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 4 - Polynomials - 4.1 Exponents and Their Properties - 4.1 Exercise Set - Page 236: 93



Work Step by Step

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