I am confused about how I am to solve simultaneous linear congruences, in the book I am reading I am only given 1 example and no general principle.

Solve

$\displaystyle x\equiv1\mbox{mod}(4)\ \ \ x\equiv2\mbox{mod}(3)\ \ \ \ x\equiv3\mbox{mod}(5)$

From the example i think i have to:

$\displaystyle n_1=4\ \ \ n_2=3\ \ \ \ n_3=5\ \ \ n=60$ and $\displaystyle c_1=15\ \ c_2=20\ \ \ c_3=12$

and so i have:

$\displaystyle 15x\equiv1\mbox{mod}(4)\rightarrow{3x\equiv1\mbox{ mod}(4)$

$\displaystyle 20x\equiv1\mbox{mod}(3)\rightarrow{2x\equiv1\mbox{ mod}(3)$

$\displaystyle 12x\equiv1\mbox{mod}(5)\rightarrow{2x\equiv1\mbox{ mod}(5)$

but after this I am lost (im not sure any of that is correct either,

Thanks very much for any help