If k#0 and k - (3 -2k^2)/k = x/k, then x =

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by Jay@ManhattanReview » Thu Dec 19, 2019 10:58 pm
BTGmoderatorDC 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

OA C

Source: Official Guide
We have \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\)

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

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

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

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Thu Dec 26, 2019 8:16 pm
BTGmoderatorDC 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



OA C

Source: Official Guide
Simplifying, we have:

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

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

3k^2 - 3 = x

Answer: C

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