Operations with algebraic fractions

​​In a nutshell

Once you know how to operate with numerical fractions, dealing with algebraic fractions is essentially the same. The key is, as always, is common factors. 

Addition and subtraction of fractions

Procedure

Example 1

Simplify the following expression $$\cfrac{8}{x} + \cfrac{7}{x+1}$$​

First find the common denominator.

$$\cfrac{8}{x} + \cfrac{7}{x+1} = \cfrac{8 \times (x+1)}{x \times (x+1)}+\cfrac{7 \times x}{x\times(x+1)}$$​​

Add the numerators and simplify.

​$$\begin{aligned}&=\cfrac{8 \times (x+1)}{x \times (x+1)}+\cfrac{7\times x}{x\times(x+1)} \\\\&= \cfrac{8\times (x+1)+7x}{x(x+1)}\\ \\&= \cfrac{8x + 8+ 7x}{x(x+1)} \\\\& = \underline{\cfrac{15x + 8}{x (x+1)}}\end{aligned}$$​​

​

Example 2

Simplify the following expression $$\cfrac{6x}{4}-\cfrac{3}{2x-7}$$​

First, find the common denominator.

$$\cfrac{6x}{4}-\cfrac{3}{2x-7} = \cfrac{6x \times (2x-7)}{4\times (2x-7)} - \cfrac{3\times 4}{4 \times (2x-7)}$$​​

Then subtract and simplify.

$$\begin{aligned}&= \cfrac{6x \times (2x-7)}{4\times (2x-7)} - \cfrac{3\times 4}{4 \times (2x-7)}\\\\ &= \cfrac{6x\times(2x-7) - 3\times 4}{4(2x-7)}\\\\ &= \cfrac{12x^2 -36x- 12}{4 (2x-7)}\end{aligned}$$​​

Cancel any common factors and simplify.

​$$\begin{aligned}&=\cfrac{\cancel {12}x^2 -\cancel {36}x- \cancel{12}}{\cancel 4\times (2x-7)}\\\\ &= \cfrac{3x^2 - 9x-3}{2x-7} \\\\&=\underline{\cfrac{3(x^2-2x-1)}{2x-7}}\end{aligned}$$​​

Multiplying fractions

​​Procedure

Example 3

Calculate $$\cfrac{x+8}{5}\times\cfrac{8}{x^2+16x+64}$$

First factorise $$x^2+16x+64$$​

​$$x^2+16x+64 = (x+8)^2$$​

Once factorised, cancle any common factors before multiplying the numerators and denominators:

$$\begin{aligned}\cfrac{x+8}{5}\times\cfrac{8}{x^2+16x+64}&=\cfrac{x+8}{5}\times\cfrac{8}{(x+8)^2} \\&=\cfrac{\cancel{x+8}}{5}\times\cfrac{8}{(x+8)^{\cancel 2}} \\&=\underline{\cfrac{8}{5(x+8)}}\end{aligned}$$​​

​

Dividing fractions

Procedure

Example 4

Calculate $$\cfrac{1}{2x^2}\div \cfrac{x+3}{4x}$$​

Change the division into a multiplication

$$\begin{aligned}\cfrac{1}{2x^2}\div \cfrac{x+3}{4x} &=\cfrac{1}{2x^2} \times \cfrac{4x}{x+3} \\\\&=\cfrac{1 \times 4x}{2x^2 \times (x+3)}\end{aligned}$$​​

​​

Now, cancel any common factors. In this case:

​$$\begin{aligned}&=\cfrac{1 \times \bcancel4\cancel{x}}{\bcancel2x^{\cancel 2} \times (x+3)} \\\\&= \cfrac{1\times 2}{x\times (x+3)}\\\\&=\underline{\cfrac{2}{x (x+3)}}\end{aligned}$$​