Can someone help me with this one?
$\displaystyle
\frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0
$
$\displaystyle \frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0$
$\displaystyle \frac {5(x-4)-1(x+2)}{(x+2)(x-4)} - \frac{5}{2x+4} = 0$
$\displaystyle \frac {5x-20-x-2}{(x+2)(x-4)}-\frac{5}{2x+4} = 0$
$\displaystyle \frac {4x-22}{(x+2)(x-4)}-\frac{5}{2x+4} = 0$
$\displaystyle \frac {(4x-22)(2x+4)-5(x+2)(x-4)}{(x+2)(x-4)(2x+4)}= 0 $ (cross multiply)
(4x-22)(2x+4)-5(x+2)(x-4) = 0
8x^2+16x-44x-88-5(x^2-4x+2x-8) = 0
8x^2+16x-44x-88-5x^2+10x+40 = 0
3x^2-18x-48 = 0
(x-8)(3x+6) = 0
x=8 or x = -2
Hope it helps.