Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.7 - Absolute Value Inequalities - Exercise Set - Page 106: 97

Answer

Solving $|x-3|=5$ is different from solving $|x-3|\lt5$ because the first one involves solving $x-3=5 \text{ or } x-3=-5$, while the second one involves solving $-5\lt x-3\lt5$.

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