If $\frac{x}{x-1} = \frac{y^2 + 2y - 1}{y^2 + 2y - 2},$ then $x$ equals
$\text{(A)} \quad y^2 + 2y - 1$
$\text{(B)} \quad y^2 + 2y - 2$
$\text{(C)} \quad y^2 + 2y + 2$
$\text{(D)} \quad y^2 + 2y + 1$
$\text{(E)} \quad -y^2 - 2y + 1$
We can cross multiply both sides of the equation and simplify.
\[x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)\]
\[xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1\]
\[-x = -y^2 - 2y + 1\]
\[x = y^2 + 2y - 1\]
The answer is $\text{A.}$
-edited by coolmath34