Answer
$(-\infty,-6]\cup[24,\infty)$
Work Step by Step
$4+|3-\displaystyle \frac{x}{3}|\geq 9$ $\qquad $... $/-4$
$|3-\displaystyle \frac{x}{3}|\geq 5 \qquad $... $/\times 3$
$|9-x|\geq 15$
... $|a-b|=|b-a|$
$|x-9|\geq 15$
... $|u| \geq c $ is equivalent to ($ u\leq -c $) or ($ u\geq c $)
$ \begin{array}{lllll}
x-9\leq-15 & /+9 & ...or... & x-9\geq 15 & /+9\\
x\leq-6 & & & x \geq 24 & \\
x\in(-\infty,-6] & & or & x\in[24,\infty) & \\
& & & &
\end{array} $
Solution set: $(-\infty,-6]\cup[24,\infty)$
