Is x + y > 0 ? (1) x - y > 0 (2) x^2 - y^2 > 0

Statement 1:

x>y

If both x and y are positive then YES

If both x and y are negative then NO

Insufficient

Statement 2:

(x+y)(x-y)>0

If x-y>0, then YES

If x-y<0, then NO

Insufficient

Statement 1&2:

x-y>0,

then x+y>0

Sufficient

C


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Maybe a better approach is theoretically.

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I am not sure your approach is correct.

You could have a gigantic positive x e a tiny negative y and still hold the conditions true.

Is it in Statement 1 or 2 or 1&2 taken together?

1)x-y>0 is insufficient as it will show x>y but if x and y are negative then x+y can't be greater than 0. (Insufficient)
2)(x+y)(x-y)>0 is insufficient as we can't say whether x+y>0 or not

Combining we can say x+y>0.. So Answer is C

The answer is C, but the approach is a bit more complex. Otherwise the right choice is by luck.

Here is a conceptual solution to the problem

We can rephrase the question by subtracting y from both sides of the inequality: Is \(x > -y\) ?
 
(1) INSUFFICIENT: If we add y to both sides, we see that x is greater than y. We can use numbers here to show that this does not necessarily mean that \(x > -y\). If \(x = 4\) and \(y = 3\), then it is true that \(x\) is also greater than \(-y\). However if \(x = 4\) and \(y = -5\), \(x\) is greater than \(y\) but it is NOT greater than \(-y\). 
 
(2) INSUFFICIENT:  If we factor this inequality, we come up \((x + y)(x – y) > 0\). 
For the product of \((x + y)\) and \((x – y)\) to be greater than zero, the must have the same sign, i.e. both negative or both positive. 
This does not help settle the issue of the sign of \(x + y\). 

(1) AND (2) SUFFICIENT: From statement 2 we know that \((x + y)\) and \((x – y)\) must have the same sign, and from statement 1 we know that \((x – y)\) is positive, so it follows that \((x + y)\) must be positive as well.
 
The correct answer is C.

It's pretty clear that individual statements aren't sufficient
We can take values to check
S1: x=3,y=1 => x+y>0
x=3,y=-4 => x+y<0

S2: x=3,y=1 => x+y>0
x=-4,y=-3 => x+y<0

Combining: divide S2 by S1:
x+y > 0 - sufficient

Ans C