$5\le 3+|2x-7|$
Simplify the expression
$3+|2x-7|\ge 5$ $3+|2x-7|-3\ge 5-3$ (subtract 3 from both sides)
$|2x-7|\ge 2$
By the absolute rule, if $\left|u\right|\ge a,a>0$ then $\left|u\right|\le -a$ and $\left|u\right|\ge a$ $2x-7\le -2$ and $2x-7\ge 2$ (by the absolute rule)
$2x-7+7\le -2+7$ and $2x-7+7\ge 2+7$ (add 7 on both sides)
$2x\le 5$ and $2x\ge 9$ $\frac{2x}{2}\le \frac{5}{2}$ and $\frac{2x}{2}\ge \frac{9}{2}$ (divide by 2)
$x\le \frac{5}{2}$ and $x\ge \frac{9}{2}$
Thus, the solution is $(-\mathrm{\infty},\frac{5}{2}]\cup [\frac{9}{2},\mathrm{\infty})$