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Year 11
Higher

Changing the subject with multiple algebraic fractions

I can change the subject of a formula involving multiple algebraic fractions.

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Year 11
Higher

Changing the subject with multiple algebraic fractions

I can change the subject of a formula involving multiple algebraic fractions.

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

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Lesson details

Key learning points

  1. It is important to identify which variable is to become the subject.
  2. Like terms containing the new subject should be collected.
  3. You should be prepared to utilise any of your algebraic manipulation skills.

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

Pupils may think that the reciprocal of $$\\frac{1}{a} + \\frac{1}{b}$$ is $$a + b$$

$$\\frac{1}{2} + \\frac{1}{4} = \\frac{3}{4}$$ which is not the reciprocal of 2 + 4 = 6. To find the reciprocal, it helps for the expression to be written as a single fraction.


To help you plan your year 11 maths lesson on: Changing the subject with multiple algebraic fractions, download all teaching resources for free and adapt to suit your pupils' needs...

Rearranging equations can produce lots of different but equivalent answers. Therefore questions can be presented as "show that" style questions. Pupils may need guidance on the level of working and presentation required for others to clearly follow their work.
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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.
What is the solution to the equation $${8\over a}=6$$?
$$a=\frac{3}{4}$$
Correct answer: $$a=\frac{4}{3}$$
$$a=1.5$$
$$a=2$$
$$a=48$$
Q2.
Make $$x$$ the subject of $${2\over 5x} = 3y$$.
$$x = {2\over 3}y$$
$$x = {3y\over 10}$$
$$x = {6y\over 5}$$
Correct answer: $$x = {2\over 15y}$$
$$x = {6\over 5y}$$
Q3.
Make $$a$$ the subject of $$x=\frac{y+5}{a-b}$$.
Correct answer: $$a=\frac{y+5}{x} + b$$
$$a=\frac{y+5}{-bx}$$
$$a=\frac{y+5+b}{x}$$
$$a=\frac{y+5-bx}{x}$$
$$a=5x+xy+b$$
Q4.
Factorise $$x^2-5x-6$$.
$$(x-5)(x-6)$$
$$(x-5)(x-1)$$
$$(x-3)(x-2)$$
$$(x-1)(x+6)$$
Correct answer: $$(x+1)(x-6)$$
Q5.
Which of these is equivalent to $${3x^2-12x\over x^2-9x+20}$$?
$$3\over x$$
$$3x\over x-4$$
Correct answer: $$3x\over x-5$$
$$x^2-2x\over 10$$
$$x^2 -4x\over x^2-3x+20$$
Q6.
Which of these is the simplest form of $${3\over 2x+4}+{7\over 5x+10}$$?
$$100x+74$$
Correct answer: $$29\over 10x+20$$
$$58\over 5x+10$$
$$10x+20\over 7x+14$$
$$29x+58\over (2x+4)(5x+10)$$

Exit quiz

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6 Questions

Q1.
In which of these is $$a$$ the subject?
$${1\over a}=bc + 2$$
$$a = {bc+2 \over 4a}$$
$$a^2 = bc + 2$$
Correct answer: $$a={5b-2\over 2c+3b}$$
$${1\over 3}a = b^2+c + 2$$
Q2.
Make $$a$$ the subject of $$5=\frac{b+2a}{a+4}$$.
$$a=\frac{b-4}{3}$$
$$a = 4 - \frac{b}{5}$$
Correct answer: $$a=\frac{b-20}{3}$$
$$a=\frac{b+2a}{5}-4$$
$$a=\frac{5a+4-b}{2}$$
Q3.
Which of these is equivalent to $$4+b=\frac{3a-4}{2b+1}$$?
$$a=\frac{10b}{3}+3$$
$$a=\frac{2b^2+8}{3}$$
$$a=\frac{b(2b+9)}{3}$$
$$a=\frac{2b^2+8b + 5}{3}$$
Correct answer: $$a=\frac{2b^2+9b + 8}{3}$$
Q4.
Make $$a$$ the subject of $$\frac{a}{4} + \frac{3}{b}= \frac{c+2}{5}$$.
$$a=\frac{4bc+8b}{15}$$
$$a=\frac{4bc-52}{5b}$$
$$a=\frac{4bc-58}{5b}$$
Correct answer: $$a=\frac{4bc+8b-60}{5b}$$
$$a=\frac{bc+2b+4c}{5}+1$$
Q5.
Make $$a$$ the subject of $$\frac{6b+3}{c}=\frac{4b+2}{a}$$.
Correct answer: $$a = {2c\over 3}$$
$$a ={3\over 2}c$$
$$a={c\over 6b+3}$$
$$a = {4bc-3\over 6}$$
$$a = {2b+1\over c}$$
Q6.
Make $$a$$ the subject of $$\frac{2}{c} = \frac{2}{b} - \frac{2}{a}$$.
$$a = b-c$$
$$a = c+b$$
Correct answer: $$a = \frac{bc}{c-b}$$
$$a = \frac{b-c}{4}$$
$$a = \frac{bc+c}{2b}$$