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

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GMATH Teacher
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Joined: 12 Oct 2010
Posts: 935
When defined, the expression (x^2-y^2)/(xy)-(xy-y^2)/(xy-x^2) is equal  [#permalink]

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03 Mar 2019, 14:28
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Difficulty:

55% (hard)

Question Stats:

55% (02:08) correct 45% (02:52) wrong based on 20 sessions

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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}

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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
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Joined: 21 Apr 2014
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Re: When defined, the expression (x^2-y^2)/(xy)-(xy-y^2)/(xy-x^2) is equal  [#permalink]

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03 Mar 2019, 16:02
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.
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Vivian Kerr | GMAT/GRE Tutor @ http://www.gmatrockstar.com | gmatrockstar[at]gmail.com | solid 5-star reviews on Yelp!

Former Kaplan and Grockit instructor. I've freelanced with Veritas Prep, Magoosh, and most of the bigger test prep companies. Now offering Skype-based private tutoring for the GMAT and GRE.

GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
When defined, the expression (x^2-y^2)/(xy)-(xy-y^2)/(xy-x^2) is equal  [#permalink]

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03 Mar 2019, 16:28
fskilnik wrote:
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}

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

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

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

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

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
When defined, the expression (x^2-y^2)/(xy)-(xy-y^2)/(xy-x^2) is equal   [#permalink] 03 Mar 2019, 16:28
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