#### Answer

$x=3$

#### Work Step by Step

We may plug the expressions involving $x$ directly into the given equation $y_{1}+y_{2}=y_{3}$, find a common denominator, and solve:
$(\frac{2x-1}{x^2+2x-8})+(\frac{2}{x+4})=(\frac{1}{x-2})$
$\frac{2x-1}{(x-2)(x+4)}+\frac{2}{x+4}=\frac{1}{x-2}$
$\frac{2x-1}{(x-2)(x+4)}+(\frac{x-2}{x-2})(\frac{2}{x+4})=(\frac{x+4}{x+4})(\frac{1}{x-2})$
$\frac{2x-1}{(x-2)(x+4)}+\frac{2x-4}{(x-2)(x+4)}=\frac{x+4}{(x-2)(x+4)}$
$\frac{(2x-1)+(2x-4)}{(x-2)(x+4)}=\frac{x+4}{(x-2)(x+4)}$
$\frac{4x-5}{(x-2)(x+4)}=\frac{x+4}{(x-2)(x+4)}$
$4x-5=x+4$
$4x-x=4+5$
$3x=9$
$x=3$