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Re: x^2y > 0 xy^2 > 0 [#permalink]
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SDC77 wrote:
B? Second inequality must be 'less than', isn't it ?



yes sir correct. I edited the typo
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Re: x^2y > 0 xy^2 > 0 [#permalink]
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From the first inequality we do know that y must be positive because x^2 is always positive and both are >0

The second y^2 is always positive so to have <0 x must be negative

x <0

y > 0

B is the answer
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Re: x^2y > 0 xy^2 > 0 [#permalink]
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Carcass wrote:
\(x^2y > 0 \)

\(xy^2 < 0\)


Quantity A
Quantity B
x
y



Given: \(x^2y > 0 \)
Since we can be certain that \(x^2\) is positive, we can safely divide both sides of the inequality by \(x^2\) to get: \(y>0\) (in other words, y is positive)

Given: \(xy^2 < 0\)
Since we can be certain that \(y^2\) is positive, we can safely divide both sides of the inequality by \(y^2\) to get: \(x<0\) (in other words, x is negative)

Answer: B

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Re: x^2y > 0 xy^2 > 0 [#permalink]
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