## Intermediate Algebra for College Students (7th Edition)

$\dfrac{-(b+c)}{a}\lt x \lt \dfrac{c-b}{a}$
Given: $|ax+b| \lt c$ As per definition of absolute value, we can write this as: or, $-c \lt ax+b \lt c$ or,$-c-b \lt ax \lt c-b$ This implies that: $\dfrac{-(b+c)}{a}\lt x \lt \dfrac{c-b}{a}$ Hence, $\dfrac{-(b+c)}{a}\lt x \lt \dfrac{c-b}{a}$