Answer
Solution set = $\{5\}$.
Work Step by Step
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-3,+3 \}\qquad (*)$
The first denominator is a difference of squares, $(x+3)(x-3)$
Multiply the equation with the LCD=$(x+3)(x-3)$
$ 8+4(x-3)=2(x+3)\qquad$ ... simplify (distribute)
$8+4x-12=2x+6$
$ 4x-4=2x+6\qquad$ ... add $4-2x$
$4x-2x=6+4$
$2x=10$
$ x=5 \qquad$ ...Checking (*), this is a valid solution
Solution set = $\{5\}$.
