Basic College Mathematics (9th Edition)

Published by Pearson
ISBN 10: 0321825535
ISBN 13: 978-0-32182-553-7

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

Answer

$z=\frac{53}{9}$

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, $z=\frac{53}{9}$.
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