Let f ( x ) = 7 + | 2 x - 1 | ....

Step 1
Given: ${\displaystyle f\left(x\right)=7+|2x-1|}$ and ${\displaystyle f\left(x\right)<16}$
We can write the inequality:
${\displaystyle 7+|2x-1|<16}$
Subtract 7 from both sides:
${\displaystyle |2x-1|<9}$
Because of the piecewise definition of the absolute value function,
${\displaystyle \left|A\right|=\{\begin{array}{l}A;A\ge 0\\ -A;A<0\end{array}}$
we can separate the inequality into two inequalities:
${\displaystyle -(2x-1)<9}$ and ${\displaystyle 2x-1<9}$
Multiply both sides of the first inequality by -1:
${\displaystyle 2x-1>-9}$ and ${\displaystyle 2x-1<9}$
Add 1 to both sides of both inequalities:
${\displaystyle 2x>-8}$ and ${\displaystyle 2x<10}$
Divide both sides of both inequalities by 2:
${\displaystyle x>-4}$ and ${\displaystyle x<5}$
This can be written as:
${\displaystyle -4<x<5}$
To check, I will verify that the end points equal 16:
${\displaystyle 7+\mid 2(-4)-1)|=7+|-9\mid =16}$
${\displaystyle 7+|2\left(5\right)-1|=7+\left|9\right|=16}$