## Algebra: A Combined Approach (4th Edition)

$\dfrac{\dfrac{8}{x+4}+2}{\dfrac{12}{x+4}-2}=-\dfrac{x+8}{x-2}$
$\dfrac{\dfrac{8}{x+4}+2}{\dfrac{12}{x+4}-2}$ Evaluate the sum indicated in the numerator and the substraction indicated in the denominator: $\dfrac{\dfrac{8}{x+4}+2}{\dfrac{12}{x+4}-2}=\dfrac{\dfrac{8+2(x+4)}{x+4}}{\dfrac{12-2(x+4)}{x+4}}=\dfrac{\dfrac{8+2x+8}{x+4}}{\dfrac{12-2x-8}{x+4}}=...$ $...=\dfrac{\dfrac{2x+16}{x+4}}{\dfrac{4-2x}{x+4}}=...$ Evaluate the division: $...=\dfrac{2x+16}{x+4}\div\dfrac{4-2x}{x+4}=\dfrac{(2x+16)(x+4)}{(x+4)(4-2x)}=...$ Take out common factor $2$ from the first parentheses in the numerator and from the second parentheses in the denominator. After that, simplify: $...=\dfrac{2(x+8)(x+4)}{2(x+4)(2-x)}=\dfrac{x+8}{2-x}=-\dfrac{x+8}{x-2}$