## Calculus: Early Transcendentals 8th Edition

$\frac{a-c}{b}\leq x\lt \frac{2a-c}{b}$
Solve the inequality $a\leq bx+c\lt 2a$ for $x$, assuming a, b, and c are positive constants. Thus, Subtract $c$: $a-c\leq bx\lt 2a-c$ Divide $b$: $\frac{a-c}{b}\leq x\lt \frac{2a-c}{b}$ Hence, $\frac{a-c}{b}\leq x\lt \frac{2a-c}{b}$