Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.7 Inequalities - 1.7 Exercises - Page 159: 36


$\color{blue}{[-12, 6]}$

Work Step by Step

Multiply $3$ to each part: \begin{array}{ccccc} &3(-5)&\le &3 \cdot \dfrac{x-3}{3} &\le &3(1) \\&-15 &\le &x-3 &\le &3 \end{array} Add $3$ to each part: \begin{array}{ccccc} \\&-15+3 &\le &x-3+3 &\le &3+3 \\&-12 &\le &x &\le &6 \end{array} Thus, the solution to the given inequality is: $\color{blue}{[-12, 6]}$
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