## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{x+3}{4x^{2}-9}\div\dfrac{x^{2}+7x+12}{2x^{2}+7x-15}=\dfrac{x+5}{(2x+3)(x+4)}$
$\dfrac{x+3}{4x^{2}-9}\div\dfrac{x^{2}+7x+12}{2x^{2}+7x-15}$ Factor the denominator of the first fraction and the numerator and the denominator of the second fraction: $\dfrac{x+3}{(2x+3)(2x-3)}\div\dfrac{(x+4)(x+3)}{(2x-3)(x+5)}=...$ Evaluate the division and simplify: $...=\dfrac{(x+3)(2x-3)(x+5)}{(2x+3)(2x-3)(x+4)(x+3)}=\dfrac{x+5}{(2x+3)(x+4)}$