## Algebra: A Combined Approach (4th Edition)

$\frac{4}{x-2}-\frac{x}{x+2}=\frac{16}{x^{2}-4}$ $\frac{4(x+2)}{(x-2)(x+2)}-\frac{x(x-2)}{(x-2)(x+2)}=\frac{16}{(x-2)(x+2)}$ $4x+8-x^{2}+2x=16$ $x^{2}-6x+8=0$ $(x-4)(x+2)=0$ x=4 or x=-2 Substitute x=-2 to the original formula and find that $\frac{x}{(-2)+2}$, $\frac{16}{(-2)^2-4}$, the denominator is 0, which is incorrect. So the answer is x=4