# Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises: 88

$\left( -\infty,\dfrac{31}{5} \right] \cup \left[ \dfrac{39}{5},\infty \right)$

#### Work Step by Step

Using the properties of inequality, the given statement, $|0.5x-3.5|+0.2\ge0.6 ,$ is equivalent to \begin{array}{l}\require{cancel} |0.5x-3.5|\ge0.6-0.2 \\\\ |0.5x-3.5|\ge0.4 .\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} 0.5x-3.5\ge0.4 \text{ OR } 0.5x-3.5\le-0.4 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 0.5x-3.5\ge0.4 \\\\ 10(0.5x-3.5)\ge(0.4)10 \\\\ 5x-35\ge4 \\\\ 5x\ge4+35 \\\\ 5x\ge39 \\\\ x\ge\dfrac{39}{5} \\\\\text{ OR }\\\\ 0.5x-3.5\le-0.4 \\\\ 10(0.5x-3.5)\le(-0.4)10 \\\\ 5x-35\le-4 \\\\ 5x\le-4+35 \\\\ 5x\le31 \\\\ x\le\dfrac{31}{5} .\end{array} Hence, the solution set is the interval $\left( -\infty,\dfrac{31}{5} \right] \cup \left[ \dfrac{39}{5},\infty \right) .$

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