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If x is not equal to zero, and x = sqrt(4xy-4y^2), then in

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Manager
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Joined: 11 May 2007
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If x is not equal to zero, and x = sqrt(4xy-4y^2), then in [#permalink] New post 16 Nov 2007, 11:19
If x is not equal to zero, and x = sqrt(4xy-4y^2), then in terms of y, x =

(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/1-2y
(E) -2y
Manager
Manager
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Joined: 01 Oct 2007
Posts: 87
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Kudos [?]: 13 [0], given: 0

Re: Equation [#permalink] New post 16 Nov 2007, 11:29
yogachgolf wrote:
If x is not equal to zero, and x = sqrt(4xy-4y^2), then in terms of y, x =

(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/1-2y
(E) -2y


Square both sides:

x^2 = 4xy - 4y^2

Bring everything to the left:

x^2 - 4xy + 4y^2 = 0

Factor:

(x - 2y)^2 = 0

So x = 2y.
CEO
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Re: Equation [#permalink] New post 16 Nov 2007, 23:32
yogachgolf wrote:
If x is not equal to zero, and x = sqrt(4xy-4y^2), then in terms of y, x =

(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/1-2y
(E) -2y


Theres an alternative to straight algebra.

Plug in 2 for y. We eventually get x^2-8x+16=0 (x-4)(x-4)=0

so x=4. Since y =2 2y=4=x.

Now this method isnt full proof as other answer choices may yield the same result. E could be a candidate b/c -2*-2=4. So try -2 in the equation for y. It turns out its not the same.

I only use this when for some reason I can't figure out how to factor something etc.. Id suggest tryin algebra first.
Director
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Re: Equation [#permalink] New post 17 Nov 2007, 00:03
yogachgolf wrote:
If x is not equal to zero, and x = sqrt(4xy-4y^2), then in terms of y, x =

(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/1-2y
(E) -2y


A. Used back solving starting w/ choice A.

x = sqrt(4xy-4y^2) --> given
let x = 2y
Is 2y = sqrt (8y^2-4y^2)?
Is 2y = sqrt (4y^2)? Yes.
VP
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Re: Equation [#permalink] New post 17 Nov 2007, 00:41
johnrb wrote:
yogachgolf wrote:
If x is not equal to zero, and x = sqrt(4xy-4y^2), then in terms of y, x =

(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/1-2y
(E) -2y


Square both sides:

x^2 = 4xy - 4y^2

Bring everything to the left:

x^2 - 4xy + 4y^2 = 0

Factor:

(x - 2y)^2 = 0

So x = 2y.


Perfect !

:)
Director
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Joined: 26 Feb 2006
Posts: 908
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Re: Equation [#permalink] New post 18 Nov 2007, 11:48
johnrb wrote:
yogachgolf wrote:
If x is not equal to zero, and x = sqrt(4xy-4y^2), then in terms of y, x =

(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/1-2y
(E) -2y


Square both sides:

x^2 = 4xy - 4y^2

Bring everything to the left:

x^2 - 4xy + 4y^2 = 0

Factor:

(x - 2y)^2 = 0

So x = 2y.


this is an equation question, and asking the value of x in terms of y. so solving the equation would be the correct approach.
Re: Equation   [#permalink] 18 Nov 2007, 11:48
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