Answer
$\left( -\infty, -\dfrac{5}{3}\right]\cup\left[ \dfrac{11}{3}, \infty\right)
$
Work Step by Step
Since for any non negative $c$, $|x|\ge c$ implies $x\ge c \text{ OR } x\le-c,$ then the solutions to the given equation, $
\left| \dfrac{3(x-1)}{4} \right|\ge2
,$ is
\begin{array}{l}\require{cancel}
\dfrac{3(x-1)}{4}\ge2
\\\\
3(x-1)\ge4(2)
\\\\
3x-3\ge8
\\\\
3x\ge8+3
\\\\
3x\ge11
\\\\
x\ge\dfrac{11}{3}
,\\\\\text{ OR }\\\\
\dfrac{3(x-1)}{4}\le-2
\\\\
3(x-1)\le4(-2)
\\\\
3x-3\le-8
\\\\
3x\le-8+3
\\\\
3x\le-5
\\\\
x\le-\dfrac{5}{3}
.\end{array}
Hence, the solution is $
\left( -\infty, -\dfrac{5}{3}\right]\cup\left[ \dfrac{11}{3}, \infty\right)
.$
