LamaSalah wrote:

Is x < y ?

(1) x^2 - y^2 < 0

(2) -x - y < 0

Statement One Alone:x^2 - y^2 < 0

We see that x^2 > y^2; however we still do not have sufficient information to determine whether x > y.

For instance, if x = -2 and y = 1, then x is less than y. However, if x = 2 and y = 1, then x is greater than y. Statement one alone is not sufficient to answer the question.

Statement Two Alone: -x - y < 0

We can multiply the inequality in statement two by -1, remembering to reverse the inequality sign, and obtain:

x + y > 0, or equivalently, 0 < x + y. However, we still cannot determine whether x > y.

For instance if x = 2 and y = 1, then x is greater than y; however if x = 1 and y = 2, then x is less than y. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:Notice that x^2 - y^2 in statement one is a difference of two squares: x^2 - y^2 = (x + y)(x - y). So x^2 - y^2 < 0 means (x + y)(x - y) < 0. From statement two, we know that x + y > 0. That means x - y < 0, since x + y is positive. Therefore, x - y must be negative in order for their product (x + y)(x - y) to be negative.

Since x - y < 0, we have that x < y.

Answer: C

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