Basic College Mathematics (10th Edition)

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

Chapter 9 - Basic Algebra - Mixed Review Exercises - Page 700: 17



Work Step by Step

We are given the equation $6z-3=3z+9$. In order to solve, we can first subtract 3z from both sides. $6z-3z-3=3z-3z+9$ $3z-3=9$ Next, add 3 to both sides. $3z-3+3=9+3$ $3z=12$ Divide both sides by 3. $\frac{3z}{3}=\frac{12}{3}$ Therefore, $z=4$.
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