system of linear congruences when moduli are not coprime $\begin{cases}x\equiv 1 \pmod{3}\\
x\equiv 2 \pmod{5}\\
x\equiv 3 \pmod{7}\\
x\equiv 4 \pmod{9}\\
x\equiv 5 \pmod{11}\end{cases}$ 
I am supposed to solve the system using the Chinese remainder theorem but $(3,5,7,9,11)\neq 1$ 
How can I transform the system so that I will be able to use the theorem?

Solving the first two equations simultaneously you get X = 7(mod 15).
Solving the third and fourth simultaneously you get X = 31(mod 63).
Solving these two results simultaneously you get X = 157(mod 315).
Solving this result with the fifth equation simultaneously,
you get the final answer X = 1732(mod 3465).
