Answer
the solutions are -4 and 6
Work Step by Step
$\frac{2x}{x+2}$-2=$\frac{x-8}{x-2}$
(x+2)(x-2)[$\frac{2x}{x+2}$-2]=$\frac{x-8}{x-2}$(x+2)(x-2)
(x+2)(x-2)$\frac{2x}{x+2}$-2(x+2)(x-2)=(x-8)(x+2)
2x(x-2)-2($x^{2}$-4)=$x^{2}$+2x-8x-16
2$x^{2}$-4x-2$x^{2}$+8=$x^{2}$-6x-16
-4x+8=$x^{2}$-6x-16
0=$x^{2}$-6x-16+4x-8
0=$x^{2}$-2x-24
0=$x^{2}$+4x-6x-24
0=x(x+4)-6(x+4)
0=(x-6)(x+4)
x-6=0 or x+4=0
x=6 or x=-4
The solutions are -4 and 6
Check
Let x=-4
$\frac{2(-4)}{-4+2}$-2=$\frac{-4-8}{-4-2}$
$\frac{-8}{-2}$-2=$\frac{-12}{-6}$
4-2=2
2=2
Let x=6
$\frac{2*6}{6+2}$-2=$\frac{6-8}{6-2}$
$\frac{12}{8}$-2=$\frac{-2}{4}$
$\frac{6}{4}$-$\frac{8}{4}$=$\frac{-2}{4}$
$\frac{-2}{4}$=$\frac{-2}{4}$
