Answer
$x\lt-12\text{ OR }x\gt6$
Refer to the graph below.
Work Step by Step
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 given inequality, $
|x+3|\gt9
,$ implies
\begin{align*}\require{cancel}
x+3&\gt9
\\\\\text{OR}\\\\
x+3&\lt-9
.\end{align*}
Using the properties of inequality, the inequality above is equivalent to
\begin{align*}\require{cancel}
x+3&\gt9
\\
x+3-3&\gt9-3
\\
x&\gt6
\\\\\text{OR}\\\\
x+3&\lt-9
\\
x+3-3&\lt-9-3
\\
x&\lt-12
.\end{align*}
Hence, the solution is $
x\lt-12\text{ OR }x\gt6
.$
Since a hollowed dot is used for the symbols $\lt$ and $\gt,$ while a solid dot is used for the symbols $\le$ and $\ge,$ then the graph of the solution above is the set of numbers to the left of $
-12
$ and to the right of $
6
$ with hollowed dots at $
-12
$ and $
6
$.
