Hi

Here is a question that stumps me:
If x not equal to -y, is (x-y)/(x+y)>1 ?

1) x>0
2) y<0

GMAT's answer is (E)

Here is how I solved it:

rephrase: is x-y > x+y ? --> is 0 > 2y --> is y < 0?

x,y cannot be zero cos x is not equal to -y should rule this out.

(2) seems to readily answer the requirement . But one corner case requires that x not equal y (else the inequality fails). Taking (1) in addition to (2), this condition is eliminated. So (C) should be the answer.

I guess I am missing something!!! Any thoughts?

Thanks very much.

Here is my two cents,
when you move X+Y over to the right side, you can't pre-assume that X+Y > 0. Also, when you phase that way, you need to divide X+Y instead of multiple. You get X-Y>1/(X+Y)

When I solved this, I phased into X^2-Y^2>1. Since you don't know if X^2>Y^2, you don't know whether their difference is larger than 1 either. So in order to solve this, we should know the relationship between X and Y, either X>Y or X<Y. Since none of them are provided in the two conditions, the answer would be E.

[editor: this is an incorrect solution; there is no way the original prompt inequality can be rephrased as x^2 - y^2 > 1.
as proof that this is not a valid rephrase, consider x = 3 and y = 2, which satisfy the "rephrase" but not the original statement.]

Hi elaine1920

I agree with your solution and explanation. I was too saturated yesterday to see something as simple as that!

Thanks.

Great. I know!! we all have those moments when come to GMAT. :)

yep.
if you have an INEQUALITY, then you cannot MULTIPLY OR DIVIDE by a quantity unless you know its sign.
the issue is that the "<"/">" would flip around upon division or multiplication by a negative number.

with equations, of course, there is no such problem.

you can also PLUG IN NUMBERS, of various signs and sizes, to test the sufficiency of these statements.

if you plug in x = 1, y = -2, and then x = 2, y = -1, you'll notice that you get opposite answers (the first "no", the second "yes").
since these plug-ins satisfy both statements, you've got proof that the answer is (e).

Ron, Elaine: I don't see how multiplying and dividing (X+Y) makes a difference since multiplying by (X+Y) on both sides is equivalent to dividing by 1/(X+Y) to cancel out (X+Y) in the denominator. We still have to keep track of the signs. I am not sure how you got your solution in the bolded, underlined portion.

Ron, Elaine: Can you please clarify?

Here are my two cents in attempting to solve this problem:
(X-Y)/(X+Y) > 1
Two cases exist: X+Y>0 OR X+Y<0
Case 1: X+Y>0
X-Y > X+Y
0>2Y
0>Y

Case 2: X+Y<0
X-Y < X+Y
0 < 2Y
0 < Y

You know nothing about X so you cannot assume anything from the simplified inequalities given above.

Start with (2) since Y < 0 is given. This implies Case 1 in which X+Y>0
(X-Y) / (POSITIVE) > 1
X - Y > 1

You know nothing about X so it could be true or false.
So Eliminate BD, leaving ACE.

(1) says X>0
We have to test both cases since our simplified inequality does not mention X in either case.
Case 1: X+Y>0
X - Y / (POSITIVE) > 1

We can pick different Ys to falsify or confirm the above inequality.

Case 2: X+Y<0
X - Y / (NEGATIVE) > 1

We can pick different Ys to falsify or confirm the above inequality.

You have no info about X so you don't know whether (X+Y) / (X-Y) is positive or negative. So A is eliminated, leaving C or E.

Combining X>0 and Y<0 yields Case (2) results, which we just saw.

Case 2: X+Y<0
X - Y / (NEGATIVE) > 1

We can pick different Ys to falsify or confirm the above inequality.

So C is out, hence producing E as the final answer.

Ideally I would PICK NUMBERS for this problem for SPEED since algebra is too complicated.
[editor: yes. that is what i would do.]

Ron, Elaine: What do you think?

here's the short version of how i would solve something horrible like this.

i would notice that we're dealing with a fractional inequality, which, worse yet, CAN'T be multiplied by the common denominator (since we don't know the sign of that denominator).

therefore, i would pick numbers.

i would just have to be careful to pick APPROPRIATE numbers. i.e., this problem contains sums and differences, as well as sign considerations, so i would pick:
* POSITIVES AND NEGATIVES (as allowed by the statements)
* DIFFERENT RELATIVE SIZES (i.e., "bigger" and "smaller" numbers) -- this is important because of addition and subtraction.

so, for statement (1), i would pick:
1, 2
2, 1
1, -2
2, -1
1, 0

for statement (2), i would pick:
1, -2
2, -1
-1, -2
-2, -1
0, -1

for "together" i would look at the two common elements, which are (1, -2) and (2, -1).

note that this is a lot of plug-ins, but you don't wind up trying them all - you STOP as soon as you get "insufficient".