Answer
$\{ -6,\frac{1}{2} \}$.
Work Step by Step
First we clear fractions. In order to do this, we factor all denominators, then calculate the Least Common Denominator (LCD) and multiply the equation by it.
Factor $y^2+5y+6$.
Rewrite the middle term $5y$ as $3y+2y$.
$=y^2+3y+2y+6$
Group terms.
$=(y^2+3y)+(2y+6)$
Factor each term.
$=y(y+3)+2(y+3)$
Factor out (y+3).
$=(y+3)(y+2)$.
Factor $y^2+y-2$.
Rewrite the middle term $y$ as $2y-y$.
$=y^2+2y-y-2$
Group terms.
$=(y^2+2y)+(-y-2)$
Factor each term.
$=y(y+2)-1(y+2)$
Factor out (y+2).
$=(y+2)(y-1)$.
Factor $y^2+2y-3$.
Rewrite the middle term $2y$ as $3y-y$.
$=y^2+3y-y-3$
Group terms.
$=(y^2+3y)+(-y-3)$
Factor each term.
$=y(y+3)-1(y+3)$
Factor out (y+3).
$=(y+3)(y-1)$
Substitute all factors into the given equation.
$\frac{3y}{(y+3)(y+2)}+\frac{2}{(y+2)(y-1)}=\frac{5y}{(y+3)(y-1)}$.... (1)
LCD of the denominators is $(y+3)(y+2)(y-1)$.
Multiply the equation by the LCD.
$(y+3)(y+2)(y-1)\left ( \frac{3y}{(y+3)(y+2)}+\frac{2}{(y+2)(y-1)}\right )=(y+3)(y+2)(y-1)\left ( \frac{5y}{(y+3)(y-1)} \right )$
Use the distributive property.
$(y+3)(y+2)(y-1)\cdot \frac{3y}{(y+3)(y+2)}+(y+3)(y+2)(y-1)\cdot\frac{2}{(y+2)(y-1)}=(y+3)(y+2)(y-1)\cdot \frac{5y}{(y+3)(y-1)} $
Cancel common factors.
$3y(y-1)+2(y+3)=5y(y+2) $
Use the distributive property again.
$3y^2-3y+2y+6=5y^2+10y $
Add like terms.
$3y^2-y+6=5y^2+10y $
Add both sides $-3y^2+y-6$.
$3y^2-y+6-3y^2+y-6=5y^2+10y -3y^2+y-6$
Simplify.
$0=2y^2+11y -6$
Rewrite the middle term $11y$ as $12y-1y$.
$0=2y^2+12y-1y -6$
Group terms.
$0=(2y^2+12y)+(-1y -6)$
Factor each term.
$0=2y(y+6)-1(y +6)$
Factor out $(y+6)$.
$0=(y+6)(2y-1)$
Set each factor equal to zero.
$y+6=0$ or $2y-1=0$
Isolate $y$.
$y=-6$ or $y=\frac{1}{2}$
The solution set is $\{ -6,\frac{1}{2} \}$.
Note: How do we know that the given equation is defined for these two values of $y$? Since rational expressions are involved we must either replace the two solutions in the original equation or find the domain of the equation. The easiest way is to determine the domain and we can easily do this from equation (1).
The "bad" values of $y$ are those for which at least one of the denominators is zero:
$y+3=0\Rightarrow y=-3$
$y+2=0\Rightarrow y=-2$
$y-1=0\Rightarrow y=1$
So the domain is $(-\infty,-3)\cup(-3,-2)\cup(-2,1)\cup(1,\infty)$.
The solutions we found, $-6$ and $\frac{1}{2}$, belong to the domain, so they are correct.
