Algebraic Fractions

What you'll learn

What an algebraic fraction is, and why some values of $x$ are not allowed.
How to simplify algebraic fractions by factorising first.
How to divide algebraic fractions using a reciprocal.
How to solve equations and ratio problems involving algebraic fractions.

1. The basic language

An algebraic fraction works like an ordinary fraction, but the top and bottom contain algebra.

Definition

Algebraic fraction

An algebraic fraction is a fraction where the numerator, the denominator, or both contain algebraic expressions. The numerator is the expression on top, and the denominator is the expression on the bottom.

For example, $x+5x2−4\frac{x+5}{x^2-4}$ is an algebraic fraction.

A very important rule still applies: the denominator cannot be zero. Any value of $x$ that makes the denominator zero is called an excluded value.

Example

Finding excluded values

Find the excluded values for $x+3x2−4\frac{x+3}{x^2-4}$ .

Focus only on the denominator.
$x^2-4$
Factorise the denominator using the difference of two squares.
$x^2-4=(x-2)(x+2)$
Set each factor equal to zero.
$\quad \text{or} \quad x+2=0$
Solve these equations.
$\quad \text{or} \quad x=-2$
So the excluded values are $x = 2$ and $x = - 2$ .

Common Mistake

Denominators cannot be zero

Even if a factor cancels later, values that made the original denominator zero are still not allowed. This matters especially when solving equations.

2. Factorising comes before cancelling

To factorise means to rewrite an expression as a product of factors. A factor is something being multiplied.

For example:

x^2+5x=x(x+5)

A common factor is a factor that appears in both the numerator and the denominator. You can cancel common factors, just like you can simplify ordinary fractions.

Key Idea

The main rule

With algebraic fractions, factorise the numerator and denominator fully first. Then cancel common factors, not individual terms.

This flowchart shows the usual simplifying process.

$Flowchart for simplifying an algebraic fraction$

Example

Simplifying by factorising

Simplify fully $x2+6xx2+8x+12\frac{x^2+6x}{x^2+8x+12}$ .

Factorise the numerator by taking out the common factor $x$ .
$x^2+6x=x(x+6)$
Factorise the denominator by finding two numbers that multiply to 12 and add to 8.
$x^2+8x+12=(x+2)(x+6)$
Rewrite the whole fraction using factors.
$x2+6xx2+8x+12=x(x+6)(x+2)(x+6)\frac{x^2+6x}{x^2+8x+12} = \frac{x(x+6)}{(x+2)(x+6)}$
Cancel the common factor $(x + 6)$ .
$x(x+6)(x+2)(x+6)=xx+2\frac{x(x+6)}{(x+2)(x+6)} = \frac{x}{x+2}$
The simplified fraction is $xx+2\frac{x}{x+2}$ .

Common Mistake

Cancelling terms instead of factors

You cannot cancel the $x$ in $x+4x+7\frac{x+4}{x+7}$ , because $x$ is part of an addition. Cancelling only works when the same whole factor is multiplied on the top and bottom.

3. Factor patterns you will use a lot

Many algebraic fraction questions are mostly testing your factorising.

You should be confident with these patterns:

Common factor: $3x^2+9x=3x(x+3)$
Difference of two squares: $x^2-25=(x-5)(x+5)$
Quadratic factorising: $x^2+7x+10=(x+5)(x+2)$
Quadratics with a coefficient of $x^2$ : $2x^2+7x+3=(2x+1)(x+3)$

Difference of two squares

The expression $a^2-b^2$ factorises to $(a - b) (a + b)$ . For example, $x^2-49=(x-7)(x+7)$ .

Example

Using the difference of two squares

Simplify fully $x+7x2−49\frac{x+7}{x^2-49}$ .

Recognise the denominator as a difference of two squares.
$x^2-49=x^2-7^2$
Factorise the denominator.
$x^2-49=(x-7)(x+7)$
Rewrite the fraction using factors.
$x+7x2−49=x+7(x−7)(x+7)\frac{x+7}{x^2-49} = \frac{x+7}{(x-7)(x+7)}$
Cancel the common factor $(x + 7)$ .
$x+7(x−7)(x+7)=1x−7\frac{x+7}{(x-7)(x+7)} = \frac{1}{x-7}$
The simplified fraction is $1x−7\frac{1}{x-7}$ .

Writing the answer in a given form

Sometimes the question tells you the final form, such as $ax+bx+c\frac{ax+b}{x+c}$ . This is a clue that a bracket factor will cancel, leaving a linear expression on top and bottom.

Example

Writing in a required form

Write $2x2+7x+3x2+4x+3\frac{2x^2+7x+3}{x^2+4x+3}$ in the form $ax+bx+c\frac{ax+b}{x+c}$ , where $a$ , $b$ and $c$ are integers.

Factorise the numerator.
$2x^2+7x+3=(2x+1)(x+3)$
Factorise the denominator.
$x^2+4x+3=(x+1)(x+3)$
Rewrite the fraction.
$2x2+7x+3x2+4x+3=(2x+1)(x+3)(x+1)(x+3)\frac{2x^2+7x+3}{x^2+4x+3} = \frac{(2x+1)(x+3)}{(x+1)(x+3)}$
Cancel the common factor $(x + 3)$ .
$(2x+1)(x+3)(x+1)(x+3)=2x+1x+1\frac{(2x+1)(x+3)}{(x+1)(x+3)} = \frac{2x+1}{x+1}$
So the form is $2x+1x+1\frac{2x+1}{x+1}$ , with $a = 2$ , $b = 1$ and $c = 1$ .

Tip

Factorising check

After factorising a quadratic, expand your brackets mentally to check. For example, $(2 x + 1) (x + 3)$ gives $2x^2+6x+x+3$ , which simplifies to $2x^2+7x+3$ .

4. Dividing algebraic fractions

To divide by a fraction, multiply by its reciprocal.

Definition

Reciprocal

The reciprocal of a fraction is found by swapping its numerator and denominator. For example, the reciprocal of $x+1x−4\frac{x+1}{x-4}$ is $x−4x+1\frac{x-4}{x+1}$ .

The method is:

Change division into multiplication.
Flip the second fraction.
Factorise everything.
Cancel common factors.

Example

Dividing algebraic fractions

Simplify fully $x+3x−5÷x2+5x+6x2−5x\frac{x+3}{x-5} \div \frac{x^2+5x+6}{x^2-5x}$ .

Change division into multiplication by the reciprocal.
$x+3x−5÷x2+5x+6x2−5x=x+3x−5×x2−5xx2+5x+6\frac{x+3}{x-5} \div \frac{x^2+5x+6}{x^2-5x} = \frac{x+3}{x-5} \times \frac{x^2-5x}{x^2+5x+6}$
Factorise the expressions that can be factorised.
$x^2-5x=x(x-5)$ $x^2+5x+6=(x+2)(x+3)$
Rewrite the multiplication using factors.
$x+3x−5×x(x−5)(x+2)(x+3)\frac{x+3}{x-5} \times \frac{x(x-5)}{(x+2)(x+3)}$
Cancel the common factors $(x + 3)$ and $(x - 5)$ .
$x+3x−5×x(x−5)(x+2)(x+3)=xx+2\frac{x+3}{x-5} \times \frac{x(x-5)}{(x+2)(x+3)} = \frac{x}{x+2}$
The simplified answer is $xx+2\frac{x}{x+2}$ .

Common Mistake

Dividing by zero

When dividing by an algebraic fraction, the fraction you divide by must not be zero. In harder questions, this can add extra excluded values.

5. Solving equations with algebraic fractions

To solve equations with algebraic fractions, your aim is to remove the denominators.

A common denominator is an expression that all the denominators can divide into. Clearing denominators means multiplying every term in the equation by a common denominator.

Key Idea

Clear the fractions

Multiply every term by the common denominator. This turns the equation into a normal algebraic equation, often a quadratic.

Example

Solving an equation with algebraic fractions

Solve $8x+2+6x+5=1\frac{8}{x+2}+\frac{6}{x+5}=1$ .

Identify the excluded values first. The denominators show that $x≠−2x\neq -2$ and $x≠−5x\neq -5$ .
Multiply every term by the common denominator $(x + 2) (x + 5)$ .
$8 (x + 5) + 6 (x + 2) = (x + 2) (x + 5)$
Expand both sides.
$8x+40+6x+12=x^2+7x+10$
Simplify and rearrange into a quadratic equal to zero.
$14x+52=x2+7x+100=x2−7x−42\begin{aligned} 14x+52&=x^2+7x+10\\ 0&=x^2-7x-42 \end{aligned}$
Factorise the quadratic.
$x^2-7x-42=(x-14)(x+3)$
Solve each factor.
$\quad \text{or} \quad x=-3$
Check against the excluded values. Both solutions are allowed.

Common Mistake

Forgetting to multiply the 1

In equations like $8x+2+6x+5=1\frac{8}{x+2}+\frac{6}{x+5}=1$ , the right-hand side also gets multiplied by the common denominator. Do not leave the 1 unchanged.

6. Ratios written with algebra

A ratio statement such as $A : B = C : D$ means the same as $AB=CD\frac{A}{B}=\frac{C}{D}$ . This is called a proportion.

To solve a proportion, cross multiply:

AB=CD⇒AD=BC\frac{A}{B}=\frac{C}{D} \quad \Rightarrow \quad AD=BC

Example

Solving an algebraic ratio problem

Given that $x + 2 : 2 x = x + 3 : 3 x - 1$ , find the possible values of $x$ .

Rewrite the ratio statement as a fraction equation.
$x+22x=x+33x−1\frac{x+2}{2x}=\frac{x+3}{3x-1}$
Cross multiply.
$(x + 2) (3 x - 1) = 2 x (x + 3)$
Expand both sides.
$3x^2+5x-2=2x^2+6x$
Rearrange into a quadratic equal to zero.
$x^2-x-2=0$
Factorise the quadratic.
$(x - 2) (x + 1) = 0$
Solve each factor.
$\quad \text{or} \quad x=-1$
Check the denominators from the fraction equation. Since $2x≠02x\neq 0$ and $3x−1≠03x-1\neq 0$ , both values are allowed.

Tip

Ratios become fractions

When you see a colon in an algebra question, think “fraction”. For example, $P : Q = R : S$ becomes $PQ=RS\frac{P}{Q}=\frac{R}{S}$ .

Exam technique

In the exam

Factorise before cancelling; never cancel across addition or subtraction.
When dividing fractions, change to multiplication and flip the second fraction before factorising and cancelling.
When solving, write down excluded values early and check your final answers against them.

Self review

Check yourself

Can you spot whether a denominator is a difference of two squares?
Can you explain why $x+4x+6\frac{x+4}{x+6}$ cannot be simplified by cancelling $x$ ?
Can you turn an algebraic ratio like $A : B = C : D$ into a fraction equation?

Algebraic Fractions

What you'll learn

What an algebraic fraction is, and why some values of $x$ are not allowed.
How to simplify algebraic fractions by factorising first.
How to divide algebraic fractions using a reciprocal.
How to solve equations and ratio problems involving algebraic fractions.

1. The basic language

An algebraic fraction works like an ordinary fraction, but the top and bottom contain algebra.

Definition

Algebraic fraction

For example, $x+5x2−4\frac{x+5}{x^2-4}$ is an algebraic fraction.

A very important rule still applies: the denominator cannot be zero. Any value of $x$ that makes the denominator zero is called an excluded value.

Example

Finding excluded values

Find the excluded values for $x+3x2−4\frac{x+3}{x^2-4}$ .

Focus only on the denominator.
$x^2-4$
Factorise the denominator using the difference of two squares.
$x^2-4=(x-2)(x+2)$
Set each factor equal to zero.
$\quad \text{or} \quad x+2=0$
Solve these equations.
$\quad \text{or} \quad x=-2$
So the excluded values are $x = 2$ and $x = - 2$ .

Common Mistake

Denominators cannot be zero

Even if a factor cancels later, values that made the original denominator zero are still not allowed. This matters especially when solving equations.

2. Factorising comes before cancelling

To factorise means to rewrite an expression as a product of factors. A factor is something being multiplied.

For example:

x^2+5x=x(x+5)

A common factor is a factor that appears in both the numerator and the denominator. You can cancel common factors, just like you can simplify ordinary fractions.

Key Idea

The main rule

With algebraic fractions, factorise the numerator and denominator fully first. Then cancel common factors, not individual terms.

This flowchart shows the usual simplifying process.

$Flowchart for simplifying an algebraic fraction$

Example

Simplifying by factorising

Simplify fully $x2+6xx2+8x+12\frac{x^2+6x}{x^2+8x+12}$ .

Factorise the numerator by taking out the common factor $x$ .
$x^2+6x=x(x+6)$
Factorise the denominator by finding two numbers that multiply to 12 and add to 8.
$x^2+8x+12=(x+2)(x+6)$
Rewrite the whole fraction using factors.
$x2+6xx2+8x+12=x(x+6)(x+2)(x+6)\frac{x^2+6x}{x^2+8x+12} = \frac{x(x+6)}{(x+2)(x+6)}$
Cancel the common factor $(x + 6)$ .
$x(x+6)(x+2)(x+6)=xx+2\frac{x(x+6)}{(x+2)(x+6)} = \frac{x}{x+2}$
The simplified fraction is $xx+2\frac{x}{x+2}$ .

Common Mistake

Cancelling terms instead of factors

You cannot cancel the $x$ in $x+4x+7\frac{x+4}{x+7}$ , because $x$ is part of an addition. Cancelling only works when the same whole factor is multiplied on the top and bottom.

3. Factor patterns you will use a lot

Many algebraic fraction questions are mostly testing your factorising.

You should be confident with these patterns:

Common factor: $3x^2+9x=3x(x+3)$
Difference of two squares: $x^2-25=(x-5)(x+5)$
Quadratic factorising: $x^2+7x+10=(x+5)(x+2)$
Quadratics with a coefficient of $x^2$ : $2x^2+7x+3=(2x+1)(x+3)$

Difference of two squares

The expression $a^2-b^2$ factorises to $(a - b) (a + b)$ . For example, $x^2-49=(x-7)(x+7)$ .

Example

Using the difference of two squares

Simplify fully $x+7x2−49\frac{x+7}{x^2-49}$ .

Recognise the denominator as a difference of two squares.
$x^2-49=x^2-7^2$
Factorise the denominator.
$x^2-49=(x-7)(x+7)$
Rewrite the fraction using factors.
$x+7x2−49=x+7(x−7)(x+7)\frac{x+7}{x^2-49} = \frac{x+7}{(x-7)(x+7)}$
Cancel the common factor $(x + 7)$ .
$x+7(x−7)(x+7)=1x−7\frac{x+7}{(x-7)(x+7)} = \frac{1}{x-7}$
The simplified fraction is $1x−7\frac{1}{x-7}$ .

Writing the answer in a given form

Sometimes the question tells you the final form, such as $ax+bx+c\frac{ax+b}{x+c}$ . This is a clue that a bracket factor will cancel, leaving a linear expression on top and bottom.

Example

Writing in a required form

Write $2x2+7x+3x2+4x+3\frac{2x^2+7x+3}{x^2+4x+3}$ in the form $ax+bx+c\frac{ax+b}{x+c}$ , where $a$ , $b$ and $c$ are integers.

Factorise the numerator.
$2x^2+7x+3=(2x+1)(x+3)$
Factorise the denominator.
$x^2+4x+3=(x+1)(x+3)$
Rewrite the fraction.
$2x2+7x+3x2+4x+3=(2x+1)(x+3)(x+1)(x+3)\frac{2x^2+7x+3}{x^2+4x+3} = \frac{(2x+1)(x+3)}{(x+1)(x+3)}$
Cancel the common factor $(x + 3)$ .
$(2x+1)(x+3)(x+1)(x+3)=2x+1x+1\frac{(2x+1)(x+3)}{(x+1)(x+3)} = \frac{2x+1}{x+1}$
So the form is $2x+1x+1\frac{2x+1}{x+1}$ , with $a = 2$ , $b = 1$ and $c = 1$ .

Tip

Factorising check

After factorising a quadratic, expand your brackets mentally to check. For example, $(2 x + 1) (x + 3)$ gives $2x^2+6x+x+3$ , which simplifies to $2x^2+7x+3$ .

4. Dividing algebraic fractions

To divide by a fraction, multiply by its reciprocal.

Definition

Reciprocal

The reciprocal of a fraction is found by swapping its numerator and denominator. For example, the reciprocal of $x+1x−4\frac{x+1}{x-4}$ is $x−4x+1\frac{x-4}{x+1}$ .

The method is:

Change division into multiplication.
Flip the second fraction.
Factorise everything.
Cancel common factors.

Example

Dividing algebraic fractions

Simplify fully $x+3x−5÷x2+5x+6x2−5x\frac{x+3}{x-5} \div \frac{x^2+5x+6}{x^2-5x}$ .

Change division into multiplication by the reciprocal.
$x+3x−5÷x2+5x+6x2−5x=x+3x−5×x2−5xx2+5x+6\frac{x+3}{x-5} \div \frac{x^2+5x+6}{x^2-5x} = \frac{x+3}{x-5} \times \frac{x^2-5x}{x^2+5x+6}$
Factorise the expressions that can be factorised.
$x^2-5x=x(x-5)$ $x^2+5x+6=(x+2)(x+3)$
Rewrite the multiplication using factors.
$x+3x−5×x(x−5)(x+2)(x+3)\frac{x+3}{x-5} \times \frac{x(x-5)}{(x+2)(x+3)}$
Cancel the common factors $(x + 3)$ and $(x - 5)$ .
$x+3x−5×x(x−5)(x+2)(x+3)=xx+2\frac{x+3}{x-5} \times \frac{x(x-5)}{(x+2)(x+3)} = \frac{x}{x+2}$
The simplified answer is $xx+2\frac{x}{x+2}$ .

Common Mistake

Dividing by zero

When dividing by an algebraic fraction, the fraction you divide by must not be zero. In harder questions, this can add extra excluded values.

5. Solving equations with algebraic fractions

To solve equations with algebraic fractions, your aim is to remove the denominators.

A common denominator is an expression that all the denominators can divide into. Clearing denominators means multiplying every term in the equation by a common denominator.

Key Idea

Clear the fractions

Multiply every term by the common denominator. This turns the equation into a normal algebraic equation, often a quadratic.

Example

Solving an equation with algebraic fractions

Solve $8x+2+6x+5=1\frac{8}{x+2}+\frac{6}{x+5}=1$ .

Identify the excluded values first. The denominators show that $x≠−2x\neq -2$ and $x≠−5x\neq -5$ .
Multiply every term by the common denominator $(x + 2) (x + 5)$ .
$8 (x + 5) + 6 (x + 2) = (x + 2) (x + 5)$
Expand both sides.
$8x+40+6x+12=x^2+7x+10$
Simplify and rearrange into a quadratic equal to zero.
$14x+52=x2+7x+100=x2−7x−42\begin{aligned} 14x+52&=x^2+7x+10\\ 0&=x^2-7x-42 \end{aligned}$
Factorise the quadratic.
$x^2-7x-42=(x-14)(x+3)$
Solve each factor.
$\quad \text{or} \quad x=-3$
Check against the excluded values. Both solutions are allowed.

Common Mistake

Forgetting to multiply the 1

In equations like $8x+2+6x+5=1\frac{8}{x+2}+\frac{6}{x+5}=1$ , the right-hand side also gets multiplied by the common denominator. Do not leave the 1 unchanged.

6. Ratios written with algebra

A ratio statement such as $A : B = C : D$ means the same as $AB=CD\frac{A}{B}=\frac{C}{D}$ . This is called a proportion.

To solve a proportion, cross multiply:

AB=CD⇒AD=BC\frac{A}{B}=\frac{C}{D} \quad \Rightarrow \quad AD=BC

Example

Solving an algebraic ratio problem

Given that $x + 2 : 2 x = x + 3 : 3 x - 1$ , find the possible values of $x$ .

Rewrite the ratio statement as a fraction equation.
$x+22x=x+33x−1\frac{x+2}{2x}=\frac{x+3}{3x-1}$
Cross multiply.
$(x + 2) (3 x - 1) = 2 x (x + 3)$
Expand both sides.
$3x^2+5x-2=2x^2+6x$
Rearrange into a quadratic equal to zero.
$x^2-x-2=0$
Factorise the quadratic.
$(x - 2) (x + 1) = 0$
Solve each factor.
$\quad \text{or} \quad x=-1$
Check the denominators from the fraction equation. Since $2x≠02x\neq 0$ and $3x−1≠03x-1\neq 0$ , both values are allowed.

Tip

Ratios become fractions

When you see a colon in an algebra question, think “fraction”. For example, $P : Q = R : S$ becomes $PQ=RS\frac{P}{Q}=\frac{R}{S}$ .

Exam technique

In the exam

Factorise before cancelling; never cancel across addition or subtraction.
When dividing fractions, change to multiplication and flip the second fraction before factorising and cancelling.
When solving, write down excluded values early and check your final answers against them.

Self review

Check yourself

Can you spot whether a denominator is a difference of two squares?
Can you explain why $x+4x+6\frac{x+4}{x+6}$ cannot be simplified by cancelling $x$ ?
Can you turn an algebraic ratio like $A : B = C : D$ into a fraction equation?

Algebraic Fractions

What you'll learn

1. The basic language

2. Factorising comes before cancelling

3. Factor patterns you will use a lot

Difference of two squares

Writing the answer in a given form

4. Dividing algebraic fractions

5. Solving equations with algebraic fractions

6. Ratios written with algebra

Recap questions

You've reached the end

Algebraic Fractions

What you'll learn

1. The basic language

2. Factorising comes before cancelling

3. Factor patterns you will use a lot

Difference of two squares

Writing the answer in a given form

4. Dividing algebraic fractions

5. Solving equations with algebraic fractions

6. Ratios written with algebra

Recap questions

You've reached the end

Algebraic Fractions

What you'll learn

1. The basic language

2. Factorising comes before cancelling

3. Factor patterns you will use a lot

Difference of two squares

Writing the answer in a given form

4. Dividing algebraic fractions

5. Solving equations with algebraic fractions

6. Ratios written with algebra

Recap questions

You've reached the end

Algebraic Fractions Revision Guide

Algebraic Fractions

What you'll learn

1. The basic language

2. Factorising comes before cancelling

3. Factor patterns you will use a lot

Difference of two squares

Writing the answer in a given form

4. Dividing algebraic fractions

5. Solving equations with algebraic fractions

6. Ratios written with algebra

Recap questions

You've reached the end

Algebraic Fractions Revision Guide