## College Algebra (6th Edition)

$\dfrac{1}{x-4}-\dfrac{5}{x+2}=\dfrac{6}{x^{2}-2x-8}$ Factor the denominator of the fraction on the right side: $\dfrac{1}{x-4}-\dfrac{5}{x+2}=\dfrac{6}{(x+2)(x-4)}$ Multiply the whole equation by $(x+2)(x-4)$: $(x+2)(x-4)\Big[\dfrac{1}{x-4}-\dfrac{5}{x+2}=\dfrac{6}{(x+2)(x-4)}\Big]$ $(x+2)-5(x-4)=6$ $x+2-5x+20=6$ Take all terms without $x$ to the right side of the equation, simplify, and solve for $x$: $x-5x=6-2-20$ $-4x=-16$ $x=\dfrac{-16}{-4}$ $x=4$ Since the original equation is undefined for $x=4$, it has no solution.