## Algebra 2 (1st Edition)

$\frac{250}{1.05}\leq x\leq\frac{250}{0.95}$
Using part a), the compound inequality is: $-0.05x\leq x-250\leq0.05x$. Thus the two parts to solve are : 1)$-0.05x\leq x-250\\-1.05x\leq-250\\x\geq\frac{250}{1.05}$ and 2) $x-250\leq0.05x\\0.95x\leq250\\x\leq\frac{250}{0.95}$ Thus $\frac{250}{1.05}\leq x\leq\frac{250}{0.95}$