jmc

Subtracting $2$ from both sides, we get $$\frac{x-1}{x-3}-2\ge 0,$$or $$\frac{x-1-2(x-3)}{x-3}=\frac{-x+5}{x-3}\ge 0.$$Negating both sides, we have $$\frac{x-5}{x-3}\le 0.$$Letting $f(x)=\frac{x-5}{x-3},$ we make a sign table with the two factors $x-5$ and $x-3$: \begin{array}{c|cc|c} &$x-5$ &$x-3$ &$f(x)$ \\ \hline$x<3$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$3<x<5$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>5$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}We see that $f(x)\le 0$ when $3<x<5,$ as well as at the endpoint $x=5.$ Therefore, the solution set is the interval $\boxed{(3,5]}.$