Answer
$\left[ \dfrac{8}{3},\infty \right)$
Work Step by Step
Using the properties of inequality, the expression $
\dfrac{3x+2}{18}-\dfrac{1+2x}{6}\le-\dfrac{1}{2}
$ simplifies to
\begin{array}{l}
18\left( \dfrac{3x+2}{18}-\dfrac{1+2x}{6} \right) \le \left( -\dfrac{1}{2} \right)18\\\\
1(3x+2)-3(1+2x)\le9(-1)\\
3x+2-3-6x\le-9\\
-3x-1\le-9\\
-3x\le-8\\
x\ge\dfrac{8}{3}
.\end{array}
In interval notation, the solution set is $
\left[ \dfrac{8}{3},\infty \right)
$.
