$\frac{1}{2}$(2x + 1) ≥ $\frac{1}{3}$x + 1$\frac{9}{10}$

Change the mixed fraction to improper fraction

1$\frac{9}{10}$ = $\frac{\mathrm{1\; x\; 10\; +\; 9}}{10}$ = $\frac{\mathrm{10\; +\; 9}}{10}$ = $\frac{19}{10}$

$\frac{1}{2}$(2x + 1) ≥ $\frac{1}{3}$x + $\frac{19}{10}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators, 1, 3 and 10. The L.C.M is 30.

30 x $\frac{1}{2}$(2x + 1) ≥ 30 x $\frac{1}{3}$x + 30 x $\frac{19}{10}$

15(2x + 1) ≥ 10x + 19 x 3

30x + 15 ≥ 10x + 57

Move all x terms to the left side and constants to the right side:

30x - 10x ≥ 57 - 15

20x ≥ 42

Divide both sides by 20

x ≥ $\frac{42}{20}$

x ≥ $\frac{21}{10}$

x ≥ 2$\frac{1}{10}$