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If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B)

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If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-k-0-and-k-3-2k-2-k-x-k-then-x-167913.html
[Reveal] Spoiler: OA

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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New post 07 Apr 2006, 11:20
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(C) :wink:

k - ( 3 - 2K^2)/k = x/k

k^2 - ( 3 - 2k^2) = x

k^2 - 3 + 2k^2 = x

x = (3k^2) - 3

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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New post 07 Apr 2006, 13:23
I dont understand how the -2k^2 became +2k^2

k^2 - ( 3 - 2k^2) = x

k^2 - 3 + 2k^2 = x
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Andre Crompton
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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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New post 07 Apr 2006, 15:10
andrecrompton wrote:
I dont understand how the -2k^2 became +2k^2

k^2 - ( 3 - 2k^2) = x

k^2 - 3 + 2k^2 = x


The brackets open and negative negative make a positive

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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New post 23 Oct 2014, 18:22
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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B) [#permalink]

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New post 24 Oct 2014, 02:34
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andrecrompton wrote:
If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2


\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\);

Multiply by k: \(k^2 - (3 -2k^2) = x\);

\(x=3k^2-3\).

Answer: C.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-k-0-and-k-3-2k-2-k-x-k-then-x-167913.html

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Re: If k≠0 and k-(3-2k^2)/k =x/k, then x= (A) -3-k^2 (B)   [#permalink] 24 Oct 2014, 02:34
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