Precalculus (6th Edition) Blitzer

The solution set of equation $\frac{2x-3}{4}=\frac{x-4}{2}-\frac{x+1}{4}$ is $\left\{ -\left. 6 \right\} \right.$.
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 $\left\{ -\left. 6 \right\} \right.$.