Chapter 8 - Section 8.2 - Solving Quadratic Equations by the Quadratic Formula - Exercise Set - Page 494: 70

$z=-15$

Multiplying both sides of the given equation, $ \dfrac{10}{z}=\dfrac{5}{z}-\dfrac{1}{3} ,$ by the $LCD= 3z $, then, \begin{array}{l}\require{cancel} 3z\left( \dfrac{10}{z} \right)=\left( \dfrac{5}{z}-\dfrac{1}{3} \right)(3z) \\\\ 3(10)=5(3)-1(z) \\\\ 30=15-z \\\\ z=15-30 \\\\ z=-15 .\end{array}