[tex]\it x\Big(1+\dfrac{1}{3}\Big)+2\leq\dfrac{10}{3} \Rightarrow x\cdot \dfrac{4}{3}+2\leq\dfrac{10}{3}|_{\cdot3} \Rightarrow 4x+6\leq10|_{-6} \Rightarrow \\ \\ \\ \Rightarrow 4x\leq4|_{:4} \Rightarrow x\leq 1\ \stackrel{x\in\mathbb{N}}{\Longrightarrow}\ \ x\in\{0,\ 1 \}[/tex]
