If y = (x^2 - y^2 ) / x-y, then what is the value of y?

1. x + y = 3
2. x - y = 2

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

Y=(X^2-Y^2)/X-Y
= (X-Y) (X+Y)/X-Y
= X+Y

Y=X+Y WHICH MEANS X=0

We can rephrase the question as Is X=0?

AD
BCE Grid

A) Talk about Y which is not needed INSUFF
B) Talk about Y which is not needed INSUFF
Combining A & B we get

x+y=3
x-y=2

Solving for X we get x=5/2

From this we can say C is sufficient to answer the question.

Hope it helps.

you need to post a source for this problem. please do so in the next couple of days, or else we'll have to delete the thread.

thank you.

Ron, the source of the problem is above.

Thank you

are you sure you've posted this question correctly?

as another poster said, you can rephrase the original statement to y = x + y, which means that x = 0. so, we have: 'if x = 0, then what is the value of y?'
technically, either (1) or (2) is sufficient, because (1) says 0 + y = 3, and (2) says 0 - y = 2. but that's where we run into problems: we have 2 statements that contradict each other, a situation that's strictly not allowed on the real test.

worse yet, if you take the two statements together, you can solve them as simultaneous equations, yielding x = 2.5 and y = 0.5. but then (x^2-y^2)/(x-y) is 6/2 = 3, which definitely doesn't equal y.

something is wrong in problemville.

but wait - you didn't include parentheses around x-y in the original. are you saying it's actually the fraction (x^2 - y^2) divided by JUST x, and then that fraction minus y? because that would be a whole different ballgame. if so, just post to that effect.