The inequality to solve is:
[tex]3\vert x-3\vert+8<20[/tex]Let's get the absolute value isolated:
[tex]\begin{gathered} 3|x-3|<20-8 \\ 3|x-3|<12 \\ |x-3|<\frac{12}{3} \\ |x-3|<4 \end{gathered}[/tex]Now we have an absolute value equation. One thing to note here is:
If
|x - a| < b
We can write:
x - a < b
and
x - a > - b
From our equation, we can thus say:
x - 3 < 4
and
x - 3 > -4
Solving 1st:
[tex]\begin{gathered} x-3<4 \\ x<4+3 \\ x<7 \end{gathered}[/tex]and solving 2nd one:
[tex]\begin{gathered} x-3>-4 \\ x>-4+3 \\ x>-1 \end{gathered}[/tex]Together, we can write the solution set as:
[tex]-1