������̳

An algebraic fraction is a fraction with algebraic terms in the numerator or the denominator or both.

Examples

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

\(\dfrac{x + 3}{x - 1}\)

\(\dfrac{12x}{5}\)

\(\dfrac{9x^2 - 1}{3x^2 - 2x - 1}\)

These topics are very useful

• Module 2 fractions
• Module 3 - algebra - factorising
• Module 4 - factorising
• Module 3 - algebra - algebraic fractions

Example

Simplify

\(\dfrac{3}{2m} + \dfrac{1}{3m}\)

Solution

Find a common denominator
The common denominator of \(2m\) and \(3m\) is \(\mathbf {6m}\)

Write each fraction as an equivalent fraction with that common denominator

\(\dfrac{3}{2m} = \dfrac{9}{6m}\quad and \quad\dfrac{1}{3m} = \dfrac{2}{6m}\)

it may help to look at the equation like this \(\color{purple}\dfrac{3}{2m} = \dfrac{9}{6m}\quad \text{and} \quad\dfrac{1}{3} = \dfrac{2}{6}\)

Answer:

Add the numerators of these equivalent fractions and keep the common denominator

\(\dfrac{9}{6m} + \frac{2}{6m} = \dfrac{11}{6m}\)

Answer
\(\mathbf {\dfrac{11}{6m}}\)

Example

Simplify

\(\dfrac{4}{x + 1} + \dfrac{3}{x - 5}\)

Solution

This example looks difficult because of the more complex terms in the denominator, but the method is just the same.
An extra final step will be needed to simplify the numerator.

Find a common denominator

The common denominator of \(x+1\) and \(x-5\) is \((x+1)(x-5)\)

When the terms in the denominator are not 'like terms', their product is the common denominator

Write each fraction as an equivalent fraction with that common denominator

\(\dfrac{4}{x + 1} = \dfrac{7}{3c} = \dfrac{4(x+5)}{(x+1)(x-5)}\) and
\(\dfrac{3}{x-5} = \dfrac{3(x+1)}{(x+1)(x-5)}\)

Add the numerators of these equivalent fractions and keep the common denominator

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

Answer:

Simplify the numerator

\(= \dfrac{4x -20 +3x + 3}{(x+1)(x-5)}\)

\(= \dfrac{7x -17}{(x+1)(x-5)}\)

Answer \(= \dfrac{7x -17}{(x+1)(x-5)}\)