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 3 - Introduction to Graphing - 3.7 Point-Slope Form and Equations of Lines - 3.7 Exercise Set - Page 220: 90



Work Step by Step

Multiplying both sides by the $LCD=6$, the solution to the given equation, $ \dfrac{1}{2}n-\dfrac{1}{3}=\dfrac{1}{6}n+\dfrac{3}{2} $, is \begin{array}{l}\require{cancel} 3(1n)-2(1)=1(1n)+3(3) \\\\ 3n-2=n+9 \\\\ 3n-n=9+2 \\\\ 2n=11 \\\\ n=\dfrac{11}{2} .\end{array}
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