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: 28

Answer

$\left[ -\dfrac{1}{2},\infty \right)$
1512743412

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ x-3(x+1)\le4x ,$ use the Distributive Property and 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} x-3(x)-3(1)\le4x \\\\ x-3x-3\le4x \\\\ x-3x-4x\le3 \\\\ -6x\le3 .\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} x\ge\dfrac{3}{-6} \\\\ x\ge-\dfrac{1}{2} .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $ \left[ -\dfrac{1}{2},\infty \right) .$
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