M15-11

Why do you assume that x and y are integers? x^2 - y^2 = 9 has infinitely many solutions for x and y.

Even if you consider only integers, which is not right, you'll have more solutions:
x = ±5 and y = -4;
x = ±3 and y = 0;
x = ±5 and y = 4.

With many of the questions related to simplifying the initial question as - Is xy<0?-, let me give two ways to solve it, thereby clearing the doubt around xy<0.

Method I

Let us not simplify the initial question. But why?
Because the initial question and the statements are in expressions x-y and x+y or \(x^2-y^2\), which is a product of those two terms.


(1) \(x^2 - y^2 = 9........(x-y)(x+y)=9=1*9\)
So x-y and x+y can have values 1 and 9 respectively or 9 and 1 respectively.
The answer will be different in both the cases.

(2) \(x - y = 2\)
Nothing about x+y

Combined
x-y=2, so (x-y)(x+y)=2(x+y)=9.
So x+y=4.5
As x-y and x+y are positive, the question is: Is 2>4.5?
The answer is NO.
Sufficient



Method II

\(|x-y|>|x+y|\) will be true only when x and y have opposite sign or xy<0.

As seen above we cannot solve the inequality by any of the statements individually.

Combined
x-y=2 and x+y=4.5
Add both the equations
2x=6.5 or x=3.25 and y = 3.25-2=1.25

So, the answer for xy<0 is no as xy=3.25*1.25
Sufficient

C

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.