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

Question

\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}

Bunuel · Accepted Answer

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.

Phaser · Answer

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".

Bunuel · Answer

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

Phaser · Answer

Ah, that explains. Thank you very much.

EvaJager · Answer

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

IanStewart · Answer

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

mbaiseasy · Answer

=(\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

PareshGmat · Answer

\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

Donnie84 · Answer

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.

shashankism · Answer

(x+y)/(x-y) 
~ (1/x+1/y)/(1/x-1/y)
=  / 
= (x+y)/(y-x)
= -(x+y)/(x-y)

Answer A

imranmah · Answer

Would you be able to explain why the "-" only affects the denominator and not the numerator? I was able to get the answer up until that point I'm just confused as to why the signs change in the denominator but not the numerator. Thank you for your help.

Bunuel · Answer

We are factoring out -1 from the denominator:

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

Nr967 · Answer

remember: x-y = -(y-x)

bumpbot · Answer

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

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