College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

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


$\left[\dfrac{11}{12}, \dfrac{13}{6}\right]$ 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)
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