Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10: 0134467795
ISBN 13: 978-0-13446-779-5

Chapter 9 - Basic Algebra - 9.6 Solving Equations - 9.6 Exercises - Page 669: 34



Work Step by Step

We are given the equation $z-\frac{7}{3}=\frac{32}{9}$. Add $\frac{7}{3}$ to both sides. $z-\frac{7}{3}+\frac{7}{3}=\frac{32}{9}+\frac{7}{3}$ $z+0=\frac{32}{9}+\frac{7}{3}=\frac{32}{9}+\frac{7\times3}{3\times3}=\frac{32}{9}+\frac{21}{9}=\frac{32+21}{9}=\frac{53}{9}$ $z+0=\frac{53}{9}$ Therefore, $z=\frac{53}{9}$.
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.