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Fofino [41]
3 years ago
15

Rearranging literal equations for z

Mathematics
1 answer:
serious [3.7K]3 years ago
5 0

Answer:

z = \frac{3n - 3}{6 - 2n}

Step-by-step explanation:

\frac{1}{(n-1)^{\frac{1}{3} } }  = (\frac{2z + 3}{4z} )^{\frac{1}{3} }

cubing on both sides we get

\frac{1}{n-1} = \frac{2z + 3}{4z}

4z = (2z+3)(n-1)

4z = 2nz + 3n - 2z - 3

4z - 2nz + 2z = 3n - 3

z(6 - 2n) = 3n - 3

z = \frac{3n - 3}{6 - 2n}

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