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

$\dfrac{1}{x+2}=\dfrac{4}{x^{2}-4}-\dfrac{1}{x-2}$ Factor the denominator of the second fraction: $\dfrac{1}{x+2}=\dfrac{4}{(x-2)(x+2)}-\dfrac{1}{x-2}$ Multiply the whole equation by $(x-2)(x+2)$: $(x-2)(x+2)\Big[\dfrac{1}{x+2}=\dfrac{4}{(x-2)(x+2)}-\dfrac{1}{x-2}\Big]$ $x-2=4-x-2$ Take all terms to the left side of the equation and simplify it by combining like terms: $x-2-4+x+2=0$ $2x-4=0$ Solve for $x$: $2x=4$ $x=2$ Substituting $x=2$ in the original equation makes both denominator on the right side $0$, this means that this equation has no solution.