## Algebra: A Combined Approach (4th Edition)

$x=2$
$\dfrac{3}{x+3}=\dfrac{12x+19}{x^{2}+7x+12}-\dfrac{5}{x+4}$ 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 $x=2$