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Bunuel
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\(\frac{x+y}{x-y} > 0\)

Multiplying numerator and denominator by x-y, we get,

\(x^2-y^2 >0\) i.e. \(x^2>y^2\) and hence |x| > |y|

1. x>y is clearly not sufficient
2|x|>|y| , this what our deduced expressions ask for!

Hence, B
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Bunuel
If x ≠ y and x and y are non-zero integers, is (x+y)/(x−y) > 0?

(1) x > y
(2) |x| > |y|

(1) x > y

If x=2, y=1 then (x+y)/(x−y) = 3/1 = 3 > 0 --> Yes.
If x=1, y=-1 then (x+y)/(x−y) = 0/2 = 0 --> No.
--> Not sufficient.

(2) |x| > |y| --> \(x^2 > y^2\) --> \(x^2-y^2 > 0\) --> \((x+y)(x-y) > 0\) --> \((x+y)\)/\((x−y)\) \(> 0\) --> Yes --> Sufficient.

Answer B.
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