## Precalculus (6th Edition)

$\dfrac{-2}{x-3}+\dfrac{3}{x+3}=\dfrac{-12}{x^{2}-9}$ Factor the denominator of the fraction on the right side of the equation: $\dfrac{-2}{x-3}+\dfrac{3}{x+3}=\dfrac{-12}{(x-3)(x+3)}$ Multiply the whole equation by $(x-3)(x+3)$: $(x-3)(x+3)\Big[\dfrac{-2}{x-3}+\dfrac{3}{x+3}=\dfrac{-12}{(x-3)(x+3)}\Big]$ $-2(x+3)+3(x-3)=-12$ Evaluate the indicated operations: $-2x-6+3x-9=-12$ Take all terms without $x$ to the right side of the equation: $-2x+3x=-12+6+9$ Simplify both sides: $x=3$ The original equation is undefined for $x=3$, so the equation has no solution.