Answer
no solution
Work Step by Step
Using the properties of inequality, the given, $
\dfrac{1}{4}|x-3|+2\lt1
,$ is equivalent to
\begin{align*}\require{cancel}
\dfrac{1}{4}|x-3|+2-2&\lt1-2
\\\\
\dfrac{1}{4}|x-3|&\lt-1
\\\\
4\left(\dfrac{1}{4}|x-3| \right)&\lt(-1)4
\\\\
|x-3|&\lt-4
.\end{align*}
The left side of the expression above always results to a nonnegative number for any value of $x.$
This is because the absolute value of a number is either $0$ or some positive value.
This value will never be less than the negative value at the right.
Hence, there are no values of $x$ that will satisfy the inequality above.
Therefore, there is no solution.
