8x y^3 + 8x^3 y = 2x^2y^2 * 8 , What is xy? (1) y > x (2)

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8x y^3 + 8x^3 y = 2x^2y^2 * 8 , What is xy? (1) y > x (2) [#permalink]

31 Mar 2011, 12:00

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8x y^3 + 8x^3 y= 2x^2y^2 * 8, What is xy?

(1) y > x

(2) x < 0

How I approached it:

We can simplify the given equation as:

8xy (x^2+y^2)= 2xy. xy. 8
=> x^2+y^2=2xy
=> x^2+y^2-2xy=0
=>(x-y)^2= 0
=> (x-y)=0
=> x=y

You see I'm getting x=y, which is in direct contradiction with Statement (1), something that cannot happen on the GMAT. Can someone please tell me where am I going wrong?

[Reveal] Spoiler: OA

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Re: Inequalities DS: Help sought [#permalink]

31 Mar 2011, 12:29

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gmatpapa wrote:

8x y^3 + 8x^3 y= 2x^2y^2 * 8, What is xy?

(1) y > x

(2) x < 0

How I approached it:

We can simplify the given equation as:

8xy (x^2+y^2)= 2xy. xy. 8
=> x^2+y^2=2xy
Never divide the variables if you don't know their value or sign
You have divided xy on both sides; RHS and LHS, which is illegal. What if either x or y is 0, then you would be dividing the polynomial with 0 and 0 in denominator is unacceptable. It results in undefined value.

=> x^2+y^2-2xy=0
=>(x-y)^2= 0
=> (x-y)=0
=> x=y

You see I'm getting x=y, which is in direct contradiction with Statement (1), something that cannot happen on the GMAT. Can someone please tell me where am I going wrong?

Sol:
I am going to follow your main idea i.e. to reduce the expression to its simplest form:

8xy^3 + 8x^3y= 2x^2y^2*8
8xy(y^2 + x^2)= 2*x^2y^2*8

Dividing by 8 on both sides; we can do this because 8 is a constant and we know its value
xy(y^2 + x^2)= 2*x^2y^2

Subtract 2*x^2y^2 from both sides
xy(y^2 + x^2) - 2*x^2y^2 = 0

Taking xy common
xy(y^2 + x^2 - 2xy) = 0
xy(y-x)^2 = 0

The expression above means either at least one of x and y is 0 OR y-x=0 i.e. x=y

1. y>x
It means x \ne y
That leaves us with only one case; At least one of x and y is 0
And xy=0
Sufficient.

2. x<0
It means x can be any negative number.
Now, {x,y} can be {-1,-1} or {-100,-100} both incurring different results when multiplied.
Insufficient.

Ans: "A"
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Re: Inequalities DS: Help sought [#permalink]

31 Mar 2011, 12:41

fluke wrote:

Never divide the variables if you don't know their value or sign
You have divided xy on both sides; RHS and LHS, which is illegal. What if either x or y is 0, then you would be dividing the polynomial with 0 and 0 in denominator is unacceptable. It results in undefined value.

Oh.. I always thought this restriction applies to only inequalities. The OE is the exact same as yours. I understood it but was wondering why cant we divide the equation with xy so we get (x-y)^2=0. Thanks for the explanation, fluke.
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Re: Inequalities DS: Help sought [#permalink]

31 Mar 2011, 19:26

The expression says :

8xy^3 + 8x^3y - 2x^2y^2*8 = 0

xy^3 + x^3y - x^2y^2 = 0

xy(y^2 + x^2 - 2xy) = 0

xy (x - y)^2 = 0

So either xy = 0 or (x-y)^2 = 0

(1) y > x

=> (x-y)^2 != 0, so xy = 0

(2) x < 0, but no information about y, so insufficient as y = 0, or y = x is possible too.

Answer - A
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Re: Inequalities DS: Help sought [#permalink]

03 Apr 2011, 07:53

solving the given expression

we have
8xy(x2+y2-2xy) = 0
=> xy(x-y)^2 =0
=> either xy =0 or x=y

1. Sufficient
y>x =>xy=0

2. Not sufficient
x<0 ,no info on y. so xy could vary depending on the selected combination

Answer is A.

Re: Inequalities DS: Help sought [#permalink] 03 Apr 2011, 07:53

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8x y^3 + 8x^3 y = 2x^2y^2 * 8 , What is xy? (1) y > x (2)

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8x y^3 + 8x^3 y = 2x^2y^2 * 8 , What is xy? (1) y > x (2)

8x y^3 + 8x^3 y = 2x^2y^2 * 8 , What is xy? (1) y > x (2)

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