Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1 - Linear Functions - 1.1 Solving Linear Equations - 1.1 Exercises: 61

Answer

$z=\frac{17}{6}$

Work Step by Step

$\frac{3}{7}(2z-5)=\frac{4}{7}(-3z+9)$ $\frac{6}{7}z-\frac{15}{7}=-\frac{12}{7}z+\frac{36}{7}$ $\frac{6}{7}z+\frac{12}{7}z=\frac{36}{7}+\frac{15}{7}$ $\frac{18}{7}z=\frac{51}{7}$ $z=\frac{51}{7}\times\frac{7}{18}$ $z=\frac{51}{18}$ (to simplify, divide both numbers by 3) $z=\frac{17}{6}$
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.