Answer
The equation has one solution, $\frac{-2}{3}$.
Work Step by Step
$\frac{z}{z+2}-\frac{1}{z}=1$
$\frac{z}{z+2}-\frac{1}{z}-1=0$
$\frac{z^2}{z(z+2)}-\frac{z+2}{z(z+2)}-\frac{z(z+2)}{z(z+2)}=0$
$\frac{z^2-z-2-z^2-2z}{z(z+2)}=0$
$-3z-2=0*z(z+2)$
$-3z-2=0$
$-3z=2$
$z=\frac{-2}{3}$
Check:
$\frac{-2}{3}\div(\frac{-2}{3}+2)-1\div\frac{-2}{3}=\frac{-2}{3}\times\frac{3}{4}+\frac{3}{2}=\frac{-1}{2}+\frac{3}{2}=1$
The equation has one solution, $\frac{-2}{3}$.
