#### Answer

$\left[ 1,\infty \right)
$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
4-6(x+3) \le -2-3(x+6)+3x
,$ use the Distributive property and the properties of inequality 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 $\ge.$
$\bf{\text{Solution Details:}}$
Using the Distributive Property, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
4-6(x)-6(3) \le -2-3(x)-3(6)+3x
\\\\
4-6x-18 \le -2-3x-18+3x
\\\\
-6x-14 \le -20
.\end{array}
Using the properties of inequality to isolate the variable results to
\begin{array}{l}\require{cancel}
-6x \le -20+14
\\\\
-6x \le -6
.\end{array}
Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-6x \le -6
\\\\
\dfrac{-6x}{-6} \ge \dfrac{-6}{-6}
\\\\
x \ge 1
.\end{array}
In interval notation, the solution set is $
\left[ 1,\infty \right)
.$
The colored graph is the graph of the solution set.