Algebra: A Combined Approach (4th Edition)

$x=5$
$\dfrac{2}{x-3}-\dfrac{4}{x+3}=\dfrac{8}{x^{2}-9}$ Factor the denominator of the fraction on the right side of the equation: $\dfrac{2}{x-3}-\dfrac{4}{x+3}=\dfrac{8}{(x-3)(x+3)}$ Multiply the whole equation by $(x-3)(x+3)$ $(x-3)(x+3)\Big[\dfrac{2}{x-3}-\dfrac{4}{x+3}=\dfrac{8}{(x-3)(x+3)}\Big]$ $2(x+3)-4(x-3)=8$ $2x+6-4x+12=8$ Take the $8$ to the left side of the equation and simplify the equation by combining like terms: $2x+6-4x+12-8=0$ $-2x+10=0$ Solve for $x$: $-2x=-10$ $x=\dfrac{-10}{-2}$ $x=5$ The initial equation is not undefine for the value of $x$ found. So the solution to this equation is $x=5$