Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 3 - Equations and Problem Solving - 3.1 - Solving First-Degree Equations - Problem Set 3.1 - Page 100: 65



Work Step by Step

Multiplying both sides by the $LCD= 5 ,$ and using the properties of equality, the value of the variable that satisfies the given equation, $ -8n=\dfrac{6}{5} ,$ is \begin{array}{l}\require{cancel} 5(-8n)=1(6) \\\\ -40n=6 \\\\ n=\dfrac{6}{-40} \\\\ n=\dfrac{\cancel{2}\cdot3}{\cancel{2}\cdot(-20)} \\\\ n=\dfrac{3}{-20} \\\\ n=-\dfrac{3}{20} .\end{array}
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