<s>[INST] solve the set of simultaneous congruences using Chinese Remainder Theorem 2x= 1 (mod 5), 3x= 9 (mod 6), 4x = 1 (mod 7), 5x = 9 (mod 11) Solve the set of simultaneous congruences using Chinese Remainder Theorem
$\begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 9 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \\
\end{cases}$
This is what I got so far:
I simplify:
$2x \equiv 1 \pmod 5$ becomes $x \equiv 3 \pmod 5$
$3x \equiv 9 \pmod 6$ has $3$ solutions since $\gcd(3,6)=3$ they are $x = 17$, $19$, or $21 \pmod 6$.
$4x \equiv 1 \pmod 7$ becomes $x \equiv 2 \pmod 7$
$5x \equiv 9 \pmod {11}$ becomes $x \equiv 4 \pmod {11}$
By CRT, and substituting the $3$ different solutions, I got:
$x \equiv 653 \pmod{2310}$, $x \equiv 1423 \pmod{2310}$ and $x \equiv 2193 \pmod {2310}$. 
But, the key answer from the book is $x \equiv 653 \pmod{770}$.
I am struggling to figure out how to get $x = 653 \pmod{770}$. I need help. Thank you. 
 [/INST] $$ \begin{cases}
2x \equiv 1 \pmod{5} \\
3x \equiv 3 \pmod{6} \\
4x \equiv 1 \pmod{7} \\
5x \equiv 9 \pmod{11} \\
\end{cases} \implies \begin{cases}
x \equiv 3 \pmod{5} \\
x \equiv 1 \pmod{2} \\
x \equiv 2 \pmod{7} \\
x \equiv 4 \pmod{11} \\
\end{cases} \implies x \equiv 653 \pmod{770} $$
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