Answer
$\text{Set Builder Notation: }
\left\{ t|t\ge-\dfrac{1}{2} \right\}
\\\text{Interval Notation: }
\left[ -\dfrac{1}{2},\infty \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Use the properties of inequality to solve the given inequality, $
t+\dfrac{2}{3}\ge\dfrac{1}{6}
.$ Write the answer in both set-builder notation and interval notation. Finally, graph the solution set.
$\bf{\text{Solution Details:}}$
The $LCD$ of the denominators, $\{
1,3,6
\},$ is $
6
$ since this is the least number that can be evenly divided (no remainder) by all the denominators. Multiplying both sides by the $LCD,$ the given inequality is equivalent to
\begin{array}{l}\require{cancel}
6\left( t+\dfrac{2}{3} \right) \ge6\left( \dfrac{1}{6} \right)
\\\\
6t+4 \ge1
\\\\
6t\ge1-4
\\\\
6t\ge-3
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
6t\ge-3
\\\\
t\ge-\dfrac{3}{6}
\\\\
t\ge-\dfrac{1}{2}
.\end{array}
Hence, the solution set is
\begin{array}{l}\require{cancel}
\text{Set Builder Notation: }
\left\{ t|t\ge-\dfrac{1}{2} \right\}
\\\text{Interval Notation: }
\left[ -\dfrac{1}{2},\infty \right)
.\end{array}
