Find solutions of linear congruences: $x\equiv 0 \pmod 2$ , $x\equiv 0 \pmod 3$, $x\equiv 1 \pmod5$, $x\equiv 6 \pmod7$ 
$x\equiv 0 \pmod 2$
  $x\equiv 0 \pmod 3$
  $x\equiv 1 \pmod5$
  $x\equiv 6 \pmod7$
  Find all the solutions of each of the following systems of linear congruences.

I know how to find solutions of three congruence equations, but
I don't know how to solve the 4 equations system...
I can't find it in my text book but it is in exercise sample... help me pls.

The big idea is that the mapping 
$$f:\mathbb Z_{210} \to 
    \mathbb Z_2 \times \mathbb Z_3 \times\mathbb Z_5 \times\mathbb Z_7$$
(where $210 = 2\cdot3\cdot5\cdot7$) defined by
$f(\bar n)= (\bar n,\bar n,\bar n,\bar n)$ is an isomorphism between additive groups.
You need to find an integer, $n$, such that 
$f(\bar n) = (\bar 0, \bar 0, \bar 1, \bar 6)$.
Because $f$ is an isomorphism, there exists
$e_1, e_2, e_3, e_4 \in \mathbb Z_{210}$ such that
\begin{align}
   f(e_1) &= (\bar 1, \bar 0, \bar 0, \bar 0)\\
   f(e_2) &= (\bar 0, \bar 1, \bar 0, \bar 0)\\
   f(e_3) &= (\bar 0, \bar 0, \bar 1, \bar 0)\\
   f(e_4) &= (\bar 0, \bar 0, \bar 0, \bar 1)
\end{align}
It follows that
$$n \equiv 0 e_1 + 0 e_2 + 1 e_3 + 6 e_4 \equiv 1 e_3 + 6 e_4 \pmod{210}.$$
We compute $e_3$
Because $e_3 \equiv 0$ modulo $2$, modulo $3$, and modulo $7$, then $e_3$ is a multiple of $42 = 2\cdot3\cdot 7$.
So $e_3 \equiv 42x \pmod{210}$ for some integer, $x$.
Because $e_3 \equiv 1 \pmod 5,$
\begin{align}
   42x &\equiv 1 \pmod 5 \\
   2x &\equiv 1 \pmod 5 \\
   x &\equiv 3 \pmod 5
\end{align}
So $e_3 \equiv 42\cdot3 \equiv 126 \pmod{210}$
We compute $e_4$
Because $e_4 \equiv 0$ modulo $2$, modulo $3$, and modulo $5$, then $e_4$ is a multiple of $30 = 2\cdot3\cdot 5$.
So $e_4 \equiv 30x \pmod{210}$ for some integer, $x$.
Because $e_4 \equiv 1 \pmod 7,$
\begin{align}
   30x &\equiv 1 \pmod 7 \\
   2x &\equiv 1 \pmod 7 \\
   x &\equiv 4 \pmod 7
\end{align}
So $e_4 \equiv 30\cdot4 \equiv 120 \pmod{210}$
We compute $n$
\begin{align}
   n
   &\equiv 1 e_3 + 6 e_4 \pmod{210}\\
   &\equiv 1 \cdot 126 + 6 \cdot 120 \pmod{210}\\
   &\equiv 126 + 90 \pmod{210} \\
   &\equiv 6 \pmod{210}
\end{align}
