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Re: xy > 0 [#permalink]
b is answer
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xy > 0 [#permalink]
1
distributing Quantity A, we get

\(2x^2 + 8xy - xy - 4y^2\)

\(2x^2 +7xy - 4y^2\)

Quantity B

\(2x^2 + 8xy - 4y^2\)


Now since \(x^2\) and \(y^2\) will be positive, we can cancel \(2x^2 - 4y^2\) on both sides to get

Quantity A = \(7xy\)
Quantity B = \(8xy\)

Since \(xy>0\)

Quantity B is greater than Quantity A
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xy > 0 [#permalink]
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