## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 1, Equations and Graphs - Section 1.7 - Solving Inequalities - 1.7 Exercises - Page 148: 32

#### Answer

$\left[\dfrac{11}{12}, \dfrac{13}{6}\right]$ Refer to the image below for the graph.

#### Work Step by Step

Multiply the LCD of $20$ to each part of the inequality to eliminate the fractions: $\begin{array}{ccccc} &20\left(-\dfrac{1}{2}\right) &\le &20\left(\dfrac{4-3x}{5}\right) &\le &20\left(\dfrac{1}{4}\right) \\&-10 &\le &4(4-3x) &\le &5 \\&-10 &\le &16-12x &\le &5 \end{array}$ Subtract $16$ to each part of the inequality to obtain: $\begin{array}{ccccc} &-10-16 &\le &16-12x-16 & \le &5-16 \\&-26 &\le &-12x &\le &-11 \end{array}$ Divide each part by $-12$ Note that this will affect the inequality symbols as they will flip to the opposite direction. $\begin{array}{ccccc} &\dfrac{-26}{-12} &\ge &\dfrac{-12x}{-12} &\ge &\dfrac{-11}{-12} \\&\dfrac{13}{6} &\ge &x &\ge &\dfrac{11}{12} \end{array}$ This inequality is equivalent to: $\dfrac{11}{12}\le x \le \dfrac{13}{6}$ Thus, the solution set is $\bf\left[\dfrac{11}{12}, \dfrac{13}{6}\right]$. To graph this solution set, plot solid dots at $\dfrac{11}{12}$ (or 0.9167) and $\dfrac{13}{6}$ (or 2.1667) then shade the region in between. (refer to the attached image in the answer part above for the graph)

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.