1981 AHSME Problems/Problem 6

Problem

If $\frac{x}{x-1} = \frac{y^2 + 2y - 1}{y^2 + 2y - 2},$ then $x$ equals

$\text{(A)} \quad y^2 + 2y - 1$

$\text{(B)} \quad y^2 + 2y - 2$

$\text{(C)} \quad y^2 + 2y + 2$

$\text{(D)} \quad y^2 + 2y + 1$

$\text{(E)} \quad -y^2 - 2y + 1$

Solution

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

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1981 AHSME Problems/Problem 6

Problem

Solution