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

$-6\le x \le4$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $19-3|x+1|\ge4 ,$ isolate first the absolute value expression. Then use the definition of less than (less than or equal to) absolute value inequality. Use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the properties of inequality to isolate the absolute value expression results to \begin{array}{l}\require{cancel} 19-3|x+1|\ge4 \\\\ -3|x+1|\ge4-19 \\\\ -3|x+1|\ge-15 .\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} -3|x+1|\ge-15 \\\\ |x+1|\le\dfrac{-15}{-3} \\\\ |x+1|\le5 .\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} -5\le x+1 \le5 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -5\le x+1 \le5 \\\\ -5-1\le x+1-1 \le5-1 \\\\ -6\le x \le4 .\end{array} Hence, the solution set $-6\le x \le4 .$