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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... _________________

Re: If x not equal to -y is (x-y)/(x+y) >1? (1) x > 0 (2) [#permalink]
26 Apr 2012, 20:23

1

This post received KUDOS

Expert's post

calreg11 wrote:

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.

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