#### Answer

$z=1$

#### Work Step by Step

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

Published by
Brooks Cole

ISBN 10:
1305115546

ISBN 13:
978-1-30511-554-5

$z=1$

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