## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$-6\lt a \lt0$ $\bf{\text{Solution Outline:}}$ To solve the given inequality, $25-2|a+3|\gt19 ,$ isolate first the absolute value expression. Then use the definition of a less than (less than or equal to) 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 to isolate the absolute value expression results to \begin{array}{l}\require{cancel} 25-2|a+3|\gt19 \\\\ -2|a+3|\gt19-25 \\\\ -2|a+3|\gt-6 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol), the inequality above is equivalent to \begin{array}{l}\require{cancel} -2|a+3|\gt-6 \\\\ |a+3|\lt\dfrac{-6}{-2} \\\\ |a+3|\lt3 .\end{array} Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -3\lt a+3 \lt3 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -3\lt a+3 \lt3 \\\\ -3-3\lt a+3-3 \lt3-3 \\\\ -6\lt a \lt0 .\end{array} Hence, the solution set $-6\lt a \lt0 .$