Answer
$-\dfrac{11}{15}\le t\le\dfrac{17}{15}$
Work Step by Step
Using the properties of inequality, the given, $
3|5t-1|+9\le23
,$ is equivalent to
\begin{array}{l}\require{cancel}
3|5t-1|+9-9\le23-9
\\
3|5t-1|\le14
\\\\
\dfrac{3|5t-1|}{3}\le\dfrac{14}{3}
\\\\
|5t-1|\le\dfrac{14}{3}
.\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 implies
\begin{array}{l}\require{cancel}
-\dfrac{14}{3}\le5t-1\le\dfrac{14}{3}
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-\dfrac{14}{3}+1\le5t-1+1\le\dfrac{14}{3}+1
\\\\
-\dfrac{14}{3}+\dfrac{3}{3}\le5t\le\dfrac{14}{3}+\dfrac{3}{3}
\\\\
-\dfrac{11}{3}\le5t\le\dfrac{17}{3}
\\\\
-\dfrac{11}{3(5)}\le\dfrac{5t}{5}\le\dfrac{17}{3(5)}
\\\\
-\dfrac{11}{15}\le t\le\dfrac{17}{15}
.\end{array}
Hence, the solution is $
-\dfrac{11}{15}\le t\le\dfrac{17}{15}
.$
The graph of the solution above is shown below.
