Answer
$\left( -\infty, \dfrac{3}{5} \right]
\cup
\left[ 1,\infty \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|8-10x| \ge 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 $\ge.$
For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot 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}
8-10x \ge 2
\\\\\text{OR}\\\\
8-10x \le -2
.\end{array}
Using the properties of inequality to isolate the variable results to
\begin{array}{l}\require{cancel}
8-10x \ge 2
\\\\
-10x \ge 2-8
\\\\
-10x \ge -6
\\\\\text{OR}\\\\
8-10x \le -2
\\\\
-10x \le -2-8
\\\\
-10x \le -10
.\end{array}
Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-10x \ge -6
\\\\
x \le \dfrac{-6}{-10}
\\\\
x \le \dfrac{3}{5}
\\\\\text{OR}\\\\
-10x \ge -10
\\\\
x \ge \dfrac{-10}{-10}
\\\\
x \ge 1
.\end{array}
In interval notation, the solution set is $
\left( -\infty, \dfrac{3}{5} \right]
\cup
\left[ 1,\infty \right)
.$
