Thomas' Calculus 13th Edition

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