## Algebra: A Combined Approach (4th Edition)

$z=-15$
$\dfrac{10}{z}=\dfrac{5}{z}-\dfrac{1}{3} \\ \dfrac{10}{z}=\dfrac{15-z}{3z} \\ 30z=z(15-z) \\ 30z=15z-z^2 \\ z^2+30z-15z=0 \\ z^2+15z=0 \\ z(z+15)=0 \\$ But the solution is only $z=-15$, because if $z=0$ the original equation would be undefined.