Is x > y^2? (1) x > y + 5 (2) x^2 - y^2 = 0

Hi, there! I'm happy to help with this.

The question: is x > y^2

Statement #1: x > y + 5

This doesn't necessarily tell us anything. If y = 1, and x = 7, then x > y^2, but if y = -6 and x = 0, then x < y^2. But itself, Statement #1 is not sufficient.

Statement #2: x^2 - y^2 = 0

This means that x^2 = y^2, which means that x = ±y. Same absolute value, but both could be positive, both could be negative, or either one could be positive and the other negative. We know that y^2 will be positive, but the x can be positive or negative, so by itself, Statement #2 is insufficient.

Combined
Now, we know that x^2 - y^2 = 0 ---> x = ±y, AND we know that x > y + 5. This leads immediate to a few conclusions
(a) x is positive and y is negative --- that's the only way they could have the same absolute value, but with x bigger than y + 5
(b) x and y must have an absolute value greater than 2.5, so that the different between positive x and negative y is more than 5

So we are comparing a positive number x, greater than 2.5, to the square of the negative number with the same absolute value. Of course, x^2 and y^2 are equal, so the question really boils down to: given that x > 2.5, is x > x^2?

For all x greater than one, the square of x is greater than x. That's because, squaring is multiplying a number by itself, and when you multiply anything by a number greater than one, it gets bigger.

Thus, if x > 2.5, when we square it, it will get bigger. Therefore, x^2 = y^2 > x for all values of x > 2.5.

Thus, combined, the statements are sufficient together. Answer = C

Does that make sense? Please let me know if you have any questions.

Mike