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If k ≠ 0, and z^2/(k + 4z + 3) = z/k, then, k = ?

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Bunuel
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If k ≠ 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink] New post   20 Dec 2019, 02:34
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If \(k\neq0\), and \(\frac{z^{2}}{k} +4z+3 = \frac{z}{k}\) , then, k=?

A \(\frac{z^{2} - 3}{4}\)

B \(\frac{z^{2} + 4z}{3}\)

C \(z (z+4) * 3\)

D \(\frac{{z - z^{2}}}{{4z+3}}\)

E \(z^{2} + 4z - 3\)
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D
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Re: If k ≠ 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink] New post   08 Mar 2020, 17:07
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KaranB1 wrote:
Bunuel

How to know what is the right way to approach such problems

KaranB1, I'm certainly no Bunuel, but hopefully this can help...

Step 1: Remove the denominators by multiplying the entire formula by k...

\(z^2\) + 4zk + 3k = z

Step 2: Get the variables with k on their own side... so subtract both sides by z^2

4zk + 3k = z - \(z^2\)

Step 3: Factor out the k

k(4z + 3) = z - \(z^2\)

Final Step: divide both sides by 4z+3 in order to get k by itself.

k = (z - \(z^2\)) / 4z+3

KaranB1, let me know if this isn't clear.
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Re: If k ≠ 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink] New post   20 Dec 2019, 02:50
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z^2/k +4z +3 = z/k
or, 4z +3 = z/k-z^2/k
or, 4z +3 =-z(z-1)/k
or, k = -z(z-1)/(4z+3)
or, K = \(\frac{{z - z^{2}}}{{4z+3}}\)

Answer D


Bunuel wrote:
If \(k\neq0\), and \(\frac{z^{2}}{k} +4z+3 = \frac{z}{k}\) , then, k=?

A \(\frac{z^{2} - 3}{4}\)

B \(\frac{z^{2} + 4z}{3}\)

C \(z (z+4) * 3\)

D \(\frac{{z - z^{2}}}{{4z+3}}\)

E \(z^{2} + 4z - 3\)
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Re: If k ≠ 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink] New post   08 Mar 2020, 11:13
GMATBusters wrote:
z^2/k +4z +3 = z/k
or, 4z +3 = z/k-z^2/k
or, 4z +3 =-z(z-1)/k
or, k = -z(z-1)/(4z+3)
or, K = \(\frac{{z - z^{2}}}{{4z+3}}\)

Answer D


Bunuel wrote:
If \(k\neq0\), and \(\frac{z^{2}}{k} +4z+3 = \frac{z}{k}\) , then, k=?

A \(\frac{z^{2} - 3}{4}\)

B \(\frac{z^{2} + 4z}{3}\)

C \(z (z+4) * 3\)

D \(\frac{{z - z^{2}}}{{4z+3}}\)

E \(z^{2} + 4z - 3\)


Bunuel

How to know what is the right way to approach such problems
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Re: If k ≠ 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink] New post   22 Nov 2020, 12:12
Bunuel wrote:
If \(k\neq0\), and \(\frac{z^{2}}{k} +4z+3 = \frac{z}{k}\) , then, k=?

A \(\frac{z^{2} - 3}{4}\)

B \(\frac{z^{2} + 4z}{3}\)

C \(z (z+4) * 3\)

D \(\frac{{z - z^{2}}}{{4z+3}}\)

E \(z^{2} + 4z - 3\)


\(\frac{z^{2}}{k} +4z+3 = \frac{z}{k}\)

= \(z^2 + 4zk + 3k = z\)

\(4zk + 3k = z - z^2\)

\(k(4z+3) = z - z^2\)

\(k = \frac{z - z^2}{4z+3}\)

Answer is D.
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Re: If k 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink] New post   16 Sep 2023, 04:18
If k≠0 and z2/k+4z+3=z/k, then, k=?

Let 1/k = t

=> z2t + 4z + 3 =zt
=> 4z + 3 = t(z -z2)
=> 4z + 3/(z -z2)= t
=> 4z + 3/(z-z2)= 1/k
=> k = (z-z2)/(4z + 3)

Hence D
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Re: If k 0, and z^2/(k + 4z + 3) = z/k, then, k = ? [#permalink]
16 Sep 2023, 04:18
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