#### Answer

$x=2$

#### Work Step by Step

$\dfrac{3}{x+3}=\dfrac{12x+19}{x^{2}+7x+12}-\dfrac{5}{x+4}$
Factor the denominator of the second fraction:
$\dfrac{3}{x+3}=\dfrac{12x+19}{(x+4)(x+3)}-\dfrac{5}{x+4}$
Multiply the whole equation by $(x+4)(x+3)$:
$(x+4)(x+3)\Big[\dfrac{3}{x+3}=\dfrac{12x+19}{(x+4)(x+3)}-\dfrac{5}{x+4}\Big]$
$3(x+4)=12x+19-5(x+3)$
$3x+12=12x+19-5x-15$
Take all terms to the right side of the equation and simplify it by combining like terms:
$0=12x+19-5x-15-3x-12$
$4x-8=0$
Solve for $x$:
$4x=8$
$x=\dfrac{8}{4}=2$
We can see that the equation is not undefined for this value of $x$. The solution is $x=2$