#### Answer

$\dfrac{-(b+c)}{a}\lt x \lt \dfrac{c-b}{a}$

#### Work Step by Step

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, $\dfrac{-(b+c)}{a}\lt x \lt \dfrac{c-b}{a}$