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 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set - Page 95: 89



Work Step by Step

Multiplying both sides by $10$, then the solution to the given equation, $ 0.7(3x+6)=1.1-(x-3) $, is \begin{array}{l} 7(3x+6)=11-10(x-3) \\\\ 21x+42=11-10x+30 \\\\ 21x+10x=11+30-42 \\\\ 31x=-1 \\\\ x=-\dfrac{1}{31} .\end{array} CHECKING: \begin{array}{l} 0.7\left(3\left( -\dfrac{1}{31} \right)+6\right)=1.1-\left(-\dfrac{1}{31}-3\right) \\\\ \dfrac{7}{10}\left(-\dfrac{3}{31}+6\right)=\dfrac{11}{10}-\left(-\dfrac{94}{31}\right) \\\\ \dfrac{7}{10}\left(\dfrac{183}{31}\right)=\dfrac{11}{10}+\dfrac{94}{31} \\\\ \dfrac{1281}{310}=\dfrac{1281}{310} \text{ (TRUE)} .\end{array} Hence, the solution is $ x=-\dfrac{1}{31} $.
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.