Answer
Solution set = $\{2\}$.
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-2,1 \}\qquad (*)$
Multiply the equation with the LCD=$(x+2)(x-1)$
$ 4x(x-1)+2(x+2)=4(x+2)(x-1)\qquad$ ... simplify (distribute)
$4x^{2}-4x+2x+4=4(x^{2}-x+2x-2)$
$ 4x^{2}-2x+4=4x^{2}+4x-8\qquad$ ... add $-4-4x^{2}-4x$
$-2x-4x=-8-4$
$-6x=-12$
$ x=2\qquad$ ...Checking (*), this is a valid solution
Solution set = $\{2\}$.
