#### Answer

$\left( -\dfrac{11}{2},5 \right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|4x+1| \lt 21
,$ use the definition of absolute value inequalities. Use the properties of inequalities to isolate the variable.
For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$
For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
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}
-21 \lt 4x+1 \lt 21
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-21-1 \lt 4x+1-1 \lt 21-1
\\\\
-22 \lt 4x \lt 20
\\\\
-\dfrac{22}{4} \lt \dfrac{4x}{4} \lt \dfrac{20}{4}
\\\\
-\dfrac{11}{2} \lt x \lt 5
.\end{array}
In interval notation, the solution set is $
\left( -\dfrac{11}{2},5 \right)
.$
The colored graph is the graph of the solution set.