#### Answer

Solution set = $\{3\}$.

#### Work Step by Step

The LHS denominator $x^{2}+2x-8$ is factored by searching
for two factors of $c=-8$ whose sum is $b=2$ ... we find $+4$ and $-2$
$x^{2}+2x-8=(x-2)(x+4)$
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-4,2 \}\qquad (*)$
Multiply the equation with the LCD=$(x-2)(x+4)$
$(2x-1)+2(x-2)=x+4\qquad$ ... simplify (distribute)
$2x-1+2x-4=x+4$
$ 4x-5=x+4\qquad$ ... add $5-x$ to both sides
$3x=9$
$ x=3 \qquad$ ...Checking (*), this is a valid solution
Solution set = $\{3\}$.