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Last edited by Bunuel on 27 May 2013, 05:19, edited 3 times in total.

Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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24 Feb 2012, 09:47

wil follow from next time
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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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15 Mar 2014, 06:07

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Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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14 Nov 2014, 06:23

1

This post was BOOKMARKED

Bunuel wrote:

8x*y^3 + 8x^3*y = 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.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.

Hi Bunuel,

I did a blunder while reducing 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3) but not sure the reason for that. i started by cancelling out the xy from both sides to get (x-y)^2 = 0 at the end. Why cant we cancel the x and y ? Please help.
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\(8xy^3+8x^3y=\frac{2x^2*y^2}{2^{-3}}\) --> \(8xy^3+8x^3y=2x^2*y^2*8\) --> 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.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.

Hi Bunuel,

I did a blunder while reducing 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3) but not sure the reason for that. i started by cancelling out the xy from both sides to get (x-y)^2 = 0 at the end. Why cant we cancel the x and y ? Please help.

If you divide (reduce) 8x*y^3 + 8x^3*y = 2x^2*y^2/2^(-3), by xy you assume, with no ground for it, that xy does not equal to zero thus exclude a possible solution (notice that xy=0 satisfies the equation). Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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10 Apr 2016, 00:43

Bunuel, In such questions,one must avoid cancelling out the variables right? I have seen myself doing that mistake in quite a few qsns but your approach helped me identify my mistake

gmatclubot

Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy?
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10 Apr 2016, 00:43

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