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 1.
Find the common denominator.
2.
Add or substract the numerators.
3.
Simplify the fraction by factorising and cancelling any common factors.
Example 1 Simplify the following expression 8 x + 7 x + 1 \cfrac{8}{x} + \cfrac{7}{x+1} x 8 + x + 1 7
First find the common denominator.
8 x + 7 x + 1 = 8 × ( x + 1 ) x × ( x + 1 ) + 7 × x x × ( x + 1 ) \cfrac{8}{x} + \cfrac{7}{x+1} = \cfrac{8 \times (x+1)}{x \times (x+1)}+\cfrac{7 \times x}{x\times(x+1)} x 8 + x + 1 7 = x × ( x + 1 ) 8 × ( x + 1 ) + x × ( x + 1 ) 7 × x
Add the numerators and simplify.
= 8 × ( x + 1 ) x × ( x + 1 ) + 7 × x x × ( x + 1 ) = 8 × ( x + 1 ) + 7 x x ( x + 1 ) = 8 x + 8 + 7 x x ( x + 1 ) = 15 x + 8 x ( x + 1 ) ‾ \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} = x × ( x + 1 ) 8 × ( x + 1 ) + x × ( x + 1 ) 7 × x = x ( x + 1 ) 8 × ( x + 1 ) + 7 x = x ( x + 1 ) 8 x + 8 + 7 x = x ( x + 1 ) 15 x + 8
Example 2 Simplify the following expression 6 x 4 − 3 2 x − 7 \cfrac{6x}{4}-\cfrac{3}{2x-7} 4 6 x − 2 x − 7 3
First, find the common denominator.
6 x 4 − 3 2 x − 7 = 6 x × ( 2 x − 7 ) 4 × ( 2 x − 7 ) − 3 × 4 4 × ( 2 x − 7 ) \cfrac{6x}{4}-\cfrac{3}{2x-7} = \cfrac{6x \times (2x-7)}{4\times (2x-7)} - \cfrac{3\times 4}{4 \times (2x-7)} 4 6 x − 2 x − 7 3 = 4 × ( 2 x − 7 ) 6 x × ( 2 x − 7 ) − 4 × ( 2 x − 7 ) 3 × 4
Then subtract and simplify.
= 6 x × ( 2 x − 7 ) 4 × ( 2 x − 7 ) − 3 × 4 4 × ( 2 x − 7 ) = 6 x × ( 2 x − 7 ) − 3 × 4 4 ( 2 x − 7 ) = 12 x 2 − 36 x − 12 4 ( 2 x − 7 ) \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} = 4 × ( 2 x − 7 ) 6 x × ( 2 x − 7 ) − 4 × ( 2 x − 7 ) 3 × 4 = 4 ( 2 x − 7 ) 6 x × ( 2 x − 7 ) − 3 × 4 = 4 ( 2 x − 7 ) 12 x 2 − 36 x − 12
Cancel any common factors and simplify.
= 12 x 2 − 36 x − 12 4 × ( 2 x − 7 ) = 3 x 2 − 9 x − 3 2 x − 7 = 3 ( x 2 − 2 x − 1 ) 2 x − 7 ‾ \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} = 4 × ( 2 x − 7 ) 12 x 2 − 36 x − 12 = 2 x − 7 3 x 2 − 9 x − 3 = 2 x − 7 3 ( x 2 − 2 x − 1 )
Multiplying fractions Procedure 1.
Factorise (where possible).
2.
Cancel any common factors.
3.
Multiply numerators and denominators.
Example 3 Calculate x + 8 5 × 8 x 2 + 16 x + 64 \cfrac{x+8}{5}\times\cfrac{8}{x^2+16x+64} 5 x + 8 × x 2 + 16 x + 64 8
First factorise x 2 + 16 x + 64 x^2+16x+64 x 2 + 16 x + 64
x 2 + 16 x + 64 = ( x + 8 ) 2 x^2+16x+64 = (x+8)^2 x 2 + 16 x + 64 = ( x + 8 ) 2
Once factorised, cancle any common factors before multiplying the numerators and denominators:
x + 8 5 × 8 x 2 + 16 x + 64 = x + 8 5 × 8 ( x + 8 ) 2 = x + 8 5 × 8 ( x + 8 ) 2 = 8 5 ( x + 8 ) ‾ \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} 5 x + 8 × x 2 + 16 x + 64 8 = 5 x + 8 × ( x + 8 ) 2 8 = 5 x + 8 × ( x + 8 ) 2 8 = 5 ( x + 8 ) 8
Dividing fractions Procedure 1.
Change the divide into multiply and flip the second fraction.
2.
Multiply the fractions.
Example 4 Calculate 1 2 x 2 ÷ x + 3 4 x \cfrac{1}{2x^2}\div \cfrac{x+3}{4x} 2 x 2 1 ÷ 4 x x + 3
Change the division into a multiplication
1 2 x 2 ÷ x + 3 4 x = 1 2 x 2 × 4 x x + 3 = 1 × 4 x 2 x 2 × ( x + 3 ) \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} 2 x 2 1 ÷ 4 x x + 3 = 2 x 2 1 × x + 3 4 x = 2 x 2 × ( x + 3 ) 1 × 4 x
Now, cancel any common factors. In this case:
= 1 × 4 x 2 x 2 × ( x + 3 ) = 1 × 2 x × ( x + 3 ) = 2 x ( x + 3 ) ‾ \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} = 2 x 2 × ( x + 3 ) 1 × 4 x = x × ( x + 3 ) 1 × 2 = x ( x + 3 ) 2
