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# Is xy>0

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Retired Moderator
Joined: 18 Jul 2008
Posts: 898

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02 Dec 2008, 09:17
Is xy>0

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

GUys, what's the easiest way to do this? I plugged in numbers and it started to get confusing, and blew past the 2 minutes.

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Manager
Joined: 02 Nov 2008
Posts: 56

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02 Dec 2008, 09:26
I get c

I subtract eq 2 from 1 and
Manager
Joined: 14 Oct 2008
Posts: 158

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02 Dec 2008, 09:38
IMO C.

From 1st equation
x > y-2 ( not sufficient )

From 2nd equation
x < 2y-6 ( not sufficient)

Combining :
(y-2) < x < (2y-6)

hence y-2 < 2y-6
y > 4

Substituting the value in 1st equations we see that x.y is always positive

we can also check by substituting values in equation (y-2) < x < (2y-6)

Hence C.
Retired Moderator
Joined: 18 Jul 2008
Posts: 898

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02 Dec 2008, 11:35
But to solve for A and also B independently, you guys plugged in numbers?
VP
Joined: 17 Jun 2008
Posts: 1477

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02 Dec 2008, 12:27
Alternatively, draw the graph of the two inequalities and find out the common area. The common area will be in quadrant A and hence xy > 0 only when both the inequalities are considered.
Manager
Joined: 14 Oct 2008
Posts: 158

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02 Dec 2008, 17:00
But to solve for A and also B independently, you guys plugged in numbers?

Yes I had plugged in numbers individually in both statements .....
Senior Manager
Joined: 05 Jun 2008
Posts: 289

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03 Dec 2008, 06:46

(1) x - y > - 2
(2) x - 2y - 2 + 6
y> 4 after puting this value in 1
X>2, which means XY>8

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Is xy>0   [#permalink] 03 Dec 2008, 06:46
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# Is xy>0

Moderator: chetan2u

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