how to solve systems of congruence with more than 2 equations $x \equiv 1 \bmod 7$
$x \equiv 2 \bmod 3$
I can solve this with the formula
$x = m_1a_2x+m_2a_1y\ $ mod $(m_1m_2)$
By doing euclid's algorithm I get, $7(1) + 3(-2) = 1$, then
$x = 7(2)(1) + 3(1)(-2)\ $ mod(21)
$x = 8$
When I have more than 2 congruences, how do I solve using this same formula ?

One option is to solve two of them, then solve the result with the next one, etc. Note that your final line there is $x\equiv 8 \bmod 21$.
For systems of more than two congruences you could try the Chinese Remainder Theorem for multiple coprime moduli, although the amount of work is similar (but the intermediate numbers are usually smaller):
Let $m_1,m_2,\ldots,m_k$ be pairwise relatively prime moduli. Then the
system of congruences:
$x \equiv c_1\pmod {m_1} \\
x \equiv c_2\pmod { m_2} \\
\vdots \\
x \equiv c_r \pmod {m_r}$
has a unique solution modulo the product $m = m_1m_2\cdots m_r$ as follows:
Take $s_i := \dfrac{m}{m_i}$.  Since the $m_i$ share no factors, $\gcd(s_i,m_i ) = 1$. Therefore, for each $i$
we can compute $t_i$, the inverse of $s_i \pmod{mi}$  
Then
$x ≡ c_1 s_1 t_1 +  c_2 s_2 t_2 +\cdots + c_r s_r t_r \pmod m $ solves
the system since
$c_1 s_1 t_1 +  c_2 s_2 t_2 +\cdots + c_r s_r t_r \equiv c_i s_i t_i\equiv c_i \pmod {m_i}$ 
Note that for two congruences, it reduces to your formula, since then $s_1 = m_2$ etc.

Example:
$x \equiv \color{blue}1 \bmod 7$
$x \equiv \color{blue}2 \bmod 3$
$x \equiv \color{blue}4 \bmod 5$
$m=7\cdot 3\cdot 5 = 105$
$(s_1,s_2,s_3) = (15,35,21)$
$t_1:\quad  15^{-1}\equiv 1^{-1} \equiv \color{red}1 \bmod 7$
$t_2:\quad  35^{-1}\equiv 2^{-1} \equiv \color{red}2 \bmod 3$
$t_3:\quad  21^{-1}\equiv 1^{-1} \equiv \color{red}1 \bmod 5$
$x= \color{blue}1\cdot 15\cdot \color{red}1 
+ \color{blue}2\cdot 35\cdot \color{red}2 
+ \color{blue}4\cdot 21\cdot \color{red}1 
= 15+140+84  \equiv 29\bmod 105$
