Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-3 - Cumulative Review - Page 228: 37



Work Step by Step

Add 13 to both sides: $4|x|−13+13=3+13 \\4|x|=16$ Divide 4 to both sides: $\dfrac{4|x|}{4}=\dfrac{16}{4} \\|x|=4$ RECALL: $|x|=a \longrightarrow x=a \text{ or } x=−a$ Thus, $|x|=4 \longrightarrow x=4 \text{ or } x=−4$ Therefore, the solution set is: $\left\{−4,4\right\}$
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