Intermediate Algebra (12th Edition)

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

Chapter 1 - Review Exercises: 41

Answer

$\left[ 3,5 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ 8 \le 3x-1 \lt 14 ,$ use the properties of inequality. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Removing the grouping symbols and using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 8+1 \le 3x-1+1 \lt 14+1 \\\\ 9 \le 3x \lt 15 \\\\ \dfrac{9}{3} \le \dfrac{3x}{3} \lt \dfrac{15}{3} \\\\ 3 \le x \lt 5 .\end{array} In interval notation, the solution set is $ \left[ 3,5 \right) .$
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.