How to graph and solve | 4 x – 3 | + 2 < 11...

The inequality is
${\displaystyle |4x-3|+2<11}$
${\displaystyle |4x-3|-9<0}$
The point to consider is
${\displaystyle 4x-3=0}$
${\displaystyle x=\frac{3}{4}}$
There are ${\displaystyle 2}$ intervals to consider
${\displaystyle (-\infty ,\frac{3}{4})}$ and ${\displaystyle (\frac{3}{4},+\infty )}$
Hence,
In the first interval
${\displaystyle -4x+3-9<0}$
${\displaystyle -4x-6<0}$
${\displaystyle 4x>-6}$
${\displaystyle x>-\frac{3}{2}}$
This is located in the interval.
In the second interval
${\displaystyle 4x-3-9<0}$
${\displaystyle 4x-12<0}$
${\displaystyle x<3}$
This is located in the interval.
The answer is ${\displaystyle x\in (-\frac{3}{2},3)}$
graph{|4x-3|-9 [-20.27, 20.27, -10.14, 10.14]}