Simplify the give expression as follows.

\(\displaystyle{3}+{\left|{2}{x}-{7}\right|}\ge{5}\)

\(\displaystyle{3}+{\left|{2}{x}-{7}\right|}-{3}\ge{5}-{3}\) (subtract 3 from both sides)

\(\displaystyle{\left|{2}{x}-{7}\right|}\ge{2}\)

by the absolute rule, if \(\displaystyle{\left|{u}\right|}\ge{a}\), \(a>0\) then \(\displaystyle{\left|{u}\right|}\le-{a}{\quad\text{and}\quad}{\left|{u}\right|}\ge{a}\)

\(\displaystyle{2}{x}-{7}\le-{2}{\quad\text{and}\quad}{2}{x}-{7}\ge{2}\) (\(\displaystyle\because\) by absolute rule)

\(\displaystyle{2}{x}-{7}+{7}\le-{2}{\quad\text{and}\quad}{2}{x}-{7}+{7}\ge{2}+{7}\) (add 7 on both sides)

\(\displaystyle\frac{{{2}{x}}}{{2}}\le\frac{{5}}{{2}}{\quad\text{and}\quad}\frac{{{2}{x}}}{{2}}\ge\frac{{9}}{{2}}\) (divie by 2 onn both sides)

\(\displaystyle{x}\le\frac{{5}}{{2}}{\quad\text{and}\quad}{x}\ge\frac{{9}}{{2}}\)

Therefore, the solution set of the given inequaliry is \(\displaystyle{\left(-\infty,\frac{{5}}{{2}}\right]}\cup{\left[\frac{{9}}{{2}},\infty\right)}\).