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# Is 2x-3y < x^2? (1) 2x-3y = -2 (2) x>2 and y>0 can

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VP
Joined: 21 Jul 2006
Posts: 1447
Is 2x-3y < x^2? (1) 2x-3y = -2 (2) x>2 and y>0 can  [#permalink]

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02 Aug 2008, 07:47
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Is 2x-3y 2 and y>0

can someone show me how to draw this on the xy-plane? i'm beginning to appreciate the usage of this approach because it makes it a HELL LOT easier to manage such problems rather than experimenting with different numbers and then drive your mind nuts!

I'll really appreciate it
thanks

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Director
Joined: 27 May 2008
Posts: 530

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02 Aug 2008, 09:35
tarek99 wrote:
Is 2x-3y < x^2?

(1) 2x-3y = -2

(2) x>2 and y>0

can someone show me how to draw this on the xy-plane? i'm beginning to appreciate the usage of this approach because it makes it a HELL LOT easier to manage such problems rather than experimenting with different numbers and then drive your mind nuts!

I'll really appreciate it
thanks

I dont think that drawing a curve is good approach here.... because we have a square function involved... let me know if you like below approach... otherwise i can draw on xy plane also...

Statement 1 : obviously suff... it tells 2x-3y = -2
we know x^2 will always be positive is if 2x-3y = -2 (a negative number) .... 2x-3y will always be less than x^2

given inequlity 2x-3y-x^2<0
2x-x^2-3y<0
(2x-x^2) - 3y < 0

Statement 2 :
for x > 2 , 2x-x^2 will always be negative
for y > 0. -3y will always be negative
so the whole expression will always be negative... Suff

Director
Joined: 10 Sep 2007
Posts: 909

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02 Aug 2008, 10:01
durgesh79 wrote:
I dont think that drawing a curve is good approach here.... because we have a square function involved... let me know if you like below approach... otherwise i can draw on xy plane also...

Statement 1 : obviously suff... it tells 2x-3y = -2
we know x^2 will always be positive is if 2x-3y = -2 (a negative number) .... 2x-3y will always be less than x^2

given inequlity 2x-3y-x^2<0
2x-x^2-3y<0
(2x-x^2) - 3y < 0

Statement 2 :
for x > 2 , 2x-x^2 will always be negative
for y > 0. -3y will always be negative
so the whole expression will always be negative... Suff

Durgesh question is asking is 2x-3y>x^2
even if we can say -3y is -ve and 2x-x^2 is -ve, there is no guarantee that -3y > 2x-x^2.
Director
Joined: 27 May 2008
Posts: 530

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02 Aug 2008, 10:10
[quote="abhijit_sen"]
Durgesh question is asking is 2x-3y>x^2
even if we can say -3y is -ve and 2x-x^2 is -ve, there is no guarantee that -3y > 2x-x^2.[/quote]

i think you missed the inequality sign there

question is 2x-3y<x^2
-3y < x^2-2x
or 3y > 2x-x^2

we have already proved that 2x-x^2 is always negative and 3y is always positive ...

a postive number > a negative number ... Suff
Director
Joined: 10 Sep 2007
Posts: 909

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02 Aug 2008, 10:37
durgesh79 wrote:
abhijit_sen wrote:
Durgesh question is asking is 2x-3y>x^2
even if we can say -3y is -ve and 2x-x^2 is -ve, there is no guarantee that -3y > 2x-x^2.[/quote]

i think you missed the inequality sign there

question is 2x-3y 2x-x^2

we have already proved that 2x-x^2 is always negative and 3y is always positive ...

a postive number > a negative number ... Suff
[/quote]
Thanks for pointing me out.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

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

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: DS: xy-plane &nbs [#permalink] 02 Aug 2008, 10:37
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