1MA1 · A4AlgebraHigherGrades 7–9

Algebraic Fractions

An algebraic fraction such as $\dfrac{x^{2} - 9}{x^{2} + 5x + 6}$ obeys exactly the same rules as $\dfrac{12}{18}$ : you simplify by cancelling common factors, find common denominators to add, and flip to divide. The difference is that the common factors are hidden until you factorise — which is why this topic pulls together everything from the expressions chapter.

55 min Algebraic Expressions

The big picture

The golden rule is that you may only cancel $\textit{factors}$ , never individual terms. $\dfrac{x + 3}{x + 5}$ cannot be simplified, however tempting it is to cross out the $x$ s, because $x$ is not a factor of the whole top and bottom.

Edexcel Higher papers typically ask you to simplify fully, to combine two fractions into one, or to solve an equation containing algebraic fractions — usually for 3 to 5 marks, with method marks for the factorised forms.

What you'll learn

Simplify algebraic fractions by factorising and cancelling common factors
Multiply and divide algebraic fractions
Add and subtract algebraic fractions using a common denominator
Solve equations involving algebraic fractions

Simplifying

Factorise the numerator and denominator fully, then cancel any bracket or factor that appears in both.

$\dfrac{4x + 8}{x^{2} - 4} = \dfrac{4(x + 2)}{(x + 2)(x - 2)} = \dfrac{4}{x - 2}$ .

Watch for brackets that are negatives of each other: $(3 - x) = -(x - 3)$ .

Example 1Simplify fully

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

1Numerator:

(2x - 1)(x + 3)

2Denominator:

x(x + 3)

3Cancel

(x + 3)

Answer

\dfrac{2x - 1}{x}

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

Simplify fully $\dfrac{x^2 + 5x + 6}{x^2 - 4}$ .

Your answer

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

Show worked solution

Solution

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

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

Tip — Before cancelling, every part of the top and bottom must be written as a product of brackets. If a plus or minus sign sits outside a bracket, you have not finished factorising.

Multiplying and dividing

To multiply, factorise everything, cancel common factors across any numerator and any denominator, then multiply what remains.

To divide, flip the second fraction and multiply.

Example 1Simplify

\dfrac{x^{2} - 16}{3x} \div \dfrac{x + 4}{6x^{2}}

1Flip:

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

2Cancel

(x + 4)

, and

\dfrac{6x^{2}}{3x} = 2x

Answer

2x(x - 4)

Adding and subtracting

Use a common denominator — usually the product of the denominators, or their lowest common multiple if they share factors.

Multiply each numerator by whatever its denominator was multiplied by, then combine numerators. Put brackets around each numerator when subtracting.

Example 1Write

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

as a single fraction in its simplest form.

1Common denominator

(x + 2)(x - 1)

2Numerator:

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

3Simplify to

x - 7

Answer

\dfrac{x - 7}{(x + 2)(x - 1)}

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

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

Your answer

Type it like (3x-5)/((x+1)(x-3)).

Show worked solution

Solution

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

Answer: $\dfrac{3x - 5}{(x + 1)(x - 3)}$

Why this works

The bracket around $2(x + 2)$ is what turns $+4$ into $-4$ . Subtracting a fraction subtracts its entire numerator, so every term must change sign.

Solving equations

To solve an equation with algebraic fractions, multiply every term by the common denominator to clear the fractions.

This usually leaves a linear or quadratic equation. Check solutions do not make any original denominator zero.

Example 1Solve

\dfrac{4}{x} + \dfrac{3}{x + 1} = 2

1Multiply by

x(x + 1)

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

7x + 4 = 2x^{2} + 2x

, so

2x^{2} - 5x - 4 = 0

3Quadratic formula:

x = \dfrac{5 \pm \sqrt{25 + 32}}{4} = \dfrac{5 \pm \sqrt{57}}{4}

Answer

x = 3.14

x = -0.64

(2 d.p.)

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

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

Your answer

Separate with a comma.

Show hint

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

Show worked solution

Solution

1 $3(x - 1) + 2x = 2x(x - 1)$ , so $5x - 3 = 2x^2 - 2x$ .
2 $2x^2 - 7x + 3 = 0$ , so $(2x - 1)(x - 3) = 0$ .

Answer: $x = \frac{1}{2}$ or $x = 3$

Tip — When clearing fractions, multiply the right-hand side too. Leaving it unmultiplied is the most common error in these questions.

Think like an examiner

Show the fully factorised forms before cancelling.
Use brackets around numerators when combining subtracted fractions.
Check that final answers cannot be simplified further.

Algebraic fractions

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

Cancel common factors only.

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

Common denominator.

\tfrac{a}{b} \div \tfrac{c}{d} = \tfrac{a}{b} \times \tfrac{d}{c}

Dividing.

Watch out for these

Cancelling

x

\dfrac{x + 3}{x}

to get

3

x

is not a factor of

x + 3

; this does not simplify.

Adding denominators:

\tfrac{1}{x} + \tfrac{1}{y} = \tfrac{2}{x + y}

Use a common denominator:

\tfrac{x + y}{xy}

Writing

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

3x - 3 - 2x - 4 = x - 7

Forgetting to multiply the right-hand side when clearing fractions.

Multiply every term on both sides.

Stretch yourself

Show that $\dfrac{x}{x - 3} - \dfrac{18}{x^{2} - 9}$ simplifies to $\dfrac{x + 6}{x + 3}$ .

Hint — Factorise $x^{2} - 9$ first: the lowest common denominator is $(x - 3)(x + 3)$ , not the product of both denominators.

Reveal solution

SolutionSince

x^{2} - 9 = (x - 3)(x + 3)

, the common denominator is

(x - 3)(x + 3)

. The first fraction becomes

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

. Combining:

\dfrac{x^{2} + 3x - 18}{(x - 3)(x + 3)}

. The numerator factorises as

(x + 6)(x - 3)

, so cancelling

(x - 3)

gives

\dfrac{x + 6}{x + 3}

, as required.

\blacksquare

Questions students ask

Why can’t I cancel terms?

Cancelling divides top and bottom by the same thing. That only works for something that multiplies the whole numerator and the whole denominator — a factor.

Should I expand the denominator?

Usually leave it factorised. It is simpler, and examiners accept it.

What values of

x

are not allowed?

Any value that makes an original denominator zero, even if the factor cancels during simplification.

Key takeaways

Factorise fully, then cancel common factors — never individual terms.
Multiply by cancelling across and multiplying; divide by flipping the second fraction.
Add and subtract with a common denominator, bracketing numerators.
Clear fractions in equations by multiplying every term by the common denominator.
Reject solutions that make a denominator zero.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 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 $\dfrac{4x^2 - 9}{2x^2 + x - 6}$ .[3]
Show mark scheme
- M1Numerator: $(2x + 3)(2x - 3)$
- M1Denominator: $(2x - 3)(x + 2)$
- A1 $\dfrac{2x + 3}{x + 2}$
Answer: $\dfrac{2x + 3}{x + 2}$

Question 2Grade 83 marks

Solve $\dfrac{6}{x + 2} + \dfrac{1}{x - 1} = 2$ .[3]
Show mark scheme
- M1Multiplies by $(x + 2)(x - 1)$ : $6(x - 1) + (x + 2) = 2(x + 2)(x - 1)$
- M1Simplifies to $2x^2 - 5x = 0$
- A1 $x = 0$ and $x = 2.5$
Answer: $x = 0$ or $x = 2.5$

Question 3Grade 93 marks

Show that $\dfrac{x^2 - 1}{x^2 + 3x + 2} \times \dfrac{x + 2}{x - 1}$ simplifies to 1.[3]
Show mark scheme
- M1Factorises $x^2 - 1$ and $x^2 + 3x + 2$
- M1Multiplies and identifies common factors
- A1Cancels fully to 1

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

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Next step

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