## Precalculus (6th Edition) Blitzer

a. $|x-4|\lt 3$ b. $|x-4|\geq 3$
a. We can express the interval as $(1,7)$ with a middle point of $x=4$ and a half distance of $\frac{7-1}{2}=3$. Thus, we can write in absolute value form as $|x-4|\lt 3$ b. We can express the interval as $(-\infty,1]\cup[7,\infty)$ with a middle point of $x=4$ and a half distance of $\frac{7-1}{2}=3$. Thus we can write in absolute value form as $|x-4|\geq 3$