## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{2x+7}{(x+3)(x+4)}$
$\frac{2}{x+3}-\frac{1}{x^{2}+7x+12}$ Factorise and replace the term $x^{2}+7x+12$: $=\frac{2}{x+3}-\frac{1}{(x+3)(x+4)}$ Find the lowest common denominator for the two fractions (i.e. $(x+3)(x+4)$) and adjust accordingly: $=\frac{2\times (x+4)}{(x+3)\times (x+4)}-\frac{1}{(x+3)(x+4)}$ $=\frac{2x+8}{(x+3)(x+4)}-\frac{1}{(x+3)(x+4)}$ Combine the fractions: $=\frac{2x+8-1}{(x+3)(x+4)}$ Simplify: $=\frac{2x+7}{(x+3)(x+4)}$