x != y, Is x-y/x+y > 1?

[zchampz]

x != y

Is x-y/x+y > 1?

1) x > 0
2) y < 0

If we plug-in the number it is clear that the answer is E. But is there any conceptual approach to solve this. I tried to do the following which results in the solution as B (which is incorrect)

x-y/x+y > 1?

If we simplify...
=> x-y > x+y
=> x-x > y+y
=> 0>2y
=> y < 0
So, can we rephrase the question as "is y < 0?"

Thanks,
Champ

[esledge]

x != y

Can you clarify your notation? I'm reading this as x! = x factorial = y. That would imply y is positive, so (2) conflicts and I know this is the wrong interpretation.

Do you mean "x does not equal y"? Or was it originally "x does not equal -y"? (so the denominator of the expression can't be zero)?

x-y/x+y > 1?

If we simplify...
=> x-y > x+y
=> x-x > y+y
=> 0>2y
=> y < 0
So, can we rephrase the question as "is y < 0?"

In any case, your error occured here. You assumed that (x+y) was positive, so when you multiplied both sides of the inequality by that denominator, you didn't have to flip the sign.

If (x+y) were negative, the rephrase would be:
x-y/x+y > 1?
=> x-y < x+y? FLIPPED SIGN HERE
=> x-x < y+y?
=> 0<2y?
=> y > 0?

So without info on x and y, we don't even know whether to ask "Is y > 0?" or "Is y < 0?"

The answer is E because x > 0 and y < 0 means that x+y = pos + neg, which could be either pos or neg depending on the (unknown) relative size of x and y.

[zchampz]

I meant x is not equal to y (x != y )...

Thanks...as you said I assumed x+y as positive.

[mschwrtz]

We're glad that Emily was able to help.

[gurvindersingh2004]

If we combine 1 and 2
and plug in numbers x=5
y=-3
then x-y=8
x+y = 2
x-y/x+y = 4 . Therefore C why E ?

I got it thanks

[RonPurewal]

If we combine 1 and 2
and plug in numbers x=5
y=-3
then x-y=8
x+y = 2
x-y/x+y = 4 . Therefore C why E ?

that's "yes" to the question.
now try plugging in x = 3 and y = -5. you'll get a "no" to the question. so, (e).