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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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02 Nov 2010, 01:48
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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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02 Nov 2010, 03:24
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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) $$\frac{y-x}{x+y}$$

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

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

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

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

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

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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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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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

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

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

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20 Sep 2012, 00:32
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Expert's post
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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10 Dec 2012, 03:59
$$=(\frac{1}{x}+\frac{1}{y})/(\frac{1}{x}-\frac{1}{y})$$
$$=(\frac{x+y}{xy})/(\frac{y-x}{xy})$$
$$=\frac{x+y}{y-x}$$
$$=\frac{{x+y}}{{-(x-y)}}$$

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

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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 $$\frac{x}{y}$$

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

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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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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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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)

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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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