Answer
$s\in(-\infty,-2]\cup[2,\infty)$
Work Step by Step
Use properties from the table 'Absolute Values and Intervals".
Applying Property $9$:
$|x|\geq a\quad\Leftrightarrow\quad x\geq a\ \ $
or
$\ \ x\leq-a, \quad $ for a = positive number
We have
$\begin{array}{ll}
2s\geq 4 & /\times\frac{1}{2}\\
s\geq 2 &
\end{array}$ $\ \ $
or
$\ \ $ $\begin{array}{ll}
2s\leq-4 & /\times\frac{1}{2}\\
s\leq-2 &
\end{array}$
In interval form, we can write this result as
$s\in(-\infty,-2]\cup[2,\infty)$
