## Intermediate Algebra (12th Edition)

$\left( 3,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $-(4+r)+2-3r\lt-14 ,$ use the properties of inequality. 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 and the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -4-r+2-3r\lt-14 \\\\ -r-3r\lt-14+4-2 \\\\ -4r\lt-12 .\end{array} Dividing both sides by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} r\gt\dfrac{-12}{-4} \\\\ r\gt3 .\end{array} The red graph is the graph of the solution set. In interval notation, the solution set is $\left( 3,\infty \right) .$