Answer
$\dfrac{\dfrac{6}{x+2}+4}{\dfrac{8}{x+2}-4}=-\dfrac{2x+7}{2x}$
Work Step by Step
$\dfrac{\dfrac{6}{x+2}+4}{\dfrac{8}{x+2}-4}$
Evaluate the sum indicated in the numerator and the substraction indicated in the denominator:
$\dfrac{\dfrac{6}{x+2}+4}{\dfrac{8}{x+2}-4}=\dfrac{\dfrac{6+4(x+2)}{x+2}}{\dfrac{8-4(x+2)}{x+2}}=\dfrac{\dfrac{6+4x+8}{x+2}}{\dfrac{8-4x-8}{x+2}}=...$
$...=\dfrac{\dfrac{4x+14}{x+2}}{\dfrac{-4x}{x+2}}=...$
Evaluate the division and simplify:
$...=\dfrac{4x+14}{x+2}\div\dfrac{-4x}{x+2}=\dfrac{(4x+14)(x+2)}{-4x(x+2)}=...$
$...=\dfrac{4x+14}{-4x}=...$
Take out common factor $2$ from the numerator and simplify one more time:
$...=\dfrac{2(2x+7)}{-4x}=-\dfrac{2x+7}{2x}$
