Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 1 - Expressions, Equations, and Inequalities - 1-6 Absolute Value Equations and Inequalities - Practice and Problem-Solving Exercises - Page 46: 26

Answer

$-\dfrac{8}{3}\lt y \lt\dfrac{10}{3}$

Work Step by Step

Using the properties of inequality, the given, $ |6y-2|+4\lt22 ,$ is equivalent to \begin{align*}\require{cancel} |6y-2|+4&\lt22 \\ |6y-2|+4-4&\lt22-4 \\ |6y-2|&\lt18 .\end{align*} Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above implies \begin{align*}\require{cancel} -18\lt 6y-2 &\lt18 .\end{align*} Using the properties of inequality, the inequality above is equivalent to \begin{align*}\require{cancel} -18+2\lt 6y-2+2 &\lt18+2 \\ -16\lt 6y &\lt20 \\\\ -\dfrac{16}{6}\lt \dfrac{6y}{6} &\lt\dfrac{20}{6} \\\\ -\dfrac{8}{3}\lt y &\lt\dfrac{10}{3} .\end{align*} Hence, the solution is $ -\dfrac{8}{3}\lt y \lt\dfrac{10}{3} .$ Since a hollowed dot is used for the symbols $\lt$ and $\gt,$ while a solid dot is used for the symbols $\le$ and $\ge,$ then the graph of the solution above is the set of numbers from $ -\dfrac{8}{3} $ to $ \dfrac{10}{3} $ with hollowed dots at $ -\dfrac{8}{3} $ and $ \dfrac{10}{3} $.
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