Bunuel wrote:

Is |x + y| > |x - y| ?

Is \(|x+y| > |x-y|\)? --> square it: is \(x^2+2xy+y^2>x^2-2xy+y^2\) --> is \(xy>0\)?

(1) |x| > |y|. We cannot get from this whether x and y have the same sign. Not sufficient.

(2) |x - y| < |x| --> square again: \(x^2-2xy+y^2<x^2\) --> \(xy>\frac{y^2}{2}\) --> since \(y^2\geq{0}\) (the square of any number is more than or equal to 0), then we have that \(xy>\frac{y^2}{2}\geq{0}\). Sufficient.

Answer: B.

Hope it's clear.

Sorry Bunuel,

but i don't understand the concept behind the solution...

I understand that an absolute value is always positive. But how can you just square it¿? It supposed to be the root square of the square, isn't it¿??