## Intermediate Algebra for College Students (7th Edition)

$\displaystyle \frac{x-3}{x(x+1)}$
Factor the denominators $x^{2}-x=x(x-1)$ $x^{2}-1=(x-1)(x+1)$ LCD = $x(x-1)(x+1)$ $\displaystyle \frac{x+3}{x^{2}-x}-\frac{8}{x^{2}-1}= \displaystyle \frac{x+3}{x(x-1)}\cdot\frac{x+1}{x+1}-\frac{8}{(x-1)(x+1)}\cdot\frac{x}{x}$ $= \displaystyle \frac{(x+3)(x+1)-8x}{x(x-1)(x+1)}$ $= \displaystyle \frac{x^{2}+4x+3-8x}{x(x-1)(x+1)}$ $= \displaystyle \frac{x^{2}-4x+3}{x(x-1)(x+1)}$ ... factors of $3$ whose sum is $-4$ ... are $-1$ and $-3$ $= \displaystyle \frac{(x-1)(x-3)}{x(x-1)(x+1)}\qquad$... cancel the common factor, $(x-1)$ = $\displaystyle \frac{x-3}{x(x+1)}$