## Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $x\leq-9$ and $x\geq7$ GRAPH In the real number line put and -9 with a close cirlce and draw a line towards $-\infty$. Then put a 7 with a close circle and draw a line pointin towards $\infty$.
$\mid\frac{x+1}{2}\mid \geq4$ So you divide it in two to take the absolute value away $\frac{x+1}{2} \leq-4$ Or $\frac{x+1}{2} \geq4$. First let’s solve one at a time $$\frac{x+1}{2} \leq-4$$ $$x+1\leq-8$$ $$x\leq-9$$ Now the second one $$\frac{x+1}{2} \geq4$$ $$x+1\geq8$$ $$x\geq7$$ So we have two solutions $x\leq-9$ and $x\geq7$ GRAPH In the real number line put and -9 with a close cirlce and draw a line towards $-\infty$. Then put a 7 with a close circle and draw a line pointin towards $\infty$.