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