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8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy?

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

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23 Feb 2012, 19:27
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$$8x*y^3 + 8x^3*y = \frac{2x^2*y^2}{2^{(-3)}}$$, What is xy?

(1) y > x
(2) x < 0
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Nov 2017, 08:27, edited 4 times in total.
Edited the question and added the OA
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8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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23 Feb 2012, 23:18
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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.

Hope it's clear.
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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 2012, 23:04
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: $$xy^3+x^3y-2x^2*y^2=0$$ --> 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.

Hope it's clear.

P.S. dvinoth86 please check the questions before posting and format them correctly. Thank you.
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Joined: 02 Sep 2009
Posts: 43898
Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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16 Mar 2012, 01:18
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: $$xy^3+x^3y-2x^2*y^2=0$$ --> 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.

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

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14 Nov 2014, 05:23
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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.

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.
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Posts: 43898
Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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14 Nov 2014, 05:28
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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: $$xy^3+x^3y-2x^2*y^2=0$$ --> 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.

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.

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: algebra-tips-and-hints-175003.html

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

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13 Aug 2015, 17:19
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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8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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22 Nov 2017, 08:24
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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: $$xy^3+x^3y-2x^2*y^2=0$$;

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.

Hope it's clear.

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Math Expert
Joined: 02 Sep 2009
Posts: 43898
Re: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? [#permalink]

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22 Nov 2017, 08:34
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: $$xy^3+x^3y-2x^2*y^2=0$$;

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.

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

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30 Nov 2017, 05:00
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: $$xy^3+x^3y-2x^2*y^2=0$$;

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.

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: 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy?   [#permalink] 30 Nov 2017, 05:00
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