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: 9


Solution set = $\displaystyle \{\frac{7}{6},\frac{25}{6}\}$

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: $8-3s=\displaystyle \pm\frac{9}{2}\qquad $... multiply with 2 $ 16-6s=\pm 9\qquad$ ... add $-16$ $-6s=-16\pm 9\qquad $divide with $(-6)$ $s=\displaystyle \frac{-16\pm 9}{-6}$ $\displaystyle \frac{-16+9}{-6}=\frac{-7}{-6}=\frac{7}{6}$ $\displaystyle \frac{-16-9}{-6}=\frac{-25}{-6}=\frac{25}{6}$ Solution set = $\displaystyle \{\frac{7}{6},\frac{25}{6}\}$
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