A4AlgebraHigherGrades 7–9

Algebraic Fractions

Fractions with letters in them — like $\dfrac{x^{2} - 9}{x^{2} + x - 6}$ — follow the same rules as ordinary fractions. The difference is that finding the common factor, or the common denominator, usually needs factorising first. AQA Higher papers test simplifying, combining and solving with algebraic fractions.

50 min AQA GCSE Maths

The big picture

Simplifying $\dfrac{15}{20}$ to $\dfrac34$ works because 5 is a factor of both. Algebraic fractions are the same: factorise top and bottom, and any bracket that appears in both can be cancelled. Individual terms joined by $+$ or $-$ can never be cancelled.

Adding and subtracting need a common denominator, and solving an equation starts by multiplying every term by that denominator. These steps are exactly the ones used with number fractions, which is the best check when unsure.

What you'll learn

Simplify algebraic fractions by factorising
Add and subtract algebraic fractions
Multiply and divide algebraic fractions
Solve equations involving algebraic fractions

Simplifying

Factorise the numerator and the denominator completely, then cancel common brackets.

$\dfrac{x^{2} - 9}{x^{2} + x - 6} = \dfrac{(x - 3)(x + 3)}{(x + 3)(x - 2)} = \dfrac{x - 3}{x - 2}$ .

Example 1Show that

\dfrac{2x^{2} + 5x - 3}{4x^{2} - 1}

simplifies to

\dfrac{x + 3}{2x + 1}

1Numerator:

(2x - 1)(x + 3)

2Denominator (difference of two squares):

(2x - 1)(2x + 1)

3Cancel

(2x - 1)

Answer

\dfrac{x + 3}{2x + 1}

, as required.

Your turnSame method, new numbers — try it before moving on.

Simplify fully $\dfrac{x^2 - 9}{x^2 + x - 6}$ .

Your answer

Type it like (x+1)/(x-4).

Show hint

The top is a difference of two squares.

Show worked solution

Solution

1Top: $(x + 3)(x - 3)$ . Bottom: $(x + 3)(x - 2)$ .
2Cancel $(x + 3)$ : $\dfrac{x - 3}{x - 2}$ .

Answer: $\dfrac{x - 3}{x - 2}$

Tip — If you cannot see a common factor, check the denominator for a difference of two squares — it is the most common hidden factorisation.

Adding and subtracting

Multiply the denominators (or use their lowest common multiple) to get a common denominator.

Adjust each numerator, then combine. Brackets around numerators prevent sign errors when subtracting.

Example 1Write

\dfrac{2}{x + 1} + \dfrac{3}{x - 2}

as a single fraction.

1Common denominator:

(x + 1)(x - 2)

2Numerator:

2(x - 2) + 3(x + 1) = 2x - 4 + 3x + 3

Answer

\dfrac{5x - 1}{(x + 1)(x - 2)}

Your turnSame method, new numbers — try it before moving on.

Write $\dfrac{3}{x + 2} - \dfrac{1}{x - 1}$ as a single fraction.

Your answer

Type it like (2x-5)/((x+2)(x-1)).

Show worked solution

Solution

1Common denominator $(x + 2)(x - 1)$ .
2Numerator: $3(x - 1) - (x + 2) = 3x - 3 - x - 2 = 2x - 5$ .

Answer: $\dfrac{2x - 5}{(x + 2)(x - 1)}$

Multiplying and dividing

To multiply, factorise everything, cancel across, and multiply what is left.

To divide, turn the second fraction upside down and multiply.

Example 1Simplify

\dfrac{x + 5}{x^{2} - 4} \times \dfrac{3x - 6}{x^{2} + 5x}

1Factorise:

\dfrac{x + 5}{(x + 2)(x - 2)} \times \dfrac{3(x - 2)}{x(x + 5)}

2Cancel

(x + 5)

and

(x - 2)

Answer

\dfrac{3}{x(x + 2)}

Why this works

Cancelling before multiplying keeps the algebra small. Multiplying everything out first produces quartic expressions that are much harder to simplify.

Solving equations

Multiply every term by the common denominator to clear the fractions, then solve the resulting equation.

Check each solution does not make a denominator zero.

Example 1Solve

\dfrac{3}{x - 1} - \dfrac{2}{x + 1} = 1

1Multiply by

(x - 1)(x + 1)

3(x + 1) - 2(x - 1) = (x - 1)(x + 1)

x + 5 = x^{2} - 1

, so

x^{2} - x - 6 = 0

(x - 3)(x + 2) = 0

Answer

x = 3

x = -2

Your turnSame method, new numbers — try it before moving on.

Solve $\dfrac{6}{x} - \dfrac{6}{x + 1} = 1$ .

Your answer

Separate the solutions with a comma.

Show hint

Multiply every term by $x(x + 1)$ .

Show worked solution

Solution

1 $6(x + 1) - 6x = x(x + 1)$ , so $6 = x^2 + x$ .
2 $x^2 + x - 6 = 0$ , so $(x + 3)(x - 2) = 0$ .

Answer: $x = 2$ or $x = -3$

Tip — Substitute answers back: at $x = 3$ , $\tfrac32 - \tfrac24 = 1$ ✓. It takes seconds and confirms the algebra.

Think like an examiner

Show fully factorised forms before cancelling.
Keep denominators factorised in the final answer.
Multiply every term, including the right-hand side, when clearing fractions.

Remember these

\tfrac{ac}{bc} = \tfrac{a}{b}

Cancel common factors.

\tfrac{a}{b} + \tfrac{c}{d} = \tfrac{ad + bc}{bd}

Adding fractions.

\tfrac{a}{b} \div \tfrac{c}{d} = \tfrac{ad}{bc}

Dividing fractions.

Watch out for these

Cancelling the

x

\dfrac{x + 4}{x}

Only factors cancel; this cannot be simplified.

Writing

\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{2}{a + b}

\dfrac{a + b}{ab}

Forgetting brackets when subtracting a numerator.

3(x + 1) - 2(x - 1)

gives

+2

, not

-2

Leaving a solution that makes a denominator zero.

Reject values where the original fractions are undefined.

Stretch yourself

Solve $\dfrac{x}{x + 3} + \dfrac{2}{x - 1} = 1$ .

Hint — Multiply by $(x + 3)(x - 1)$ ; the $x^{2}$ terms will cancel.

Reveal solution

SolutionMultiplying through:

x(x - 1) + 2(x + 3) = (x + 3)(x - 1)

. So

x^{2} - x + 2x + 6 = x^{2} + 2x - 3

, which gives

x + 6 = 2x - 3

and

x = 9

. Check:

\tfrac{9}{12} + \tfrac28 = \tfrac34 + \tfrac14 = 1

✓.

\blacksquare

Questions students ask

When can I cancel?

Only when the same factor multiplies the whole numerator and the whole denominator.

Should I expand the denominator in my answer?

No — leaving it factorised is simpler and accepted.

Why check solutions?

Clearing fractions can introduce values that make an original denominator zero.

Key takeaways

Factorise fully before simplifying.
Cancel factors, never individual terms.
Add and subtract with a common denominator.
Multiply by cancelling across; divide by flipping the second fraction.
Solve by multiplying every term by the common denominator.

Exam-style questions

Practise it the way it is examined

3 original questions · 10 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 73 marks

Simplify fully $\dfrac{3x^2 - 12}{x^2 + 3x - 10}$ .[3]
Show mark scheme
- M1Numerator factorised: $3(x + 2)(x - 2)$
- M1Denominator factorised: $(x + 5)(x - 2)$
- A1 $\dfrac{3(x + 2)}{x + 5}$
Answer: $\dfrac{3(x + 2)}{x + 5}$

Question 2Grade 84 marks

Solve $\dfrac{2}{x - 3} + \dfrac{5}{x + 2} = 1$ . Give your answers to 2 decimal places.[4]
Show mark scheme
- M1Multiplies through: $2(x + 2) + 5(x - 3) = (x - 3)(x + 2)$
- M1Forms $x^2 - 8x + 5 = 0$
- M1Correct use of the quadratic formula
- A1 $x = 7.32$ and $x = 0.68$
Answer: $x = 7.32$ or $x = 0.68$

Question 3Grade 93 marks

Show that $\dfrac{x + 1}{x^2 - 1} \div \dfrac{x^2 + 2x}{x - 1}$ simplifies to $\dfrac{1}{x(x + 2)}$ .[3]
Show mark scheme
- M1Inverts and multiplies
- M1Factorises $x^2 - 1 = (x + 1)(x - 1)$ and $x^2 + 2x = x(x + 2)$
- A1Cancels fully to $\dfrac{1}{x(x + 2)}$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

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