#### Answer

$-4\lt x \lt 5$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find all $x$ such that $f(x)
\lt16
,$ with $f(x)=
7+|2x-1|
,$ solve the inequality
\begin{array}{l}\require{cancel}
7+|2x-1|\lt16
.\end{array}
Then graph the solution set.
In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
Isolating the absolute value expression results to
\begin{array}{l}\require{cancel}
7+|2x-1|\lt16
\\\\
|2x-1|\lt16-7
\\\\
|2x-1|\lt9
.\end{array}
Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-9\lt 2x-1 \lt9
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-9\lt 2x-1 \lt9
\\\\
-9+1\lt 2x-1+1 \lt9+1
\\\\
-8\lt 2x \lt 10
\\\\
-\dfrac{8}{2}\lt \dfrac{2x}{2} \lt \dfrac{10}{2}
\\\\
-4\lt x \lt 5
.\end{array}
Hence, the solution set $
-4\lt x \lt 5
.$