Answer
$\displaystyle \frac{29}{6(x+4)}, \qquad x\neq-4$
Work Step by Step
Factor each denominator.
$\displaystyle \frac{5}{2x+8}+\frac{7}{3x+12}=\frac{5}{2(x+4)}+\frac{7}{3(x+4)}$
The denominators are different. Find a common denominator.
LCD=$2(x+4)\cdot 3=6(x+4)$
$=\displaystyle \frac{5}{2(x+4)}\times\frac{3}{3}+\frac{7}{3(x+4)}\times\frac{2}{2}$
$=\displaystyle \frac{15}{6(x+4)}+\frac{14}{6(x+4)}$
$=\displaystyle \frac{15+14}{6(x+4)}$
= $\displaystyle \frac{29}{6(x+4)}, \qquad x\neq-4$
