<s>[INST] Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem, 
I started with the 'highest' divisor: $11$. Since $(3, 11) =1$ there is a unique solution. $x= 6 \cdot 3 ^{\phi (11) -1} \equiv 6 \cdot 3^4\pmod{11}$ But to be honest, I have no clue how to continue.
Perhaps, cancel out the last equation to: $x \equiv 2 \pmod{11}$?
 [/INST] This is an old-fashioned solution.
$$3x \equiv 2 \mod 5 \qquad 3x \equiv 4 \mod 7 \qquad 3x \equiv 6 \mod{11}$$
\begin{array}{|r|rrr|}
\hline
 385 &  5 &  7 & 11 \\
\hline
  77 &  2 &  0 &  0 \\
  55 &  0 &  6 &  0 \\
  35 &  0 &  0 &  2 \\
\hline
 231 &  1 &  0 &  0 \\
 -55 &  0 &  1 &  0 \\
-175 &  0 &  0 &  1 \\
\hline
\end{array}
$$3x \equiv 2(231) -4(55) -6(175) \equiv -808 + 385^{\#} \equiv -423 \pmod {385}$$
$$x \equiv -141 \equiv 244 \pmod {385}$$
($\#:$ Adding $385$ makes the answer, $-423$, a multiple of $3$.)
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