Answer
$\{ \frac{3}{7} \}$.
Work Step by Step
First we will clear fractions. In order to do this we have to determine the Least Common Denominator of all the denominators and multiply the equation by it. We factor each denominator.
Factor $y^2-4$.
$y^2-2^2$
Use the algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(y+2)(y-2)$
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 $y^2+3y+2$.
Rewrite the middle term $3y$ 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).
$=(y+2)(y+1)$.
Substitute all factors into the given equation.
$\frac{y-1}{(y+2)(y-2)}+\frac{y}{(y-2)(y+1)}=\frac{2y-1}{(y+2)(y+1)}$... (1)
The LCD of the denominators is $(y+2)(y+1)(y-2)$.
Multiply the equation by LCD.
$(y+2)(y+1)(y-2)\left ( \frac{y-1}{(y+2)(y-2)}+\frac{y}{(y-2)(y+1)}\right )=(y+2)(y+1)(y-2)\left ( \frac{2y-1}{(y+2)(y+1)} \right )$
Use the distributive property.
$(y+2)(y+1)(y-2)\cdot \frac{y-1}{(y+2)(y-2)}+(y+2)(y+1)(y-2)\cdot\frac{y}{(y-2)(y+1)}=(y+2)(y+1)(y-2)\cdot \frac{2y-1}{(y+2)(y+1)} $
Cancel common factors.
$(y+1)(y-1)+y(y+2)=(2y-1)(y-2) $
Use the distributive property again.
$y^2-y+y-1+y^2+2y=2y^2-4y-y+2 $
Add like terms.
$2y^2+2y-1=2y^2-5y+2 $
Add both sides $-2y^2+5y-2$.
$2y^2+2y-1-2y^2+5y-2=2y^2-5y+2 -2y^2+5y-2$
Simplify.
$7y-3=0$
Isolate $y$.
$y=\frac{3}{7}$
The solution set is $\{ \frac{3}{7} \}$.
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:
$y+2=0\Rightarrow y=-2$
$y-2=0\Rightarrow y=2$
$y+1=0\Rightarrow y=-1$
So the domain is $(-\infty,-2)\cup(-2,-1)\cup(-1,2)\cup(2,\infty)$.
The solution we found, $\frac{3}{7}$, belongs to the domain, so it is correct.
