Wrong applying of simple Chinese Remainder Theorem problem What am I doing wrong?
So for the following equations
$$
\begin{align}
(*) \left\{
  \begin{array}{l}
    2x\equiv 3\pmod 5 \\
    4x\equiv 2\pmod 6 \\
    3x\equiv 2\pmod 7
  \end{array} \right.
\end{align}
$$
and $N =\mathrm{lcm}\langle5,6,7\rangle = 210$, giving
$N_1 = \frac{210}{5} = 42, \enspace N_2 = \frac{210}{6} = 35, \enspace N_3 = \frac{210}{7} = 30 $.
$$
\begin{align}
42z_1&\equiv 1\pmod 5\Rightarrow\enspace\enspace\;2z_1\equiv 1\pmod 5\Rightarrow\enspace &&\overline{z_1}=\overline{3}\\
35z_2&\equiv 1\pmod 6\Rightarrow\enspace-1z_2\equiv 1\pmod 6\Rightarrow\enspace &&\overline{z_2}=\overline{-1}\\
30z_3&\equiv 1\pmod 7\Rightarrow\enspace\enspace\;2z_3\equiv 1\pmod 7\Rightarrow\enspace &&\overline{z_1}=\overline{4}
\end{align}
$$
So the solution should be
$$
\begin{align}
\overline{x} &= \overline{3\times42\times3} + \overline{2\times35\times(-1)}+\overline{2\times30\times4}\\ 
&= \overline{378-70+240}\\
&= \overline{548}\\
&= \overline{128}
\end{align}
$$
Which is clearly wrong, so I'm wondering which additional steps I need to take to get to the correct answer.
Thanks in advance.
 A: You seem to have solved $$\left\{\begin{array}{c} x\equiv3 \pmod{5}\\x\equiv 2\pmod{6}\\x\equiv 2 \pmod{7}\end{array}\right.$$
But you need to take into account the coefficients on $x$ in each congruence.
$2x\equiv 3\pmod{5}$ means $x\equiv 4\pmod{5}$
$4x\equiv 2\pmod{6}$ means $x\equiv 2 \mbox{ or } 5 \pmod{6}$
$3x\equiv 2\pmod{7}$ means $x\equiv 3\pmod{7}$
A: \begin{align}
    2x &\equiv 3\pmod 5 \\
    4x &\equiv 2\pmod 6 \\
    3x &\equiv 2\pmod 7
\end{align}
First, you need to solve each congruence for $x$.
\begin{align}
    2x &\equiv 3 \pmod 5 \\
    x  &\equiv 4 \pmod 5 \\
   \hline
    4x &\equiv 2 \pmod 6 \\
    2x &\equiv 1 \pmod 3 \\
    x  &\equiv 2 \pmod 3 \\
    \hline
    3x &\equiv 2 \pmod 7 \\
    x  &\equiv 3 \pmod 7
\end{align}
Note that the solution set (modulo 6) to $4x \equiv 2 \pmod 6$ is
$\{2,5\}$ and that both solutions are included in $x \equiv 2 \pmod 3.$
summarizing, we get
\begin{align}
    x  &\equiv 4 \pmod 5 \\
    x  &\equiv 2 \pmod 3 \\
    x  &\equiv 3 \pmod 7
\end{align}
One way to compute the solution is this.
\begin{align}
   x &\equiv 4 \pmod 5 \\
   x &= 5A + 4 \\
   \hline
   x &\equiv 2 \pmod 3 \\
   5A+4 &\equiv 2 \pmod 3 \\
   2A &\equiv 1 \pmod 3 \\
   A &= 2 + 3B\\
   x &= 15B + 14 \\
   \hline
   x  &\equiv 3 \pmod 7 \\
   15B + 14  &\equiv 3 \pmod 7 \\
   B  &\equiv 3 \pmod 7 \\
   B &= 7C + 3 \\
   x &= 105C + 59 \\
   x &\equiv 59 \pmod{105}
\end{align}
Your way of solving is this way.
$N =\mathrm{lcm}\{5,3,7\} = 105$.
$N_1 = 3 \cdot 7z_1 = 21z_1,\enspace
 N_2 = 5 \cdot 7z_2 = 35z_2, \enspace
 N_3 = 5 \cdot 3z_3 = 15z_3 $.
$21z_1 \equiv 1 \pmod 5 \implies
z_1 \equiv 1 \pmod 5
\implies  N_1 = 21$
$35z_2 \equiv 1 \pmod 6 \implies 
z_2 \equiv -1 \pmod 6 \implies
N_2 = -35$
$15z_3 \equiv 1 \pmod 7 \implies
   z_3 \equiv 1 \pmod 7 \implies
N_3 = 15$
So the solution is
\begin{align}
   x &\equiv 4 \cdot 21 - 2\cdot 35 + 3 \cdot 15 \pmod{105}\\
   x &\equiv 59 \pmod{105}
\end{align}
