Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.5 - Linear Inequalities in One Variable - 1.5 Exercises - Page 101: 63

Answer

$\left[ -9,\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the concepts of inequalities to translate the given description, \begin{array}{l}\require{cancel}\text{ one third of a number is added to 6, giving a result of at least 3 ,}\end{array} into symbols. Then solve using the properties of inequality. Express the solution set in interval notation. $\bf{\text{Solution Details:}}$ In symbols, the given description translates to \begin{array}{l}\require{cancel} 6+\dfrac{1}{3}x\ge3 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 3\left( 6+\dfrac{1}{3}x \right)\ge3(3) \\\\ 18+x\ge9 \\\\ x\ge9-18 \\\\ x\ge-9 .\end{array} In interval notation, the solution set is $ \left[ -9,\infty \right) .$
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