## Intermediate Algebra: Connecting Concepts through Application

$b\le\dfrac{22}{25} \text{ OR } b\ge\dfrac{38}{25}$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|2.5b-3|-1.4\ge-0.6 ,$ isolate first the absolute value expression. Then use the definition of absolute value inequality. Use the properties of inequality to isolate the variable. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given is equivalent to \begin{array}{l}\require{cancel} |2.5b-3|-1.4\ge-0.6 \\\\ |2.5b-3|\ge-0.6+1.4 \\\\ |2.5b-3|\ge0.8 .\end{array} 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} 2.5b-3\ge0.8 \\\\\text{OR}\\\\ 2.5b-3\le-0.8 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} 2.5b-3\ge0.8 \\\\ 10(2.5b-3)\ge10(0.8) \\\\ 25b-30\ge8 \\\\ 25b\ge8+30 \\\\ 25b\ge38 \\\\ b\ge\dfrac{38}{25} \\\\\text{OR}\\\\ 2.5b-3\le-0.8 \\\\ 10(2.5b-3)\le10(-0.8) \\\\ 25b-30\le-8 \\\\ 25b\le-8+30 \\\\ 25b\le22 \\\\ b\le\dfrac{22}{25} .\end{array} Upon checking, any value of the variable in the solution set satisfies the original inequality. Any value of the variable not in the solution set does not satisfy the original inequality. Hence, the solution set is $b\le\dfrac{22}{25} \text{ OR } b\ge\dfrac{38}{25} .$