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: 52



Work Step by Step

Multiplying both sides by the $LCD= 2 $, then the solution to the given equation, $ \dfrac{1}{2}+4m=3m-\dfrac{5}{2} $, is \begin{array}{l} 1(1)+2(4m)=2(3m)-1(5) \\\\ 1+8m=6m-5 \\\\ 8m-6m=-5-1 \\\\ 2m=-6 \\\\ m=-\dfrac{6}{2} \\\\ m=-3 .\end{array} CHECKING: \begin{array}{l} \dfrac{1}{2}+4(-3)=3(-3)-\dfrac{5}{2} \\\\ \dfrac{1}{2}-12=-9-\dfrac{5}{2} \\\\ \dfrac{1}{2}-\dfrac{24}{2}=-\dfrac{18}{2}-\dfrac{5}{2} \\\\ -\dfrac{23}{2}=-\dfrac{23}{2} \text{ (TRUE)} .\end{array} Hence, the solution is $ m=-3 $.
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.