Is 2x-3y<x^2 ? 1) 2x-3y = -2 2) x>2 and y>0 : DS Archive
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# Is 2x-3y<x^2 ? 1) 2x-3y = -2 2) x>2 and y>0

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Is 2x-3y<x^2 ? 1) 2x-3y = -2 2) x>2 and y>0 [#permalink]

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23 Sep 2005, 16:24
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Is 2x-3y<x^2 ?

1) 2x-3y = -2
2) x>2 and y>0
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-Vikram

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Re: DS: Inequality [#permalink]

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23 Sep 2005, 17:42
Quote:
Is 2x-3y<x^2?
1) 2x-3y = -2
2) x>2 and y>0

this is a good one.

from i, 2x-3y = -2. this is sufficient because the left part of the inequality is negative and the right part is 0 or greater than 0 because square of any value cannot be negative.

from ii, x>2 and y>0. this is also sufficient because x and y could have any values but 2x-3y is always smaller than x^2. in only case 2x-3y would be greater than x^2 is when y is negative but it clearly is not. so this is also sufficient.

if needed, will explain more....
Re: DS: Inequality   [#permalink] 23 Sep 2005, 17:42
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# Is 2x-3y<x^2 ? 1) 2x-3y = -2 2) x>2 and y>0

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