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 9 - Inequalities and Problem Solving - 9.1 Inequalities and Applications - 9.1 Exercise Set - Page 580: 9


equivalent equations

Work Step by Step

$\bf{\text{Solution Outline:}}$ Solve the solution sets of the given statements, \begin{array}{l}\require{cancel} \dfrac{3}{5}a+\dfrac{1}{5}=2 \\\text{and}\\ 3a+1=10 .\end{array} If the solution sets are the same, then they are equivalent. Otherwise, they are not equivalent. $\bf{\text{Solution Details:}}$ Using the properties of equality, the first equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{3}{5}a+\dfrac{1}{5}=2 \\\\ 5\left( \dfrac{3}{5}a+\dfrac{1}{5} \right)=5(2) \\\\ 3a+1=10 .\end{array} Since the equation above is the same as the second given equation, then the given are $\text{ equivalent equations .}$
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.