jmc

Let $f(x)$ be the quantity on the left-hand side. Constructing a sign table, we get \begin{array}{c|ccc|c} &$2x-7$ &$x-3$ &$x$ &$f(x)$ \\ \hline$x<0$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-$\\ [.1cm]$0<x<3$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$3<x<\frac{7}{2}$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>\frac{7}{2}$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}It follows that $f(x)>0$ when $0<x<3$ or $x>\frac{7}{2}.$ Since the inequality is nonstrict, we must include the values of $x$ such that $f(x)=0,$ which are $x=3$ and $x=\frac{7}{2}.$ Thus, the solution set is $$x\in \boxed{(0,3]\cup [\frac{7}{2},\infty )}.$$