Question

Subtract$\frac{4}{n+4}-\frac{8n}{n^2-16}$.

Short answer

The difference is$\frac{-4}{n-4}$.

Step-by-step solution

Given Information  

The Given expression is$\frac{4}{n+4}-\frac{8n}{n^2-16}$.

Find the LCD of denominators.    

Factor the denominators.

$n^2-16=(n+4)(n-4)$

$n+4=(n+4)$

LCD$=(n+4)(n-4)$

Rewrite the expression in factored form.

$=\frac{4}{n+4}-\frac{8n}{(n+4)(n-4)}$

Rewrite each fraction as an equivalent fraction with the LCD.    

$=\frac{4(n-4)}{(n+4)(n-4)}-\frac{8n}{(n+4)(n-4)}$

$=\frac{4n-16}{(n+4)(n-4)}-\frac{8n}{(n+4)(n-4)}$

Subtract and simplify the numerator.   

Take common denominator.

$=\frac{4n-16-8n}{(n+4)(n-4)}$

Simplify.

$=\frac{-4n-16}{(n+4)(n-4)}$

$=\frac{-4(n+4)}{(n+4)(n-4)}$

$=\frac{-4}{n-4}$