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If x # -y , is (x-y)/(x+y) > 1 ? 1) x>0 2) y<0

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If x # -y , is (x-y)/(x+y) > 1 ? 1) x>0 2) y<0 [#permalink] New post 25 Dec 2007, 10:15
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I have a doubt in OG-11 question 139. This belongs to DS.

The question is :-
Q) If x # -y , is (x-y)/(x+y) > 1 ?
1) x>0
2) y<0

Can anybody please explain what does x # -y stand for?
can you provide some examples.
Manager
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 [#permalink] New post 25 Dec 2007, 10:33
I understand X is not to Y but does t mean that when x > 0 then y < 0.
OG assumes the above.
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 [#permalink] New post 25 Dec 2007, 11:01
The only reason X Not equals to Y is mentioned because if X = -Y the function has an indeterminate value because of a divide by 0.

It has no correlation to if X >0 or Y < 0 or the other way around by the fact that X is not equals to -Y.
  [#permalink] 25 Dec 2007, 11:01
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If x # -y , is (x-y)/(x+y) > 1 ? 1) x>0 2) y<0

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