Answer
$x\le-3 \text{ OR } x\ge4$
Work Step by Step
Using the properties of inequality, the given, $
3|2x-1|\ge21
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{3|2x-1|}{3}\ge\dfrac{21}{3}
\\\\
|2x-1|\ge7
.\end{array}
Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above implies
\begin{array}{l}\require{cancel}
2x-1\ge7
\\\\\text{OR}\\\\
2x-1\le-7
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
2x-1\ge7
\\
2x-1+1\ge7+1
\\
2x\ge8
\\
\dfrac{2x}{2}\ge\dfrac{8}{2}
\\
x\ge4
\\\\\text{OR}\\\\
2x-1\le-7
\\
2x-1+1\le-7+1
\\
2x\le-6
\\
\dfrac{2x}{2}\le-\dfrac{6}{2}
\\
x\le-3
.\end{array}
Hence, the solution is $
x\le-3 \text{ OR } x\ge4
$.
The graph of the solution above is shown below.
