Intermediate Algebra (12th Edition)

$\left[ 1,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $4-6(x+3) \le -2-3(x+6)+3x ,$ use the Distributive property and the properties of inequality to isolate the variable. 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, the inequality above is equivalent to \begin{array}{l}\require{cancel} 4-6(x)-6(3) \le -2-3(x)-3(6)+3x \\\\ 4-6x-18 \le -2-3x-18+3x \\\\ -6x-14 \le -20 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} -6x \le -20+14 \\\\ -6x \le -6 .\end{array} Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} -6x \le -6 \\\\ \dfrac{-6x}{-6} \ge \dfrac{-6}{-6} \\\\ x \ge 1 .\end{array} In interval notation, the solution set is $\left[ 1,\infty \right) .$ The colored graph is the graph of the solution set.