Answer
$-\dfrac{8}{3}\lt y \lt\dfrac{10}{3}$
Work Step by Step
Using the properties of inequality, the given, $
|6y-2|+4\lt22
,$ is equivalent to
\begin{align*}\require{cancel}
|6y-2|+4&\lt22
\\
|6y-2|+4-4&\lt22-4
\\
|6y-2|&\lt18
.\end{align*}
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 implies
\begin{align*}\require{cancel}
-18\lt 6y-2 &\lt18
.\end{align*}
Using the properties of inequality, the inequality above is equivalent to
\begin{align*}\require{cancel}
-18+2\lt 6y-2+2 &\lt18+2
\\
-16\lt 6y &\lt20
\\\\
-\dfrac{16}{6}\lt \dfrac{6y}{6} &\lt\dfrac{20}{6}
\\\\
-\dfrac{8}{3}\lt y &\lt\dfrac{10}{3}
.\end{align*}
Hence, the solution is $
-\dfrac{8}{3}\lt y \lt\dfrac{10}{3}
.$
Since a hollowed dot is used for the symbols $\lt$ and $\gt,$ while a solid dot is used for the symbols $\le$ and $\ge,$ then the graph of the solution above is the set of numbers from $
-\dfrac{8}{3}
$ to $
\dfrac{10}{3}
$ with hollowed dots at $
-\dfrac{8}{3}
$ and $
\dfrac{10}{3}
$.
