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1) is insufficient. if x = 2, and y = -2, X = |y|; if x = 3 and y = -2, x>|y|
2) is sufficient. If y <0 then x> -(y) because y <0 so x >|y|. An example. y = -13. This means that x > -(-13); x > 13, so x >|y|.

Last edited by Makky07 on 05 Nov 2003, 10:19, edited 1 time in total.

I misstated my reply to ndidi--
In ndidi's example, s/he uses the plug method, and he s/he states:
"if x = 2, and y = -2"
but that violates the parameters of the problem, which states that "x + y > 0". If you plug those numbers, you get 2+(-2)>0, which is untrue.