## College Algebra (6th Edition)

$x=7$
$\dfrac{6}{x+3}-\dfrac{5}{x-2}=\dfrac{-20}{x^{2}+x-6}$ Factor the denominator of the fraction on the right side: $\dfrac{6}{x+3}-\dfrac{5}{x-2}=\dfrac{-20}{(x+3)(x-2)}$ Multiply the whole fraction by $(x+3)(x-2)$: $(x+3)(x-2)\Big[\dfrac{6}{x+3}-\dfrac{5}{x-2}=\dfrac{-20}{(x+3)(x-2)}\Big]$ $6(x-2)-5(x+3)=-20$ $6x-12-5x-15=-20$ Take all terms without $x$ to the right side of the equation, simplify and solve for $x$: $6x-5x=-20+15+12$ $x=7$