Answer
$\frac{2x+7}{(x+3)(x+4)}$
Work Step by Step
$\frac{2}{x+3}-\frac{1}{x^{2}+7x+12}$
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:
$=\frac{2x+7}{(x+3)(x+4)}$
