## Intermediate Algebra (12th Edition)

$\left( -\infty,20 \right] \cup \left[ 30,\infty \right)$
Using the properties of inequality, the given statement, $|0.1x-2.5|+0.3\ge0.8 ,$ is equivalent to \begin{array}{l}\require{cancel} |0.1x-2.5|\ge0.8-0.3 \\\\ |0.1x-2.5|\ge0.5 .\end{array} Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ OR $x\lt-a$, then the inequality above is equivalent to \begin{array}{l}\require{cancel} 0.1x-2.5\ge0.5 \text{ OR } 0.1x-2.5\le-0.5 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 0.1x-2.5\ge0.5 \\\\ 10(0.1x-2.5)\ge(0.5)10 \\\\ x-25\ge5 \\\\ x\ge5+25 \\\\ x\ge30 \\\\\text{ OR }\\\\ 0.1x-2.5\le-0.5 \\\\ 10(0.1x-2.5)\le(-0.5)10 \\\\ x-25\le-5 \\\\ x\le-5+25 \\\\ x\le20 .\end{array} Hence, the solution set is the interval $\left( -\infty,20 \right] \cup \left[ 30,\infty \right) .$