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If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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Updated on: 21 May 2018, 00:30
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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
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Originally posted by dvinoth86 on 23 Feb 2012, 20:27.
Last edited by Bunuel on 21 May 2018, 00:30, edited 5 times in total.
Edited the question and added the OA




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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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24 Feb 2012, 00:18
\(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(3)}}\), 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: \(xy^3+x^3y2x^2*y^2=0\); Factor out \(xy\): \(xy(y^2+x^22xy)=0\); \(xy(yx)^2=0\); Either \(xy=0\) or \(yx=0\) (\(x=y\)). (1) y > x > \(y\neq{x}\), which means that \(xy=0\). Sufficient. (2) x < 0. Clearly not sufficient. Answer: A. Hope it's clear.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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16 Mar 2012, 00:04
if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand 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: \(xy^3+x^3y2x^2*y^2=0\) > factor out \(xy\): \(xy(y^2+x^22xy)=0\) > \(xy(yx)^2=0\) > either \(xy=0\) or \(yx=0\) (\(x=y\)).
(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.



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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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16 Mar 2012, 02:18
devinawilliam83 wrote: if the question stem says y=x and option A says y>x.. How can er say xy=0 . am unable to understand 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: \(xy^3+x^3y2x^2*y^2=0\) > factor out \(xy\): \(xy(y^2+x^22xy)=0\) > \(xy(yx)^2=0\) > either \(xy=0\) or \(yx=0\) (\(x=y\)).
(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. The stem DOES NOT say that \(x=y\), it says: EITHER \(xy=0\) OR \(x=y\). (1) says \(y > x\), which means that \(x=y\) is not possible, hence \(xy=0\). Hope it's clear.
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If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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24 Dec 2012, 06:32



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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/ 2^3,what is the value of x?
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25 Dec 2012, 13:18
Bunuel wrote: If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{3}}\), what is the value of xy?
\(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{3}}\) > \(8xy^3 + 8x^3y=8*2x^2y^2=0\) > reduce by 8 and rearrange: \(xy^3+x^3y2x^2y^2\) > factor out xy: \(xy(y^2+x^22xy)=0\) > \(xy(yx)^2=0\) > \(xy=0\) or \(yx=0\).
(1) y > x. Since \(y>x\), then \(yx\neq{0}\), thus \(xy=0\). Sufficient.
(2) x < 0. Not sufficient.
Answer: A. This solution does not answer to the question. Question asks about value xy. if y>x then xy can't be equal 0. It can't be discussed. Really I don't understand your solution please explain



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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/ 2^3,what is the value of x?
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26 Dec 2012, 03:43
akshin wrote: Bunuel wrote: If \(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{3}}\), what is the value of xy?
\(8xy^3 + 8x^3y=\frac{2x^2y^2}{2^{3}}\) > \(8xy^3 + 8x^3y=8*2x^2y^2=0\) > reduce by 8 and rearrange: \(xy^3+x^3y2x^2y^2\) > factor out xy: \(xy(y^2+x^22xy)=0\) > \(xy(yx)^2=0\) > \(xy=0\) or \(yx=0\).
(1) y > x. Since \(y>x\), then \(yx\neq{0}\), thus \(xy=0\). Sufficient.
(2) x < 0. Not sufficient.
Answer: A. This solution does not answer to the question. Question asks about value xy. if y>x then xy can't be equal 0. It can't be discussed. Really I don't understand your solution please explainYou should read a solution carefully. From the stem we have that either \(xy=0\) or \(yx=0\). (1) says that y > x, so y  x > 0, which means that \(yx\neq{0}\), thus \(xy=0\).
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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14 Nov 2014, 06:23
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: \(xy^3+x^3y2x^2*y^2=0\) > factor out \(xy\): \(xy(y^2+x^22xy)=0\) > \(xy(yx)^2=0\) > either \(xy=0\) or \(yx=0\) (\(x=y\)).
(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 (xy)^2 = 0 at the end. Why cant we cancel the x and y ? Please help.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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14 Nov 2014, 06:28
Ankur9 wrote: 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: \(xy^3+x^3y2x^2*y^2=0\) > factor out \(xy\): \(xy(y^2+x^22xy)=0\) > \(xy(yx)^2=0\) > either \(xy=0\) or \(yx=0\) (\(x=y\)).
(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 (xy)^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.Check more tips on Algebra here: algebratipsandhints175003.htmlHope it helps.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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13 Aug 2015, 18:19
I factored out the xy from both sides and that screwed up my equation. Note to self: dont factor out things when it may lead to division by zero!



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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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22 Nov 2017, 09:24
St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0. and hence xy = 0. this sums to Answer D. Where am I getting it wrong. Bunuel wrote: \(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(3)}}\), 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: \(xy^3+x^3y2x^2*y^2=0\);
Factor out \(xy\): \(xy(y^2+x^22xy)=0\);
\(xy(yx)^2=0\);
Either \(xy=0\) or \(yx=0\) (\(x=y\)).
(1) y > x > \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.
Answer: A.
Hope it's clear.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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22 Nov 2017, 09:34
Barui wrote: St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0. and hence xy = 0. this sums to Answer D. Where am I getting it wrong. Bunuel wrote: \(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(3)}}\), 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: \(xy^3+x^3y2x^2*y^2=0\);
Factor out \(xy\): \(xy(y^2+x^22xy)=0\);
\(xy(yx)^2=0\);
Either \(xy=0\) or \(yx=0\) (\(x=y\)).
(1) y > x > \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.
Answer: A.
Hope it's clear. You can always check your theoretical deduction by simple number plugging. Fro (2) plug x = y = 1 or x = y = 2.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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30 Nov 2017, 06:00
Thanks I understood now... Bunuel wrote: Barui wrote: St2: if x < 0, and y not equal to 0 , then LHS is negative but RHS is positive, which is not possible so y has to be 0. and hence xy = 0. this sums to Answer D. Where am I getting it wrong. Bunuel wrote: \(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(3)}}\), 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: \(xy^3+x^3y2x^2*y^2=0\);
Factor out \(xy\): \(xy(y^2+x^22xy)=0\);
\(xy(yx)^2=0\);
Either \(xy=0\) or \(yx=0\) (\(x=y\)).
(1) y > x > \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.
Answer: A.
Hope it's clear. You can always check your theoretical deduction by simple number plugging. Fro (2) plug x = y = 1 or x = y = 2.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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26 Mar 2018, 00:28
This is an excellent question. The trick is in simplyfing the question. Once you paraphrase the question into simple terms, it becomes, xy*(xy)^2=0.8x∗y3+8x3∗y=2x2∗y22(−3)8x∗y3+8x3∗y=2x2∗y22(−3), What is xy?(1) y > x ==>. Sufficient. Because, if y is greater than 1)x then x can be equal to Y. so only one equation left. xy=0. (2) x < 0. Clearly not sufficient. Answer: A.
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Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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13 Apr 2019, 04:55
Bunuel wrote: \(8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(3)}}\), 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: \(xy^3+x^3y2x^2*y^2=0\);
Factor out \(xy\): \(xy(y^2+x^22xy)=0\);
\(xy(yx)^2=0\);
Either \(xy=0\) or \(yx=0\) (\(x=y\)).
(1) y > x > \(y\neq{x}\), which means that \(xy=0\). Sufficient.
(2) x < 0. Clearly not sufficient.
Answer: A.
Hope it's clear. This may be me being extremely dim but can someone explain why \(xy\): \(xy(y^2+x^22xy)=0\); factors to \(xy(yx)^2=0\) and not \(xy(xy)^2=0\) I've written it out a lot and it seems to have the same product?




Re: If 8xy^3 + 8x^3*y=2x^2*y^2/2^3, what is the value of x?
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