Algebraic Fractions (AQA GCSE Further Maths)

Revision Note

Test Yourself

Author

Amber

Expertise

Maths

Simplifying Algebraic Fractions

What is an algebraic fraction?

An algebraic fraction is a fraction with an algebraic expression on the top (numerator) and/or the bottom (denominator)

How do you simplify an algebraic fraction?

Factorise fully top and bottom
Cancel common factors (including common brackets)

Exam Tip

If you are asked to simplify an algebraic fraction and have to factorise the top or bottom, it is very likely that one of the factors will be the same on the top and the bottom – you can use this to help you factorise difficult quadratics!

Worked example

Simplify $\frac{4 x + 6}{2 x^{2} - 7 x - 15}$

Factorise the top, by using 2 as a common factor

$\frac{2 (2 x + 3)}{2 x^{2} - 7 x - 15}$

Factorise the bottom using your preferred method
Using the fact that the top factorised to $(2 x + 3)$ may help!

$\frac{2 (2 x + 3)}{(2 x + 3) (x - 5)}$

The common factors on the top and bottom reduce to 1 (cancel out)

$\frac{2 (2 x + 3)}{(2 x + 3) (x - 5)}$

$= \frac{2}{(x - 5)}$

Adding & Subtracting Algebraic Fractions

How do I add (or subtract) two algebraic fractions?

The rules are the same as fractions with numbers:

Find the lowest common denominator (LCD)
- The LCD of x - 2 and x + 5 is found by multiplying them together: LCD = (x - 2)(x + 5)
  - this is the same as with numbers, where the LCD of 2 and 9 is 2 × 9 = 18
- The LCD of x and 2x is not found by multiplying them together, as 2x already includes an x, so the LCD is just 2x
  - this is the same as with numbers, where the LCD of 2 and 4 is just 4, not 2 × 4 = 8
- The LCD of x + 2 and (x + 2)(x - 1) is just (x + 2)(x - 1), as this already includes an (x + 2)
- The LCD of x + 1 and (x + 1)² is just (x + 1)², as this already includes an (x + 1)
- The LCD of (x + 3)(x - 1) and (x + 4)(x - 1) is three brackets: (x + 3)(x - 1)(x + 4), without repeating the (x - 1)
Write each fraction over this lowest common denominator
Multiply the numerators of each fraction by the same amount as the denominators
Write as a single fraction over the lowest common denominator (by adding or subtracting the numerators, taking care to use brackets when subtracting)
Check at the end to see if the top factorises and cancels

Exam Tip

Leaving the top and bottom of the fraction in factorised form will help you see if anything cancels at the end.

Worked example

(a) Express $\frac{x}{x + 4} - \frac{3}{x - 1}$ as a single fraction

The lowest common denominator is $(x + 4) (x - 1)$
Write each fraction over this common denominator, remember to multiply the top of the fractions too

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

Simplify the numerators

$\frac{x^{2} - x}{(x + 4) (x - 1)} - \frac{3 x + 12}{(x - 1) (x + 4)}$

Combine the fractions, as they have the same denominator

$\frac{x^{2} - x - (3 x + 12)}{(x + 4) (x - 1)} = \frac{x^{2} - x - 3 x - 12}{(x + 4) (x - 1)} = \frac{x^{2} - 4 x - 12}{(x + 4) (x - 1)}$

Factorise the top

$\frac{(x + 2) (x - 6)}{(x + 4) (x - 1)}$

There are no terms which would cancel here, so this is the final answer

$\frac{(x + 2) (x - 6)}{(x + 4) (x - 1)}$

(b) Express $\frac{x - 4}{2 (x - 3)} - \frac{x - 1}{2 x}$ as a single fraction

The lowest common denominator is $2 x (x - 3)$ (You could also use $4 x (x - 3)$ but this wouldn't be the lowest common denominator)
Write each fraction over this common denominator, remember to multiply the top of the fractions too

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

Simplify the numerators

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

Combine the fractions, as they have the same denominator

$\frac{x^{2} - 4 x - (x^{2} - 4 x + 3)}{2 x (x - 3)} = \frac{x^{2} - 4 x - x^{2} + 4 x - 3}{2 x (x - 3)} = \frac{- 3}{2 x (x - 3)}$

There is nothing else that can be factorised on the numerator, so this is the final answer

$\frac{- 3}{2 x (x - 3)}$

Multiplying & Dividing Algebraic Fractions

How do I multiply algebraic fractions?

Simplify both fractions first by fully factorising, then cancelling any common brackets on top or bottom (from either fraction)
Multiply the tops together
Multiply the bottoms together
Check for any further factorising and cancelling

How do I divide algebraic fractions?

Flip ("reciprocate") the second fraction and replace ÷ with ×
- So $\div \frac{a}{b}$ becomes $\times \frac{b}{a}$

Then follow the same rules for multiplying two fractions

Worked example

Divide $\frac{x + 3}{x - 4}$ by $\frac{2 x + 6}{x^{2} - 16}$ , giving your answer as a simplified fraction

Division is the same as multiplying by the reciprocal (the fraction flipped)

$\frac{x + 3}{x - 4} \div \frac{2 x + 6}{x^{2} - 16} = \frac{x + 3}{x - 4} \times \frac{x^{2} - 16}{2 x + 6}$

It can often help to factorise first, as there may be factors that cancel out

$\frac{x + 3}{x - 4} \times \frac{x^{2} - 16}{2 x + 6} = \frac{x + 3}{x - 4} \times \frac{(x - 4) (x + 4)}{2 (x + 3)}$

Multiply the numerators and denominators, and cancel any terms that are the same on the top and bottom

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

$= \frac{x + 4}{2}$

Solving Algebraic Fractions

How do I solve an equation that contains algebraic fractions?

There are two methods for solving equations that contain algebraic fractions
One method is to deal with the algebraic fractions by adding or subtracting them first and then solving the equation
- Follow the rules for solving a linear equation containing a fraction on one or both sides
  - Remove the fractions first by multiplying both sides by everything on the denominator
  - Remember to put brackets around any expression that you multiply by
The second method is to begin by multiplying everything in the fraction by each of the expressions on the denominator
- This will remove the denominators of the fractions, leaving you with either a linear or a quadratic equation to solve
- Multiplying everything in the fraction by the common denominator is a way of carrying out this process in one go
For example, to solve the equation $\frac{4}{x - 3} + \frac{5}{x + 1} = 5$ you will need to multiply every term in the equation by both $(x - 3)$ and $(x + 1)$
- STEP 1
  Multiply every term by $(x - 3)$

$\begin{array}{rcl} \frac{4}{x - 3} (x - 3) + \frac{5}{x + 1} (x - 3) & = & 5 (x - 3) \\ \frac{4}{(x - 3)} (x - 3) + \frac{5 (x - 3)}{x + 1} & = & 5 (x - 3) \\ 4 + \frac{5 (x - 3)}{x + 1} & = & 5 (x - 3) \end{array}$

- STEP 2
  Multiply every term by $\begin{array}{rcl} (x + 1) \end{array}$

$\begin{array}{rcl} 4 (x + 1) + \frac{5 (x - 3)}{x + 1} (x + 1) & = & 5 (x - 3) (x + 1) \\ 4 (x + 1) + \frac{5 (x - 3)}{(x + 1)} (x + 1) & = & 5 (x - 3) (x + 1) \\ 4 (x + 1) + 5 (x - 3) & = & 5 (x - 3) (x + 1) \end{array}$

- STEP 3
  Expand the brackets on both sides and simplify

$\begin{array}{rcl} 4 x + 4 + 5 x - 15 & = & 5 (x^{2} + x - 3 x - 3) \\ 9 x - 11 & = & 5 (x^{2} - 2 x - 3) \\ 9 x - 11 & = & 5 x^{2} - 10 x - 15 \end{array}$

- STEP 4
  Rearrange the equation so that it is in a form that can be solved

$\begin{array}{rcl} 9 x - 11 & = & 5 x^{2} - 10 x - 15 \end{array}$
$\begin{array}{rcl} (9 x - 11) - (9 x - 11) \end{array}$
$\begin{array}{rcl} 0 & = & 5 x^{2} - 10 x - 9 x - 15 + 11 \\ 0 & = & 5 x^{2} - 19 x - 4 \end{array}$

- STEP 5
  Solve the equation
  - You can swap the sides if it makes solving the equation easier

$\begin{array}{rcl} 5 x^{2} - 19 x - 4 & = & 0 \\ (5 x + 1) (x - 4) & = & 0 \\ x & = & - \frac{1}{5} or x = 4 \end{array}$

Exam Tip

Multiplying by both denominators at once can speed up the process, but be careful with the algebra if choosing this technique

Worked example

$\frac{2}{p + 3} - \frac{5}{p} = 6 p$

Show that this equation can be written as $2 p^{3} + 6 p^{2} + p + 5 = 0$ .

To clear the fractions, we multiply both sides by the denominators.
We can do this one denominator at a time. We can start by multiplying by $(p + 3)$ .

$2 - \frac{5 (p + 3)}{p} = 6 p (p + 3)$

Now multiply by $p$

$2 (p) - 5 (p + 3) = 6 p (p + 3) (p)$

Now expand brackets

$2 p - 5 (p + 3) = 6 p^{2} (p + 3) 2 p - 5 p - 15 = 6 p^{3} + 18 p^{2}$

Collect like terms

$- 3 p - 15 = 6 p^{3} + 18 p^{2}$

Add the terms on the left hand side to the right hand side, to complete the question

$0 = 6 p^{3} + 18 p^{2} + 3 p + 15$

Each coefficient is a multiple of 3 so you can divide each side by 3.

$2 p^{3} + 6 p + p + 5 = 0$

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

Downloadable PDFs
Unlimited Revision Notes
Topic Questions
Past Papers
Model Answers
Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Test yourself Next topic

Did this page help you?

Algebraic Fractions (AQA GCSE Further Maths)

Revision Note

What is an algebraic fraction?

How do you simplify an algebraic fraction?

How do I add (or subtract) two algebraic fractions?

How do I multiply algebraic fractions?

How do I divide algebraic fractions?

How do I solve an equation that contains algebraic fractions?

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Amber