# How do you solve abs( x + 3) <2?

$x < - 1 \mathmr{and} x > - 5$
In interval notation solution is $\left(- 5 , - 1\right)$
|x+3| < 2 ; x+3 < 2 and x+3 > -2 :. x< -1 and x> -5 In interval notation solution is $\left(- 5 , - 1\right)$[Ans]