Answer
$(-\displaystyle \infty,-\frac{1}{3}]\cup[3,\infty)$
Work Step by Step
3 times a number is subtracted from 4 is: $4-3x$
the absolute value of the difference $|4-3x|$
is at least 5$\qquad...\qquad |4-3x|\geq 5$
$|4-3x|=|3x-4|\qquad $, because $|a-b|=|b-a|$
So, we solve
$|3x-4|\geq 5$
$|u| \gt c $ is equivalent to ($ u\lt -c $) or ($ u\gt c $)
$ \begin{array}{lllll}
3x-4\leq-5 & /+4 & ...or... & 3x-4\geq 5 & /+4\\
3x\leq-1 & /\div 3 & & 3x \geq 9 & /\div 3\\
x\leq-1/3 & & & x\geq 3 & \\
x\in(-\infty,-1/3] & & or & x\in[3,\infty) & \\
& & & &
\end{array} $
Solution set: $(-\displaystyle \infty,-\frac{1}{3}]\cup[3,\infty)$
