{3a+3b}=b$$ ?
Correct answer: Factorise and look for common factors.
Add the 2 fractions.
Multiply both sides by $$6a+3b$$.
Subtract $$b$$ from both sides.
Q6.
Make $$a$$ the subject of $$\frac{4(2a+b)}{3(2a+b)}+\frac{a+1}{3(a+b)}=b$$.
$$a=\frac{3b^2-2b-1}{5-3b}$$
Correct answer: $$a=\frac{3b^2-4b-1}{5-3b}$$
$$a=\frac{3b^2-2b-1}{9-6b}$$
$$a=\frac{9b^2-4b-1}{9-7b}$$