## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{2}{(x+3)(x-1)}$
Join the two fractions $\frac{1}{x-1}+\frac{1}{x+3}$: Find the LCD of the two fractions (i.e. $(x+3)(x-1)$) and adjust accordingly. This then becomes: $\frac{1\times(x+3)}{(x+3)(x-1)}+\frac{1\times(x-1)}{(x+3)(x-1)}$ Combine the two fractions: $\frac{(x+3)+(x-1)}{(x+3)(x-1)}$ Collect like terms: $\frac{2x+2}{(x+3)(x-1)}$ So the fraction is now: $=\frac{\frac{2x+2}{(x+3)(x-1)}}{x+1}$ Apply the fraction rule: $\frac{\frac{b}{c}}{a}=\frac{b}{c\times a}$ This then becomes: $=\frac{2x+2}{(x+3)(x-1)(x+1)}$ Factor the numerator: $=\frac{2(x+1)}{(x+3)(x-1)(x+1)}$ Cancel out the common factor: $=\frac{2}{(x+3)(x-1)}$