If x does not equal y, Is x-y/x+y > 1?

(1) x >0

(2) y<0

I am not looking for the answer per say, but I was having trouble simplifying.

Can it be done this way:

x/ x+y - y/x+y then:

x's cancel and y's cancel so you get

1/y - 1/x >1 or

1/y-x >1

Thank you.

dk,

Fractions cannot be reduced in the manner you describe when dealing with the operations of addition and sutraction. Try substituting numbers for the x and y and you will quickly realize this. For example, 2/(2+3) equals 2/5 rather than 1/3, the answer that would result if the 2's were simply eliminated.

My approach to this particular problem was one of plugging in numbers. Beginning with statement (1) and our AD/BCE grid, x>0 but we have no idea about the value of y. Eliminate AD from the AD/BCE grid.

Statement (2) tells us that y<0, but we have no idea about the value of x. Eliminate B from the AD/BCE grid.

Together we know that x>0 and y<0, but we don't know about the absolute value of either number. For example, if x=2 and y=-1, the value of (x-y)/(x+y) = 3. However, if x=2 and y=-3, the value of (x-y)(x+y) = 5/-1, or -5. Since 3>1 but
-5<1, the correct answer is E.

so is it incorrect to rephrase this statement to:

Is (x-y) > (x+y) ?

If this rephrasal is appropriate....is there a value for x that one could find that would make (x-y) less than (x+y) when y is negative?

(I) If y is negative, and x positive then x-y will be greater than x+y, since you're subtracting from a negative number - and by definition adding
(II) If y is negative, and x negative, then x-y will be greater than x+y, because of the same principle as above...

I.e. I used this rephrasal and got B.

I am clearly doing something incorrectly here - is my rephrasal off-base? Or are my two postulations incorrect - If so, can you provide counter-examples?
'
Thanks.

It is incomplete (not incorrect exactly) to rephrase the question as (x-y) > (x+y) because we don't know whether (x+y) is positive or negative.

Starting with:
x-y/x+y > 1
In order to move (x+y) to the right hand side of the equation, you have to multiply. When you multiply or divide an inequality by a negative, you have to switch the sign. Because we don't know whether (x+y) is positive or negative, we have to test BOTH possibilities:

If x+y is positive, then x-y > x+y
If x+y is negative, then x-y < x+y

You can keep going this way, but this suddenly makes the problem a lot more complicated, which is why Dan tried real numbers with this one.