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Difficulty:
95%
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Question Stats:
43% (02:14) correct
57%
(02:37)
wrong
based on 680
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If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{-3}}\), what is the value of xy?
(1) y > x
(2) x < 0
(1) y > x
(2) x < 0
24 Feb 2012, 00:18
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\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}\), What is xy?
\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);
\(8xy^3+8x^3y=2x^2*y^2*8\);
Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);
Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);
Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);
\(xy(y-x)^2=0\);
Either \(xy=0\) or \(y-x=0\) (\(x=y\)).
(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.
Answer: A.
Hope it's clear.
\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\);
\(8xy^3+8x^3y=2x^2*y^2*8\);
Reduce by 8: \(xy^3+x^3y=2x^2*y^2\);
Rearrange: \(xy^3+x^3y-2x^2*y^2=0\);
Factor out \(xy\): \(xy(y^2+x^2-2xy)=0\);
\(xy(y-x)^2=0\);
Either \(xy=0\) or \(y-x=0\) (\(x=y\)).
(1) y > x --> \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.
Answer: A.
Hope it's clear.
24 Dec 2012, 06:32
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