[tex]1 - \dfrac{y-1}{y} = \dfrac{1}{y^2 + 2y}[/tex]
[tex]\dfrac y y - \dfrac{y-1}{y} = \dfrac{1}{y^2 + 2y}[/tex]
[tex]\dfrac{1}{y} = \dfrac{1}{y^2 + 2y}[/tex]
[tex]y = y^2 + 2y[/tex]
[tex]y^2 + y = 0[/tex]
[tex]y(y+1)=0[/tex]
[tex]y = 0 \textrm{ or } y = -1[/tex]
We rule out y=0 because of the denominator y in the original equation.
Answer: B: y=-1
