Answer
$\left( -\infty, -30 \right)
\cup
\left( 10,\infty \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|0.1x+1| \gt 2
,$ use the definition of absolute value inequalities. Then use the properties of inequality to isolate the variable.
For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
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 is equivalent to
\begin{array}{l}\require{cancel}
0.1x+1 \gt 2
\\\\\text{OR}\\\\
0.1x+1 \lt -2
.\end{array}
Using the properties of inequality to isolate the variable results to
\begin{array}{l}\require{cancel}
0.1x+1 \gt 2
\\\\
0.1x \gt 2-1
\\\\
0.1x \gt 1
\\\\
10(0.1x) \gt 10(1)
\\\\
x \gt 10
\\\\\text{OR}\\\\
0.1x+1 \lt -2
\\\\
0.1x \lt -2-1
\\\\
0.1x \lt -3
\\\\
10(0.1x) \lt 10(-3)
\\\\
x \lt -30
.\end{array}
In interval notation, the solution set is $
\left( -\infty, -30 \right)
\cup
\left( 10,\infty \right)
.$
