## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\text{Set Builder Notation: } \left\{ t|t\ge-\dfrac{1}{2} \right\} \\\text{Interval Notation: } \left[ -\dfrac{1}{2},\infty \right)$
$\bf{\text{Solution Outline:}}$ 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}