Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.1 - Real Numbers and the Real Line - Exercises A.1 - Page AP-6: 8

Answer

Solution set = $\displaystyle \{-\frac{1}{2},-\frac{9}{2}\}$

Work Step by Step

Use properties from the table 'Absolute Values and Intervals", Here: property 5: $|x|=a \Leftrightarrow x=\pm a,\ \quad $(for any positive number a) Thus: $2t+5=\pm 4\qquad $... add $-5$ $2t=-5\pm 4\qquad $... divide with 2, $t=\displaystyle \frac{-5\pm 4}{2}$ $\displaystyle \frac{-5+4}{2}=-\frac{1}{2}$ $\displaystyle \frac{-5-4}{2}=-\frac{9}{2}$ Solution set = $\displaystyle \{-\frac{1}{2},-\frac{9}{2}\}$
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