Answer
$\left[ -\dfrac{1}{2},1 \right]$
Work Step by Step
Using the properties of inequality, the given expression, $
6+|4x-1|\le9
,$ is equivalent to
\begin{array}{l}\require{cancel}
|4x-1|\le9-6
\\\\
|4x-1|\le3
.\end{array}
Since for any $a\gt0$, $|x|\le a$ implies $-a\le x\le a$, then the given inequality, $
|4x-1|\le3
,$ is equivalent to
\begin{array}{l}\require{cancel}
-3\le 4x-1\le3
\\\\
-3+1\le 4x-1+1\le3+1
\\\\
-2\le 4x\le4
\\\\
-\dfrac{2}{4}\le \dfrac{4}{4}x\le\dfrac{4}{4}
\\\\
-\dfrac{1}{2}\le x\le1
.\end{array}
Hence, the solution set is $
\left[ -\dfrac{1}{2},1 \right]
.$
See graph below.