Algebraic Fractions GCSE Questions and Worked Answers

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Start with sharing an amount

What you need to know about algebraic fractions

Suppose a ribbon is x centimetres long and you cut it into three equal pieces. Each piece is x ÷ 3 centimetres long, written as the fraction x/3. Here x stands for a number we have not specified. If x is 12, each piece is 4 cm. The fraction bar means divide the whole top expression by the whole bottom expression.

The fraction rules still apply

A letter changes the value, not the fraction rules

The top of a fraction is its numerator; the bottom is its denominator. Brackets keep an entire sum together. A denominator cannot be zero because division by zero is undefined.

Amountx centimetresx is a number, not a new operation

Share equallydivide the amount by 3the denominator counts equal shares here

Substitute12 ÷ 3 = 4ordinary arithmetic checks the notation

Why cancellation works

In $\dfrac{6\times5}{9\times5}$ , both the whole numerator and whole denominator have a factor of $5$ . Dividing both by $5$ preserves the fraction's value. A factor is something multiplied to make an expression. A term is a part added or subtracted.

\frac{6x}{9x}=\frac{6\times x}{9\times x}=\frac23\qquad(x\ne0)

The condition $x\ne0$ means x must not equal zero. The original fraction is undefined there, even though the simplified expression looks harmless.

But $\dfrac{x+3}{x}$ contains a sum on top. The x is not a factor of the whole numerator, so you cannot cancel it to leave 3. At $x=6$ , the actual value is $9/6=1.5$ , not 3.

To expose a shared factor, factorise: rewrite a sum as a product. For example, $x^2+4x=x(x+4)$ because multiplying out the bracket gives the original sum.

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Simplify a fraction

What the problem asks: Write (x² + 4x)/x in its simplest form.

How to solve it: Factorise the numerator as x(x + 4). Divide the entire top and bottom by the shared non-zero factor x to get x + 4, retaining x ≠ 0.

Add or subtract fractions

What the problem asks: Write 2/x + 1/3 as one fraction.

How to solve it: Use denominator 3x: multiply the first fraction by 3/3 and the second by x/x. These are both 1 when x ≠ 0, so values are preserved. Then combine the numerators.

Multiply or divide fractions

What the problem asks: Simplify (3/x) × (x/8), or divide one fraction by another.

How to solve it: Multiply numerators and denominators, cancelling shared factors. For division, multiply by the reciprocal (the divisor with top and bottom swapped). The divisor itself must not be zero.

Solve a fraction equation

What the problem asks: Find x when 6/(x + 1) = 2.

How to solve it: Exclude x = −1, then multiply both sides by x + 1. This removes the denominator and gives 6 = 2(x + 1). Solve, then check the result in the original equation.

A reliable routine

Simplify by factorising and cancelling

Use this method when a fraction's numerator and denominator share multiplied factors. It preserves value because both are divided by the same non-zero amount. Addition, subtraction and solving equations need the separate methods shown in their examples.

Record any values that make an original denominator zero.
Factorise the whole numerator and denominator where possible.
Cancel only identical factors multiplying the entire numerator and denominator.
Simplify the remaining numerical fraction.
Retain the original exclusions and check using an allowed numerical value.

Check: The main manipulation and variable-denominator work on this page is Higher content. Be comfortable with ordinary fractions, expanding brackets and factorising first.

Fully worked

Algebraic fractions GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Cancel a common factor

Higher only3 marks

Question

Simplify $\dfrac{x^2+5x}{3x}$ and state any excluded value.

The denominator is zero at $x=0$ , so exclude it. Factorise the numerator:

\frac{x^2+5x}{3x}=\frac{x(x+5)}{3x}

Both whole expressions have factor $x$ . Divide top and bottom by it.

\boxed{\frac{x+5}{3},\quad x\ne0}

Check at $x=1$ : both forms give $2$ .

Example 2

Factorise both expressions

Higher only4 marks

Question

Simplify $\dfrac{x^2-16}{x^2+7x+12}$ . State all original denominator restrictions.

The top is a difference of two squares. For the bottom, 3 and 4 multiply to 12 and add to 7.

x^2-16=(x-4)(x+4)

x^2+7x+12=(x+3)(x+4)

The denominator requires $x\ne-3,-4$ .

\frac{(x-4)(x+4)}{(x+3)(x+4)}

Cancel the shared factor $(x+4)$ .

\boxed{\frac{x-4}{x+3},\quad x\ne-3,-4}

Tip: the cancelled factor still creates an original restriction.

Example 3

Add with unlike denominators

Higher only3 marks

Question

Write $\dfrac3x+\dfrac25$ as one fraction.

A common denominator is $5x$ , with $x\ne0$ . Multiply the numerator and denominator of the first fraction by $5$ , and those of the second fraction by $x$ . Each fraction keeps its value because its top and bottom are multiplied by the same non-zero number.

\frac3x=\frac{15}{5x}

\frac25=\frac{2x}{5x}

Now combine equal-sized parts.

\boxed{\frac{15+2x}{5x},\quad x\ne0}

Do not add the denominators.

Example 4

Subtract an entire numerator

Higher only4 marks

Question

Simplify $\dfrac2{x-1}-\dfrac1{x+2}$ .

Exclude $x=1,-2$ . Use the product $(x-1)(x+2)$ as a common denominator.

\frac{2(x+2)-(x-1)}{(x-1)(x+2)}

The minus applies to both terms in the second bracket.

\frac{2x+4-x+1}{(x-1)(x+2)}

\boxed{\frac{x+5}{(x-1)(x+2)},\quad x\ne1,-2}

Keeping the denominator factored makes its restrictions visible.

Example 5

Divide by another fraction

Higher only4 marks

Question

Simplify $\dfrac{3x}{x+2}\div\dfrac6{x-1}$ .

The original denominators exclude $x=-2,1$ . The divisor is non-zero everywhere it is defined.

Dividing by a fraction is multiplying by its reciprocal.

\frac{3x}{x+2}\times\frac{x-1}{6}

Cancel the numerical factor $3$ from $3$ and $6$ .

\boxed{\frac{x(x-1)}{2(x+2)},\quad x\ne-2,1}

Tip: flip only the fraction you are dividing by.

Example 6

Solve a variable-denominator equation

Higher only4 marks

Question

Solve $\dfrac5{x-2}=\dfrac3{x+1}$ .

Exclude $x=2,-1$ . Multiply both sides by $(x-2)(x+1)$ ; each denominator cancels from its whole side.

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

5x+5=3x-6

2x+5=-6

2x=-11

\boxed{x=-\frac{11}{2}}

It is not excluded. Substitution gives $-2/3$ on both original sides.

10 original questions · total 30 marks

Algebraic fractions GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 40 minutes · Show factorisation, keep brackets around whole numerators and record excluded values. Check equation answers in the original equation. · answers start collapsed

1

Simplify coefficients and powers

Higher only2 marks

Simplify $\dfrac{12x^2}{18x}$ , stating the restriction.

Show worked answer

Divide the coefficients by $6$ and cancel one shared factor $x$ :

\frac{12x^2}{18x}=\frac{2x}{3}

\boxed{\frac{2x}{3},\quad x\ne0}

2

Factorise before cancelling

Higher only3 marks

Simplify $\dfrac{x^2+3x}{x}$ with the original restriction.

Show worked answer

\frac{x^2+3x}{x}=\frac{x(x+3)}x

Cancel the common non-zero factor.

\boxed{x+3,\quad x\ne0}

3

Use a difference of squares

Higher only3 marks

Simplify $\dfrac{x^2-25}{x+5}$ .

Show worked answer

\frac{x^2-25}{x+5}=\frac{(x-5)(x+5)}{x+5}

\boxed{x-5,\quad x\ne-5}

The original denominator still rules out $-5$ .

4

Add two simple fractions

Higher only3 marks

Write $\dfrac2x+\dfrac14$ as one fraction.

Show worked answer

Use denominator $4x$ .

\frac2x+\frac14=\frac8{4x}+\frac{x}{4x}

\boxed{\frac{x+8}{4x},\quad x\ne0}

5

Subtract with a shared denominator

Higher only2 marks

Simplify $\dfrac{3x+2}{x+4}-\dfrac{x-1}{x+4}$ .

Show worked answer

Keep the denominator and subtract the entire numerator.

\frac{3x+2-(x-1)}{x+4}

\boxed{\frac{2x+3}{x+4},\quad x\ne-4}

6

Multiply and cancel

Higher only3 marks

Simplify $\dfrac{2x}{3}\times\dfrac9{4x}$ .

Show worked answer

\frac{2x}{3}\times\frac9{4x}=\frac{18x}{12x}

Cancel $6x$ , valid for $x\ne0$ .

\boxed{\frac32,\quad x\ne0}

7

Exclude a zero divisor

Higher only4 marks

Simplify $\dfrac{x+1}{x-2}\div\dfrac{x+1}{3}$ and state all excluded values.

Show worked answer

The first denominator excludes $x=2$ . The second fraction is a divisor, so it cannot equal zero: exclude $x=-1$ too.

\frac{x+1}{x-2}\times\frac3{x+1}

\boxed{\frac3{x-2},\quad x\ne2,-1}

8

Solve a single-fraction equation

Higher only3 marks

Solve $\dfrac8{x+1}=2$ .

Show worked answer

Exclude $x=-1$ . Multiply both sides by $x+1$ .

8=2(x+1)

4=x+1

\boxed{x=3}

Check: $8/(3+1)=2$ .

9

Solve an equation leading to a quadratic

Higher only5 marks

Solve $\dfrac6x=x-1$ .

Show worked answer

Exclude $x=0$ . Multiply both sides by $x$ .

6=x(x-1)

6=x^2-x

x^2-x-6=0

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

A product is zero when one factor is zero.

\boxed{x=3\text{ or }x=-2}

Both are allowed. For $x=3$ , substitution in the original equation gives $6/3=3-1$ , so both sides equal $2$ . For $x=-2$ , it gives $6/(-2)=-2-1$ , so both sides equal $-3$ .

10

Explain an invalid cancellation

Higher only2 marks

A student says $\dfrac{x+6}{x}=6$ . Explain the error using $x=3$ .

Show worked answer

The numerator is a sum, so x cannot be cancelled from just one term.

\frac{3+6}{3}=\frac93=3

This is not 6. The valid split is $1+6/x$ , for $x\ne0$ .

Examiner-style feedback

Common algebraic fractions mistakes

Cancelling across addition

A factor must multiply the entire numerator and denominator. Rewrite as products before crossing anything out.

Losing a minus sign

When subtracting a fraction with a two-term numerator, bracket both terms before expanding the subtraction.

Forgetting the divisor cannot be zero

In fraction division, restrictions come from original denominators and from values making the whole divisor zero.

Solving when only simplification is asked

An expression has no equals sign to another value. Simplifying rewrites it; solving an equation finds values of x.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

The fraction bar divides two whole expressions.
Cancel factors, never isolated added terms.
Use common denominators for addition and subtraction.
Keep original exclusions and check solutions.

Quick answers

Algebraic fractions FAQ

Are algebraic fractions Higher tier?

Manipulating algebraic fractions and working with variable denominators is Higher-tier content. Basic numerical fraction skills and simple algebra are prerequisites.

Can I cancel x from x + 4?

Not on its own: x + 4 is a sum. You can cancel x from x(x + 4) when x is also a non-zero factor of the whole denominator.

Why keep a restriction after simplifying?

Simplification does not make the original expression defined where it involved division by zero.

Build connected skills

What to revise next

Prerequisite

Fractions

Review ordinary fraction arithmetic.

Revise fractions

Quadratic solving

Quadratic formula

Solve equations that do not factorise neatly.

Revise quadratic formula

Linked algebra

Simultaneous equations

Use substitutions and eliminate unknowns.

Revise simultaneous equations

Content standards

Curriculum and rights review

Reviewed 7 September 2026 against GCSE A4, A17 and A18. Original practice and worked solutions. Suggested marks are Pass an Exam estimates, not official mark schemes.

Official specification references

AQA GCSE algebra: manipulation and equations