Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 8 - Systems of Linear Equations and Problem Solving - 8.4 Systems of Equations in Three Variables - 8.4 Exercise Set - Page 538: 32



Work Step by Step

Solving the system of equations using substitution, we find: $$ 2c-3\cdot \frac{-16c-25}{42}=1 \\ c=-.25 \\ b=\frac{-16\left(-\frac{1}{4}\right)-25}{42}=-\frac{1}{2} \\ a=\frac{2+3\left(-\frac{1}{2}\right)}{2}=\frac{1}{4} $$
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.