## College Algebra 7th Edition

$z=1$
$z+\displaystyle \frac{4}{z+1}=3$ We multiply through by $z+1$ and solve by factoring: $z(z+1)+4=3(z+1)$ $z^2+z+4=3z+3$ $z^2+z+4-3z-3=0$ $z^2-2z+1=0$ $(z-1)(z-1)=0$ $z=1$