## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{3x^2-2x-3}{(x-1)(x+1)}, \quad x\ne -1,1$
Make the expressions similar using their LCD which is $(x-1)(x+1)$, to obtain: $$\frac{x}{x+1}+\frac{2x-3}{x-1}=\frac{x(x-1)}{(x-1)(x+1)}+\frac{(2x-3)(x+1)}{(x-1)(x+1)}, \quad x\ne-1, 1.$$ Now, since the denominators are the same, add the numerators to obtain: \begin{align*} \frac{x(x-1)}{(x-1)(x+1)}+\frac{(2x-3)(x+1)}{(x-1)(x+1)}&=\frac{x^2-x}{(x-1)(x+1)}+\frac{2x^2+2x-3x-3}{(x-1)(x+1)}\\ \\&=\frac{x^2-x}{(x-1)(x+1)}+\frac{2x^2-x-3}{(x-1)(x+1)}\\ \\&=\frac{x^2-x+2x^2-x-3}{(x-1)(x+1)} \\ \\&=\frac{3x^2-2x-3}{(x-1)(x+1)},\quad x\ne-1, 1\end{align*}