Is 1/(x - y) < y - x (1) 1/x < 1/y (2) 2x = 3y

Is 1 / (x-y) < y - x?

1. 1 / x < 1 / y
2. 2x = 3y

Hi Buneul,

Can you please explain this part?
if both are negative cross multiply and flip the sigh twice to get y<x again)

First of all if it were realistic GMAT question it would most likely state that \(xy\neq{0}\) and \(x\neq{y}\) (as x, y and x-y are in denominators).

Is \(\frac{1}{x-y}<y-x\)? --> is \(\frac{1}{x-y}+x-y<0\) --> is \(\frac{1+(x-y)^2}{x-y}<0\)? as the nominator (\(1+(x-y)^2\)) is always positive then the question basically becomes whether denominator (\(x-y\)) is negative --> is \(x-y<0\)? or is \(x<y\)?

(1) \(\frac{1}{x}<\frac{1}{y}\) --> if both unknowns are positive or both unknowns are negative then \(y<x\) (if both are positive cross multiply to get \(y<x\) and if both are negative cross multiply and flip the sigh twice to get \(y<x\) again) and the answer will be NO but if \(x<0<y\) given inequality also holds true and in this case the answer will be YES (if \(x\) is any negative number and \(y\) is any positive number then \(\frac{1}{x}=negative<positive=\frac{1}{y}\)). Not sufficient.

(2) \(2x=3y\) --> \(x\) and \(y\) have the same sign, next: \(\frac{x}{y}=\frac{3}{2}\): if both \(x\) and \(y\) are positive (for example 3 and 2 respectively) then \(0<y<x\) and the answer will be NO but if both \(x\) and \(y\) are negative (for example -3 and -2 respectively) then \(x<y<0\) and the answer will be NO. Not sufficient.

(1)+(2) As from (2) \(x\) and \(y\) have the same sign then from (1) \(y<x\) and the answer to the question is NO. Sufficient.

Answer: C.

First of all if it were realistic GMAT question it would most likely state that \(xy\neq{0}\) and \(x\neq{y}\) (as x, y and x-y are in denominators).

Is \(\frac{1}{x-y}<y-x\)? --> is \(\frac{1}{x-y}+x-y<0\) --> is \(\frac{1+(x-y)^2}{x-y}<0\)? as the nominator (\(1+(x-y)^2\)) is always positive then the question basically becomes whether denominator (\(x-y\)) is negative --> is \(x-y<0\)? or is \(x<y\)?

(1) \(\frac{1}{x}<\frac{1}{y}\) --> if both unknowns are positive or both unknowns are negative then \(y<x\) (if both are positive cross multiply to get \(y<x\) and if both are negative cross multiply and flip the sigh twice to get \(y<x\) again) and the answer will be NO but if \(x<0<y\) given inequality also holds true and in this case the answer will be YES (if \(x\) is any negative number and \(y\) is any positive number then \(\frac{1}{x}=negative<positive=\frac{1}{y}\)). Not sufficient.

(2) \(2x=3y\) --> \(x\) and \(y\) have the same sign, next: \(\frac{x}{y}=\frac{3}{2}\): if both \(x\) and \(y\) are positive (for example 3 and 2 respectively) then \(0<y<x\) and the answer will be NO but if both \(x\) and \(y\) are negative (for example -3 and -2 respectively) then \(x<y<0\) and the answer will be NO. Not sufficient.

(1)+(2) As from (2) \(x\) and \(y\) have the same sign then from (1) \(y<x\) and the answer to the question is NO. Sufficient.

Answer: C.

Hi Bunuel.
When taking both statements together, I substituted x=3/2 and y=2/3, or x=-3/2 and y=-2/3
When taking into account statement 1, I noticed that we can't have the negative value for the numbers, as then statement 1 won't hold true.
Is this true? Am I missing something?

Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

So both gives the answer.

Can you help.