Answer
The solution set is $\displaystyle \{-\frac{11}{6}\}$.
Work Step by Step
STEP 1: Restrictions. There must be no zeros in the denominator, so
$y\neq-3, y\neq 4,y\neq-6$
STEP 2: Fractions.
The LCM of all denominators is $(y+3)(y-4)(y+6)$
After multiplying both sides with the LCM,
$2(y-4)(y+6) + 3(y+3)(y+6) = 5(y+3)(y-4)$
STEP 3: Remove all parentheses and simplify.
$2(y^{2}+2y-24)+3(y^{2}+9y+18)=5(y^{2}-y-12)$
$2y^{2}+4y-48+3y^{2}+27y+54=5y^{2}-5y-60$
$5y^{2}+31y+6=5y^{2}-5y-60$
STEP 4: Variables on one side, all other terms on the other...
Add $-5y^{2}+5y-6$ to both sides,
$36y=-66$
STEP 5: Simplify and solve.
$36y=-66\qquad/\div 36$
$x=-\displaystyle \frac{66}{36}$
$x=-\displaystyle \frac{11}{6}$
No restriction prevents $-\displaystyle \frac{11}{6}$ being a solution, so the
solution set is $\displaystyle \{-\frac{11}{6}\}$.
