Question

Solve for x: 5 over quantity x squared minus 4 plus 2 over x equals 2 over quantity x minus 2.

Answer 1

$\text{The domain}\\\\x\neq0\ \wedge\ x\neq-2\ \wedge\ x\neq2$

$\dfrac{5}{x^2-4}+\dfrac{2}{x}=\dfrac{2}{x-2}\qquad\text{subtract}\ \dfrac{2}{x-2}\ \text{from obth sides}\\\\\dfrac{5}{x^2-2^2}+\dfrac{2}{x}-\dfrac{2}{x-2}=0\\\\\dfrac{5}{(x-2)(x+2)}+\dfrac{2}{x}-\dfrac{2}{x-2}=0\\\\\dfrac{5x}{x(x-2)(x+2)}+\dfrac{2(x-2)(x+2)}{x(x-2)(x+2)}-\dfrac{2x(x+2)}{x(x-2)(x+2)}=0\\\\\dfrac{5x+2(x^2-4)-2x(x+2)}{x(x-2)(x+2)}=0\\\\\dfrac{5x+2x^2-8-2x^2-4x}{x(x-2)(x+2)}=0\\\\\dfrac{x-8}{x(x-2)(x+2)}=0\iff x-8=0\to\boxed{x=8}\in D$