Modular Arithmetic: Congruence System

Question

How to solve this congruence system: \begin{cases} x≡2 &\bmod3 \\ 2x≡1& \bmod5 \\ 3x≡3 &\bmod6 \end{cases}

Answer 1

$x\equiv 2 \pmod{3}$ is equivalent to say $x=2+3k$ for some $k\in \mathbb{Z}$. Substitute this into the second congruence and you have $$2(2+3k)\equiv 1 \pmod{5}$$ or $$k\equiv 2 \pmod{5}$$ Which means $k=2+5\ell$ and that $x=2+3(2+5\ell)=8+15\ell$ for $\ell\in \mathbb{Z}$. Finally, put this into the last congruence and then $$3(8+15\ell)\equiv 3 \pmod{6}$$ which is simplified as $$\ell\equiv1 \pmod{2}$$ or well $\ell=1+2t$ for some integer $t$. Therefore $x=8+15(1+2t)=23+30t$ or equivalently $$x\equiv 23 \pmod{30}$$ and this is the solution.