Answer
$=2, \displaystyle \quad x\neq-\frac{4}{3}$
Work Step by Step
The fractions have a common denominator, we add the numerators...
$\displaystyle \frac{3x+2}{3x+4}+\frac{3x+6}{3x+4}=\frac{3x+2+3x+6}{3x+4}$
$=\displaystyle \frac{6x+8}{3x+4}$
... factor what we can...
$=\displaystyle \frac{2(3x+4)}{(3x+4)}$
... exclude values that yield 0 in the denominator:
$x\displaystyle \neq-\frac{4}{3}$
... and, cancel the common factor
$=2, \displaystyle \quad x\neq-\frac{4}{3}$
