Answer
$\left( -\infty,-\dfrac{11}{3} \right]\cup[-3,\infty)$
Work Step by Step
Since for any $a\gt0$, $|x|\ge a$ implies $x\ge a$ OR $x\le -a$, then the given inequality, $
|1+0.3x|\ge0.1
,$ is equivalent to
\begin{array}{l}\require{cancel}
1+0.3x\ge0.1
\\\\
0.3x\ge0.1-1
\\\\
0.3x\ge-0.9
\\\\
x\ge-\dfrac{0.9}{0.3}
\\\\
x\ge-3
,\\\\\text{OR}\\\\
1+0.3x\le-0.1
\\\\
0.3x\le-0.1-1
\\\\
0.3x\le-1.1
\\\\
x\le-\dfrac{1.1}{0.3}
\\\\
x\le-\dfrac{11}{3}
.\end{array}
Hence, the solution set is $
\left( -\infty,-\dfrac{11}{3} \right]\cup[-3,\infty)
.$
See graph below.