x+$\frac{2x-1}{5}$ = 3+$\frac{3-x}{4}$

⇔ $\frac{5x}{5}$  + $\frac{2x-1}{5}$ = $\frac{12}{4}$  + $\frac{3-x}{4}$

⇔ $\frac{5x+2x-1}{5}$  = $\frac{12+3-x}{4}$ 

⇔ $\frac{7x-1}{5}$ = $\frac{12+3-x}{4}$ 

⇔ $\frac{4(7x-1)}{5×4}$ = $\frac{5(15-x)}{4×5}$ 

⇔ 28x-4 = 75-5x

⇔ 33x = 79

⇔ x = $\frac{79}{33}$ 

Vậy x= $\frac{79}{33}$