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

Condition (1): (x^2 - y^2 > 1)

Analysis: (x^2 - y^2) can be analyzed as ((x + y)(x - y) > 1).
Consider the following cases:
If (x + y > 0), then (x - y) must be greater than a positive number, i.e. (x > y).
If (x + y < 0), then (x - y) must be a negative number with an absolute value greater than 1, i.e. (x < y).
Conclusion: Condition (1) is not sufficient to conclude that (x + y > 0).
Condition (2): (\frac{x}{y} + 1 > 0)

Analysis: (\frac{x}{y} + 1 > 0) is equivalent to (\frac{x + y}{y} > 0)
Consider the following cases:
If (y > 0), let (\frac{x + y}{y} > 0) then (x + y > 0).
If (y < 0), let (\frac{x + y}{y} > 0) then (x + y < 0).
Conclusion: Condition (2) is not sufficient to conclude (x + y > 0).
Combining (1) and (2):

Assume (y > 0). From (2) we have (x + y > 0). This is consistent with some cases of (1) (for example: (x > 1, y = 0)).
Assume (y < 0). From (2) we have (x + y < 0). This also holds for some cases of (1) (e.g., (x < -1, y = 0)).

Even if we combine both conditions (1) and (2), we still cannot determine with certainty that (x + y > 0).

So the answer is (E) - Both conditions are not sufficient.