## College Algebra (6th Edition)

$\dfrac{5}{x+2}+\dfrac{3}{x-2}=\dfrac{12}{(x+2)(x-2)}$ Multiply the whole equation by $(x+2)(x-2)$: $(x+2)(x-2)\Big[\dfrac{5}{x+2}+\dfrac{3}{x-2}=\dfrac{12}{(x+2)(x-2)}\Big]$ $5(x-2)+3(x+2)=12$ $5x-10+3x+6=12$ Take all terms without $x$ to the right side of the equation and solve for $x$: $5x+3x=12+10-6$ $8x=16$ $x=\dfrac{16}{8}$ $x=2$ Since the original equation is undefined for $x=2$, it has no solution.