abcdefg wrote:Is x + y > 0 ?

(1) x - y > 0

(2) x^2 - y^2 > 0

[spoiler] Statement 2 I could easily answer by x^2 - y^2 => (x + y) (x - y).
So (x + y) ( x - y) > 0 which means either (x + y) is pos and (x - y) is pos, or (x + y) is neg AND (x - y) is neg. so insuff.

The original statement can be rephrased as "is x > -y?" and statement 1 can be rephrased as "x > y". But these 2 statements can't both be true so statement 1 is sufficient to answer the question. [/spoiler].

Am I right? Thanks.

Statement 1 only tells you that x is greater than y. But x can be -2 and y -3 so x+y=-5 or x can be 3 and y 2 and x+y = 5 insufficient.
Statement 2 says that x^2 is greater than y^2 but again X could be -3 and y -2 or x could be 3 and y 2=insufficient

C) if x is greater than y and if x^2 is greater than y^2 then both should be positive and their sum greater than cero

ABC,

you are on the right track, but making it more difficult than necessary.

Is x + y > 0?

A.) x - y > 0

Obviously insufficient. X = 4, Y = -5

B.) x^2 - y^2 > 0
Insufficient. X = -5, Y = 4

Both Together:
Factor x^2 - y^2 = (x+y)(x-y) > 0

Now since we know that x - y > 0, x + y MUST be > 0 because
positive*positive = positive

If x + y was negative, then (x-y)(x+y) would be negative, which violates the second constraint.

Takes a while to write out but problems like this can be done completely in your head to save time on Math section.