#### Answer

$x=2$

#### Work Step by Step

We plug the expressions involving $x$ into the given equation $y_{1}+y_{2}=y_{3}$, look for a common denominator, and solve:
$(\frac{5}{x+4})+(\frac{3}{x+3})=\frac{12x+19}{x^{2}+7x+12}$
$(\frac{5}{x+4})+(\frac{3}{x+3})=\frac{12x+19}{(x+3)(x+4)}$ (Notice that the denominator on the right-hand side of the equation factors nicely.)
$(\frac{x+3}{x+3})(\frac{5}{x+4})+(\frac{x+4}{x+4})(\frac{3}{x+3})=\frac{12x+19}{(x+3)(x+4)}$
$\frac{5(x+3)}{(x+3)(x+4)}+\frac{3(x+4)}{(x+3)(x+4)}=\frac{12x+19}{(x+3)(x+4)}$
$\frac{5x+15+3x+12}{(x+3)(x+4)}=\frac{12x+19}{(x+3)(x+4)}$
$\frac{8x+27}{(x+3)(x+4)}=\frac{12x+19}{(x+3)(x+4)}$
$(x+3)(x+4)(\frac{8x+27}{(x+3)(x+4)})=(x+3)(x+4)(\frac{12x+19}{(x+3)(x+4)})$
$8x+27=12x+19$
$27-19=12x-8x$
$4x=8$
$x=2$