Question

please just tell me what I'm doing wrong, don't tell me the answer, i need to figure it out.

Answer 1

we have the expression

$\frac{x}{x^2-16}-\frac{3}{x-4}$

REmember that

x^2-16=(x+4)(x-4)

substitute

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

Multiply by (x+4)/(x+4) the second term

$\begin{gathered} \frac{x}{(x+4)(x-4)}-\frac{3}{x-4}_{}\cdot\frac{x+4}{x+4} \\ \\ \frac{x}{(x+4)(x-4)}-\frac{3x+12}{(x-4)(x+4)}_{} \end{gathered}$

Adds fractions with the same denominator

$\begin{gathered} \frac{x-3x-12}{(x+4)(x-4)} \\ \\ \frac{-2x-12}{(x+4)(x-4)} \end{gathered}$

Factor minus

$-\frac{2x+12}{(x+4)(x-4)}$

Factor 2

$-\frac{2(x+6)}{(x+4)(x-4)}$