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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]

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

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]

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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]

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New post 02 Nov 2010, 03:58
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]

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New post 02 Nov 2010, 04:04
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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:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Variable Inversion [#permalink]

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New post 02 Nov 2010, 04:09
Ah, that explains. Thank you very much.

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Re: If x does not equal y... [#permalink]

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New post 20 Sep 2012, 00:21
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]

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New post 20 Sep 2012, 00:32
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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]

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New post 10 Dec 2012, 03:59
\(=(\frac{1}{x}+\frac{1}{y})/(\frac{1}{x}-\frac{1}{y})\)
\(=(\frac{x+y}{xy})/(\frac{y-x}{xy})\)
\(=\frac{x+y}{y-x}\)
\(=\frac{{x+y}}{{-(x-y)}}\)

Answer: A
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If x does not equal y, and xy does not equal 0, then when x [#permalink]

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New post 30 Oct 2014, 01:32
\(\frac{x+y}{x-y}\)

Divide numerator & denominator by x

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

Replacing x with \(\frac{1}{x}\) & y with \(\frac{1}{y}\) means replacing \(\frac{y}{x}\) with \(\frac{x}{y}\)

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

Answer = A
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Re: If x does not equal y, and xy does not equal 0, then when x [#permalink]

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New post 14 Aug 2016, 18:39
Take x = 2 and y = 1.

Value of original expression = (2 + 1)/(2 - 1) = 3

Value of original expression as per the suggested transformation = (0.5 + 1)/(0.5 - 1) = -3. This is value of original expression * -1.

Only choice (A) fits.

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Re: If x does not equal y, and xy does not equal 0, then when x [#permalink]

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New post 18 Oct 2017, 11:38
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) \(\frac{y-x}{x+y}\)



(x+y)/(x-y)
~ (1/x+1/y)/(1/x-1/y)
= [(x+y)/xy]/[(y-x)/xy]
= (x+y)/(y-x)
= -(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] 18 Oct 2017, 11:38
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