GCSE Maths revision · Higher: number and algebra

Algebraic fractions

Higher only

The idea

Algebraic fractions work like number fractions. To simplify, factorise top and bottom and cancel common factors. To add, use a common denominator.

Factorise everything you can.
Cancel whole brackets that appear on top and bottom.
To solve an equation with fractions, multiply every term by the common denominator, then reject any answer that makes a denominator zero.

Watch out Cancel factors, never terms: $\frac{x + 6}{x}$ does not simplify to $6$ .

See it worked

Worked exampleNon-calculator4 marks

Simplify fully.

(a) Simplify fully $\dfrac{3x^2 + 6x}{x^2 - 4}$

(b) Simplify fully $\dfrac{2x}{5} \div \dfrac{4x^2}{15}$

(a) Factorise. $\frac{3x(x + 2)}{(x + 2)(x - 2)} = \frac{3x}{x - 2}$
(b) Multiply by the reciprocal. $\frac{2x}{5} \times \frac{15}{4x^2} = \frac{30x}{20x^2} = \frac{3}{2x}$

Answer (a) $\dfrac{3x}{x - 2}$ ; (b) $\dfrac{3}{2x}$

Your turn

Question 1Non-calculator2 marks

Simplify ( $x^{2}$ $-$ 16)/(x + 4), where x $\ne$ $-4.$

Question 2Calculator allowed3 marks

Simplify [( $x^{2}$ $-$ 9)/( $x^{2}$ + x $-$ 6)] $\div$ [(x + 3)/(x $-$ 1)], where x is not $-3$ , 1 or 2. Give your answer as a single fraction in factorised form.

Question 3Non-calculator4 marks

Simplify $\frac{x-5}{x+7}\div\frac{x-5}{x}$ , where $x\ne0,5,-7$ .

(a) Give your answer as a single fraction.

(b) Find the value of the simplified expression when $x = 14$ .

In past papers

59 official questions in 55 papers since 2017. The latest:

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