## Intermediate Algebra (12th Edition)

$\left[ -9,\infty \right)$
$\bf{\text{Solution Outline:}}$ Use the concepts of inequalities to translate the given description, \begin{array}{l}\require{cancel}\text{ one third of a number is added to 6, giving a result of at least 3 ,}\end{array} into symbols. Then solve using the properties of inequality. Express the solution set in interval notation. $\bf{\text{Solution Details:}}$ In symbols, the given description translates to \begin{array}{l}\require{cancel} 6+\dfrac{1}{3}x\ge3 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 3\left( 6+\dfrac{1}{3}x \right)\ge3(3) \\\\ 18+x\ge9 \\\\ x\ge9-18 \\\\ x\ge-9 .\end{array} In interval notation, the solution set is $\left[ -9,\infty \right) .$