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 100: 29

Answer

$\left( 3,\infty \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ -(4+r)+2-3r\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 $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the Distributive Property and the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -4-r+2-3r\lt-14 \\\\ -r-3r\lt-14+4-2 \\\\ -4r\lt-12 .\end{array} Dividing both sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} r\gt\dfrac{-12}{-4} \\\\ r\gt3 .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $ \left( 3,\infty \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.