## Algebra 2 (1st Edition)

$|x-45|\lt15$
If $a\lt x\lt b$, we can turn this into an absolute value inequality in the following way: Take the mean of $a$ and $b$: $\frac{a+b}{2}$. This will be equidistant from $a$ and $b$. Thus if we take x's difference from this, it must be within $|a-\frac{a+b}{2}|=|b-\frac{a+b}{2}|$. Thus the inequality will be: $|x-\frac{a+b}{2}|\lt|a-\frac{a+b}{2}|$. Hence here the inequality is: $|x-\frac{30+60}{2}|\lt|30-\frac{30+60}{2}|\\|x-45|\lt15$