# Algebraic Fractions

A LevelAQAEdexcelOCR

## Algebraic Fractions

An algebraic fraction is a fraction involving algebraic expressions. You can manipulate them in the same way as normal fractions, however it is a little bit harder – therefore it is helpful to simplify the fractions first.

There are 3 skills you need to learn for algebraic fractions.

Make sure you are happy with the following topics before continuing.

A LevelAQAEdexcelOCR

## Skill 1: Simplifying Algebraic Fractions

To simplify algebraic fractions, you need to factorise the denominator and numerator (if necessary), and then look for common factors in the numerator and denominator and cancel.

Example: Simplify the following fraction: $\dfrac{8x+12}{4x^2 - 9}$

Factorise the numerator and the denominator:

$\dfrac{8x+12}{4x^2 - 9} = \dfrac{4(2x+3)}{(2x+3)(2x-3)}$

Then, cancel any common factors:

$\dfrac{\textcolor{red}{4}\cancel{(2x+3)}}{\cancel{(2x+3)} \textcolor{red}{(2x-3)}} = \textcolor{red}{\dfrac{4}{2x-3}}$

A LevelAQAEdexcelOCR

## Skill 2: Multiplying and Dividing Algebraic Fractions

To multiply algebraic fractions, you multiply the numerators together and multiply the denominators together – the same way you would for normal fractions.

To divide by an algebraic fraction, you flip the fraction and then multiply.

Example: Simplify the following expression: $\textcolor{red}{\dfrac{x^2-2x-3}{2x}} \div \dfrac{\textcolor{limegreen}{x+1}}{\textcolor{blue}{x^2}}$

Step 1: Factorise the numerator and denominator:

$\textcolor{red}{\dfrac{x^2-2x-3}{2x}} \div \dfrac{\textcolor{limegreen}{x+1}}{\textcolor{blue}{x^2}} = \textcolor{red}{\dfrac{(x+1)(x-3)}{2x}} \div \dfrac{\textcolor{limegreen}{x+1}}{\textcolor{blue}{x^2}}$

Step 2: Flip the second fraction and turn the divide into a multiply:

$\textcolor{red}{\dfrac{(x+1)(x-3)}{2x}} \div \dfrac{\textcolor{limegreen}{x+1}}{\textcolor{blue}{x^2}} = \textcolor{red}{\dfrac{(x+1)(x-3)}{2x}} \times \dfrac{\textcolor{blue}{x^2}}{\textcolor{limegreen}{x+1}}$

Step 3: Multiply the numerator together and the denominators together (combine into one fraction):

$\textcolor{red}{\dfrac{(x+1)(x-3)}{2x}} \times \dfrac{\textcolor{blue}{x^2}}{\textcolor{limegreen}{x+1}} = \dfrac{x^2(x+1)(x-3)}{2x(x+1)}$

Step 4: Cancel down any common factors:

$\dfrac{x ^{\cancel{2}} \cancel{(x+1)}(x-3)}{2\cancel{x} \cancel{(x+1)}} = \dfrac{x(x-3)}{2} \,\,\, \left( = \dfrac{x^2 - 3x}{2} \right)$

A LevelAQAEdexcelOCR

@mmerevise

## Skill 3: Adding and Subtracting Algebraic Fractions

You use the same methods for normal fractions to add and subtract algebraic fractions.

Example: Simplify $\dfrac{2x}{x+1} + \dfrac{5}{x^2} - \dfrac{1}{x}$

Step 1: Find a common denominator – multiply the numerator and denominator of each fraction by the denominators of the other fractions:

$\left( \dfrac{2x}{x+1} \times \textcolor{blue}{\dfrac{x^2}{x^2}} \times \textcolor{limegreen}{\dfrac{x}{x}} \right) + \left( \dfrac{5}{x^2} \times \textcolor{red}{\dfrac{x+1}{x+1}} \times \textcolor{limegreen}{\dfrac{x}{x}} \right) - \left( \dfrac{1}{x} \times \textcolor{red}{\dfrac{x+1}{x+1}} \times \textcolor{blue}{\dfrac{x^2}{x^2}} \right) = \dfrac{2x^4}{x^3(x+1)} + \dfrac{5x(x+1)}{x^3(x+1)} - \dfrac{x^2(x+1)}{x^3(x+1)}$

Step 2: Now we can put everything over a common denominator, and then add the numerators together:

$\dfrac{2x^4}{x^3(x+1)} + \dfrac{5x(x+1)}{x^3(x+1)} - \dfrac{x^2(x+1)}{x^3(x+1)} = \dfrac{2x^4 + 5x(x+1) - x^2(x+1)}{x^3(x+1)}$

Step 3: Simplify the fraction as much as possible:

$\dfrac{2x^4 + 5x(x+1) - x^2(x+1)}{x^3(x+1)} = \dfrac{2x^4 - x^3 + 4x^2 + 5x}{x^3(x+1)}$

Notice that $x$ is a common factor in the numerator and denominator, so the fraction can be cancelled further:

$\dfrac{2x^{\cancel{4}} - x^{\cancel{3}} + 4x^{\cancel{2}} + 5\cancel{x}}{x^{\cancel{3}}(x+1)} = \dfrac{2x^3 - x^2 + 4x + 5}{x^2(x+1)}$

A LevelAQAEdexcelOCR
A LevelAQAEdexcelOCR

## Example 1: Simplifying Algebraic Fractions

Simplify the following:

a) $\dfrac{27x^5y^3}{9x^4y}$

b) $\dfrac{2 + \dfrac{1}{3x}}{6x^2 + x}$

[4 marks]

a) Here, you do not need to factorise, you just need to cancel down common factors:

$\dfrac{\cancel{27}x^{\cancel{5}}y^{\cancel{3}}}{\cancel{9} x^{\cancel{4}} \cancel{y}} = 3xy^2$

b) Here, you need to multiply the numerator and denominator by the same factor to cancel the fraction in the numerator:

$\dfrac{2 + \dfrac{1}{3x}}{6x^2 + x} \times \dfrac{3x}{3x} = \dfrac{6x+1}{18x^3 + 3x^2}$

Then, you can factorise and simplify as normal:

$\dfrac{6x+1}{18x^3 + 3x^2} = \dfrac{\cancel{6x+1}}{3x^2 \cancel{(6x+1)}} = \dfrac{1}{3x^2}$

A LevelAQAEdexcelOCR

## Example 2: Multiplying Algebraic Fractions

Simplify the following algebraic fraction fully:

$\dfrac{3x+6}{5xy} \times \dfrac{x}{x+2}$

[3 marks]

Multiply the numerators together and the denominators together:

$\dfrac{3x+6}{5xy} \times \dfrac{x}{x+2} = \dfrac{x(3x+6)}{5xy(x+2)}$

Notice that $3x+6 = 3(x+2)$

Then, simplify the fraction, by cancelling down any common factors:

$\dfrac{x(3x+6)}{5xy(x+2)} = \dfrac{3\cancel{x} \cancel{(x+2)}}{5\cancel{x}y \cancel{(x+2)}} = \dfrac{3}{5y}$

A LevelAQAEdexcelOCR

## Algebraic Fractions Example Questions

$\dfrac{ax^2 + ay}{az} = \dfrac{a(x^2 + y)}{az} = \dfrac{x^2 + y}{z}$
$\dfrac{2 - \dfrac{1}{x^2}}{2x^2-1} \times \dfrac{x^2}{x^2} = \dfrac{2x^2 - 1}{2x^4 - x^2} = \dfrac{2x^2 - 1}{x^2(2x^2-1)} = \dfrac{1}{x^2}$
\begin{aligned} \dfrac{3x}{x+4} \div \dfrac{x^2}{2x+8} &= \dfrac{3x}{x+4} \times \dfrac{2x+8}{x^2} \\[1.1em] &= \dfrac{3x}{x+4} \times \dfrac{2(x+4)}{x^2} \\[1.1em] &= \dfrac{6x(x+4)}{x^2(x+4)} \\[1.1em] &= \dfrac{6}{x} \end{aligned}
\begin{aligned} \dfrac{4x}{x+1} - \dfrac{1}{2x} &= \dfrac{4x(2x)}{2x(x+1)} - \dfrac{x+1}{2x(x+1)} \\[1.1em] &= \dfrac{8x^2 - x - 1}{2x(x+1)}\end{aligned}
$\dfrac{x^2-x-12}{x+2} \times \dfrac{x^2 - 4}{x^2 + 3x} = \dfrac{(x+3)(x-4)}{x+2} \times \dfrac{(x+2)(x-2)}{x(x+3)} = \dfrac{(x-4)(x-2)}{x}$

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