Answer
The solution set of equation $\frac{2x-3}{4}=\frac{x-4}{2}-\frac{x+1}{4}$ is $\left\{ -\left. 6 \right\} \right.$.
Work Step by Step
Consider the provided equation:
$\frac{2x-3}{4}=\frac{x-4}{2}-\frac{x+1}{4}$
Multiply both sides by $4$
$\begin{align}
& \frac{2x-3}{4}\cdot 4=\frac{x-4}{2}\cdot 4-\frac{x+1}{4}\cdot 4 \\
& 2x-3=2\left( x-4 \right)-\left( x+1 \right) \\
& 2x-3=2x-8-x-1 \\
\end{align}$
Take the terms containing x on the left hand side and constant terms on the right side
$\begin{align}
& 2x-2x+x=-8-1+3 \\
& x=-6
\end{align}$
Check:
For $x=-6$
Put $x=-6$ in $\frac{2x-3}{4}=\frac{x-4}{2}-\frac{x+1}{4}$
$\begin{align}
& \frac{2\left( -6 \right)-3}{4}\overset{?}{\mathop{=}}\,\frac{-6-4}{2}-\frac{-6+1}{4} \\
& \frac{-12-3}{4}\overset{?}{\mathop{=}}\,\frac{-10}{2}-\frac{-5}{4} \\
& \frac{-15}{4}\overset{?}{\mathop{=}}\,\frac{-20-\left( -5 \right)}{4} \\
& \frac{-15}{4}=\frac{-15}{4}
\end{align}$
Which is true.
Therefore, the solution set of the equation $\frac{2x-3}{4}=\frac{x-4}{2}-\frac{x+1}{4}$ is $\left\{ -\left. 6 \right\} \right.$.
