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if x+ and y- (1,-2) =x-y/x+y = -3 (2,-1) =x-y/x+y = +3

if x- and y+ (-1,2) =x-y/x+y = -3 (-2,1) =x-y/x+y = 3

A) says x>0 does not tell us anything about Y insufficient B) says y<0 does not tell us anything about X insufficient

togather x+ and y- scnerio 3, it can be positve or negative so insufficient!

Hence E

Thats great explanation...thanks Chirag and hass_mba
How long did it take you guys to do it?
I couldn't do this on the test due to time constraints and got it wrong...
_________________

if you reorganize the question it becomes x - y < x + y ==> 0 < 2y ==> 0 < y.... Why is this logic not possible?

Given: \(\frac{x-y}{x+y}>1\). When you are then writing \(x-y>x+y\), you are actually multiplying both sides of inequality by \(x+y\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x+y>0\) you should write \(x-y>x+y\) BUT if \(x+y<0\), you should write \(x-y<x+y\) (flip the sign when multiplying by negative expression).

COMPLETE SOLUTION:

If \(x\neq{-y}\) is \(\frac{x-y}{x+y}>1\)?

Is \(\frac{x-y}{x+y}>1\)? --> Is \(0>1-\frac{x-y}{x+y}\)? --> Is \(0>\frac{x+y-x+y}{x+y}\)? --> Is \(0>\frac{2y}{x+y}\)?

(1) \(x>0\) --> Not sufficient.

(2) \(y<0\) --> Not sufficient.

(1)+(2) \(x>0\) and \(y<0\) --> numerator (y) is negative, but we can not say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.