hard question from good source

[tanviet]

Pls, tell me how to solve quick

is xy>0?

1, x-y>2

2, x-2y<-6

[thephoenix]

imo e

[Brent@GMATPrepNow]

Pls, tell me how to solve quick

is xy>0?
1, x-y>2
2, x-2y<-6

This question is testing your knowledge of how positive and negative numbers behave.
First recognize that if xy>0 then x and y must be either both positive or both negative.
So, the question can be rewritten as "Do x and y have the same sign?" (both pos or both neg).

1) We can rewrite this as y+2 < x
Aside: I always rewrite my inequalities so that the greater value is on the right. Then the values are listed in ascendiing order.
Notice that, since we also know that y < y+2, we can add this to the inequality above to get: y < y+2 < x
From here we can see that x is greater than y.
This is not enough information to answer our question.

2) We can rewrite this as x < 2y-6
We can see that this isn't enough to answer our question (y could be positive and x could be negative, or they both could be positive)

(1&2)
We can combine both inequalities to get: y < y+2 < x < 2y-6
From this, we can conclude that y < 2y-6
From here we know that y must be positive. If y were negative, it would be impossible for 2y-6 to be greater than y.
Proof: Assume that y is negative and notice that this leads us to a contradiction. First, if y is negative, then 2y<y. And since 2y-6 < 2y, we would have to conclude that 2y-6 < 2y < y. In other words, it would have to be true that 2y-6 < y.
Our inequality y < y+2 < x < 2y-6 tells us the opposite is true.
So, y must be positive.
From statement (1), we know that y<x, so x must also be positive.
If x and y are both positive, we can answer our question.

[money9111]

Brent you make deriving the answer look so simple! I guess it is once you put it like that haha... I'm hoping to pick up the skills like there, because as I was reading your explanation I wasn't nervous about understand it. I'm beginning to find myself getting nervous about this exam when I don't understand the explanations to problems.

[Stuart@KaplanGMAT]

[
From this, we can conclude that y < 2y-6
From here we know that y must be positive. If y were negative, it would be impossible for 2y-6 to be greater than y.
Proof: Assume that y is negative and notice that this leads us to a contradiction. First, if y is negative, then 2y<y. And since 2y-6 < 2y, we would have to conclude that 2y-6 < 2y < y. In other words, it would have to be true that 2y-6 < y.
Our inequality y < y+2 < x < 2y-6 tells us the opposite is true.
So, y must be positive.

Alternatively:

y < 2y - 6

(subtract y from both sides)

0 < y - 6

(add 6 to both sides)

6 < y

therefore y is positive

[money9111]

even easier!

[sreak1089]

x > y + 2 --> (1)
x < 2y - 6 --> (2)

Multiply 2nd equation by -1, we get:

-x > 6 - 2y --> (3)
x > y + 2 --> (4)

Adding (3) & (4), we get:

0 > 8 - y
y > 8
x > y +2 > 10

Hence, xy > 0 . Ans is C.