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: $x \equiv 2 \pmod{11}$?

It just jumped out at me that if I define $y=x+1$ the first two become 
$$3y \equiv 0 \pmod 5\\3y \equiv 0 \pmod 7$$
which clearly calls for $y$ to be a multiple of $35$.  Now we have $$3y \equiv 9 \pmod {11}\\y \equiv 3 \pmod {11}$$
We note that $35 \equiv 2 \pmod {11}$ so $7 \cdot 35 = 245 \equiv 7\cdot 2 \equiv 14\equiv 3 \pmod{11}$ and $y=245, x=244$ is a solution.
