## Thomas' Calculus 13th Edition

$x\displaystyle \leq-\frac{1}{3}$
Apply rules from the table "Rules for Inequalities" $2x-\displaystyle \frac{1}{2}\geq 7x+\frac{7}{6}\quad$ ... multiply with LCD=6, (apply rule 3) $12x-3\geq 42x+7\quad$... add $-12x-7$ (rules 1 and 2) $-3-7\geq 42x-12x$ $-10\geq 30x\quad$.... multiply with $c=\displaystyle \frac{1}{30}\quad$ (positive, rule 3) $-\displaystyle \frac{1}{3}\geq x$ $x\displaystyle \leq-\frac{1}{3}$ This inequality describes a closed interval, $x\displaystyle \in(-\infty,-\frac{1}{3}]$ See table "Types of intervals."