Answer
Solution set = $\emptyset$.
(no solution)
Work Step by Step
The RHS denominator $x^{2}+4x+3$ is factored by searching
for two factors of $c=3$ whose sum is $b=4$ ... we find $+1$ and $+3$
$x^{2}+4x+3=(x+1)(x+3)$
First, we exclude those values of x that yield a zero in any of the denominators.
$x\not\in\{-3,-1 \}\qquad (*)$
Multiply the equation with the LCD=$(x+1)(x+3)$
$ 2(x+1)-5(x+3)=3x+5\qquad$ ... simplify (distribute)
$2x+2-5x-15=3x+5$
$-3x-13=3x+5\qquad$ ... add $13-3x$ to both sides
$-6x=18$
$ x=-3 \qquad$ ...Checking (*), this is not a valid solution
Solution set = $\emptyset$.
(no solution)
