## Algebra 2 (1st Edition)

$x\leq-26$ or $x\geq-10$ The solutions are all real numbers less than or equal to $-26$ or greater than or equal to $-10$.
$\left|\frac{1}{2}x+9\right|\geq 4\qquad$.... when the form of inequality is $|ax+b|\geq c$, then the equivalent forms are $ax+b\geq c$ or $ax+b\leq-c$. $\displaystyle \frac{1}{2}x+9\geq 4$ or $\displaystyle \frac{1}{2}x+9\leq-4\qquad$....solve each inequality separately. $\displaystyle \frac{1}{2}x+9\geq 4\qquad$ ...add $-9$ to each side. $\displaystyle \frac{1}{2}x+9-9\geq 4-9\qquad$ ...simplify. $\displaystyle \frac{1}{2}x\geq-5\qquad$ ...multiply each side of the inequality with $2$. $x\geq-10$ $\displaystyle \frac{1}{2}x+9\leq-4\qquad$ ...add $-9$ to each side. $\displaystyle \frac{1}{2}x+9-9\leq-4-9\qquad$ ...simplify. $\displaystyle \frac{1}{2}x\leq-13\qquad$ ...multiply each side of the inequality with $2$. $x\leq-26$