#### Answer

$-6\le x \le4$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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
.$