Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 695: 22


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, $\frac{-2}{3}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.