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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] New post 07 Apr 2006, 11:10
If k≠0 and k-(3-2k^2)/k =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
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 [#permalink] New post 07 Apr 2006, 11:20
(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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 [#permalink] 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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 [#permalink] 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
  [#permalink] 07 Apr 2006, 15:10
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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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