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If x does not equal y, and xy does not equal 0, then when x

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If x does not equal y, and xy does not equal 0, then when x [#permalink] New post 02 Nov 2010, 00:48
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69% (01:52) correct 31% (00:46) wrong based on 122 sessions
\frac{x+y}{x-y}

If x does not equal y, and xy does not equal 0, then when x is replaced by \frac{1}{x} and y is replaced by \frac{1}{y} everywhere in the expression above, the resulting expression is equivalent to:

A) -\frac{x+y}{x-y}

B) \frac{x-y}{x+y}

C) \frac{x+y}{x-y}

D) \frac{x^2-y^2}{xy}

E) \frac{y-x}{x+y}
[Reveal] Spoiler: OA
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Re: Variable Inversion [#permalink] New post 02 Nov 2010, 02:24
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Phaser wrote:
\frac{x+y}{x-y}

If x does not equal y, and xy does not equal 0, then when x is replaced by \frac{1}{x} and y is replaced by \frac{1}{y} everywhere in the expression above, the resulting expression is equivalent to:

A) -\frac{x+y}{x-y}

B) \frac{x-y}{x+y}

C) \frac{x+y}{x-y}

D) \frac{x^2-y^2}{xy}

E) \frac{y-x}{x+y}

Please explain your answers if you can. OA will be provided in a few hours.


When we replace we'll get: \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{y-x}=-\frac{x+y}{x-y}.

Answer: A.
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Re: Variable Inversion [#permalink] New post 02 Nov 2010, 02:58
Bunuel wrote:
Phaser wrote:
\frac{x+y}{x-y}

If x does not equal y, and xy does not equal 0, then when x is replaced by \frac{1}{x} and y is replaced by \frac{1}{y} everywhere in the expression above, the resulting expression is equivalent to:

A) -\frac{x+y}{x-y}

B) \frac{x-y}{x+y}

C) \frac{x+y}{x-y}

D) \frac{x^2-y^2}{xy}

E) \frac{y-x}{x+y}

Please explain your answers if you can. OA will be provided in a few hours.


When we replace we'll get: \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{y-x}=-\frac{x+y}{x-y}.

Answer: A.


Yes, that is correct. But can you explain how the denominator ends up as y-x instead of x-y? Because that is what I did when I got this question on the MGMAT CAT and ended up with the wrong answer "A".
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Re: Variable Inversion [#permalink] New post 02 Nov 2010, 03:04
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Phaser wrote:
Bunuel wrote:
Phaser wrote:
\frac{x+y}{x-y}

If x does not equal y, and xy does not equal 0, then when x is replaced by \frac{1}{x} and y is replaced by \frac{1}{y} everywhere in the expression above, the resulting expression is equivalent to:

A) -\frac{x+y}{x-y}

B) \frac{x-y}{x+y}

C) \frac{x+y}{x-y}

D) \frac{x^2-y^2}{xy}

E) \frac{y-x}{x+y}

Please explain your answers if you can. OA will be provided in a few hours.


When we replace we'll get: \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{x+y}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{y-x}=-\frac{x+y}{x-y}.

Answer: A.


Yes, that is correct. But can you explain how the denominator ends up as y-x instead of x-y? Because that is what I did when I got this question on the MGMAT CAT and ended up with the wrong answer "A".


\frac{x+y}{y-x}=\frac{x+y}{-(x-y)}=-\frac{x+y}{x-y}
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COLLECTION OF QUESTIONS:
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Re: Variable Inversion [#permalink] New post 02 Nov 2010, 03:09
Ah, that explains. Thank you very much.
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Re: If x does not equal y... [#permalink] New post 19 Sep 2012, 23:21
egiles wrote:
x + y
--------
x – y


If x does not equal y, and xy does not equal 0, then when x is replaced by 1/x and y is replaced by 1/y everywhere in the expression above, the resulting expression is equivalent to...

a)- (x+y)
-------
x-y

b) x - y
--------
x + y

c) x + y
--------
x - y

d) x^2 - y^2
---------------
xy

e) y - x
-------
x + y


[Reveal] Spoiler:
When I do the algebra, I get d. Here are the steps I take. Where am I going wrong?...

1/x + 1/y (1/x + 1/y)(x - y) 1 - y/x + x/y - 1 x/y - y/x x^2 - y^2
------------- = = = = ----------------
1/x - 1/y xy


Hard to read your answer due to the formatting.

\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x}-\frac{1}{y}}=\frac{\frac{y+x}{xy}}{\frac{y-x}{xy}}=\frac{x+y}{xy}\cdot{\frac{xy}{y-x}}=\frac{x+y}{y-x}=\frac{x+y}{-(x-y)}=-\frac{x+y}{x-y}

Answer A
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Re: If x does not equal y... [#permalink] New post 19 Sep 2012, 23:32
I think the formatting of your solution (inside the spoiler tags) became illegible, but I may be able to guess at the mistake. When we replace 'x' with '1/x', and 'y' with '1/y', we have the following:

\frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x} - \frac{1}{y} }

I think you then rewrote this as \left( \frac{1}{x} + \frac{1}{y} \right) (x - y), and if so, that isn't right. If that's what you did, you were probably trying to apply the rule "when you divide by a fraction, multiply by its reciprocal". There are two problems with trying to do that right away: we aren't dividing by a simple fraction here, so we can't use that rule (in the denominator, we're subtracting one fraction from another), and regardless, x-y is not the reciprocal of (1/x) - (1/y). If those terms were reciprocals of each other, their product would be 1, and if you multiply them out, you'll see their product is not 1.

As a first step, since we have fractions in both our numerator and denominator, I'd just multiply the top and bottom by xy. We then need to multiply by -1 on top and bottom only because of how the question designer chose to write the correct answer choice:

\begin{align}
\frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x} - \frac{1}{y} } &= \left( \frac{ \frac{1}{x} + \frac{1}{y} }{ \frac{1}{x} - \frac{1}{y} } \right) \left( \frac{xy}{xy} \right) \\
&= \frac{y + x}{y - x} \\
&= \left( \frac{x + y}{-x + y} \right) \left( \frac{-1}{-1} \right) \\
&= \frac{-(x+y)}{x-y}
\end{align}

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Re: If x does not equal y, and xy does not equal 0, then when x [#permalink] New post 10 Dec 2012, 02:59
=(\frac{1}{x}+\frac{1}{y})/(\frac{1}{x}-\frac{1}{y})
=(\frac{x+y}{xy})/(\frac{y-x}{xy})
=\frac{x+y}{y-x}
=\frac{{x+y}}{{-(x-y)}}

Answer: A
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Re: If x does not equal y, and xy does not equal 0, then when x   [#permalink] 10 Dec 2012, 02:59
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