how to solve 3/x+4 minus 8/x-4 =11/x^2-16

Start by getting a common denominator with the terms on the left side of the equal sign.

$\frac{3(x-4)}{(x+4)(x-4)}-\frac{8(x+4)}{(x-4)(x+4)}$

Can you take it from there?

Hello, DeziB!

$\text{Solve for }x\!:\;\frac{3}{x+4} - \frac{8}{x-4} \:=\:\frac{11}{x^2-16}$

We are given: . $\frac{3}{x+4} - \frac{8}{x-4} \;=\;\frac{11}{(x-4)(x+4)}$

Multiply by $(x\!-\!4)(x\!+\!4)\!:\;\;3(x-4) - 8(x+4) \:=\:11$

Got it?