When defined, the expression (x^2-y^2)/(xy)-(xy-y^2)/(xy-x^2) is equal

Question

GMATH  practice exercise (Quant Class 3)

When defined, the expression  \,{{{x^2} - {y^2}} \over {xy}} - {{xy - {y^2}} \over {xy - {x^2}}}\,  is equal to:

\eqalign{
 & \left( A \right)\,\,x{y^{ - 1}} \cr 
 & \left( B \right)\,\left( {{x^2} - 2{y^2}} \right){\left( {xy} \right)^{ - 1}} \cr 
 & \left( C \right)\,\,{x^2} \cr 
 & \left( D \right)\,\,x - 2y \cr 
 & \left( E \right)\,\,x + 2y \cr}

GMATRockstar · Answer

Holy algebraic nightmare, Batman!

Just kidding! But for real, the GMAT isn't likely to force you to do this level of algebraic manipulation. My advice to students if they see something that is even 50% as complicated as this is to quickly plug in numbers. The GMAT is about flexibility and strategy, not rote calculation.

fskilnik · Answer

? = {{{x^2} - {y^2}} \over {xy}} - {{xy - {y^2}} \over {xy - {x^2}}}\,\,\,

\frac{{xy - {y^2}}}{{xy - {x^2}}} = \frac{{y\left( {x - y} \right)}}{{x\left( {y - x} \right)}} = \underleftrightarrow {\frac{{ - y\left( {y - x} \right)}}{{x\left( {y - x} \right)}}} = \frac{{ - y}}{x}

? = \frac{{{x^2} - {y^2}}}{{xy}} + \frac{y}{x} = \underleftrightarrow {\frac{{{x^2} - {y^2}}}{{xy}} + \frac{{y \cdot y}}{{x \cdot y}} = \frac{{{x^2}}}{{xy}}} = \frac{x}{y}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( A \right)

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

When defined, the expression (x^2-y^2)/(xy)-(xy-y^2)/(xy-x^2) is equal

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