How to solve it in a generalised way? Recently I encountered a sum which is as follows:

Let a number when divided by 9 gives remainder 1 , when divided by 11 gives remainder 2 and when divided by 13 gives remainder 3. so Find ot that number ?

Is there a general method to solve this sum ?
Also if we want to generalize this sum by placing three numbers with 9,11 and 13. How can we do it ? (I think it might be broad question) But How to generalize it ? Hints or hints toward a particular method will be appreciated. Thanks. 
My attempt:
I have a quite idea that it might be able to solve using chinese remainder theorem. But I am not able to successfully use it. 
 A: Your system is has congruences $x \equiv a_i \pmod m_i$:
\begin{align}
x & \equiv 1 \pmod{9} \\
x & \equiv 2 \pmod{11} \\
x & \equiv 3 \pmod{13}
\end{align}
We apply the Chinese remainder theorem (link):
Then $\DeclareMathOperator{lcm}{lcm}M = \lcm(9,11,13) = 1287$.
$M_1 = M/m_1 = 1287/9=143$, 
$M_2 = M/m_2 = 1287/11 = 117$, $M_3 = M/m_3 = 1287/13 = 99$.
Then we use the extended Euclidean algorithm to solve
$$
s_i m_i + t_i M_i = 1
$$
which gives $(s_1, t_1) = (16,-1)$, $(s_2, t_2) = (32,-3)$ and $(s_3, t_3) = (-38,5)$.
For $e_i = t_i M_i$ we then get $e_1 = -143$, $e_2 = -351$, $e_3 = 495$.
A solution is
$$
x = \sum_i a_i e_i = 640
$$
All solutions are
$$
x = 640 + k \, 1287 \quad (k \in \mathbb{Z})
$$
A: The problem has the form
$$x\equiv
\begin{cases}
a\mod A\\
b\mod B\\
c\mod C\\
\end{cases}$$
so, the Chinese remainder theorem seems clearly called for.  However, in this case, the numbers $a,b,c$ are in arithmetic progression, as are the numbers $A,B,C$.  That makes it possible to simplify things a bit before doing any explicit arithmetic with actual numbers.
Writing $a=b-d$ and $c=b+d$ for one progression and $A=B-D$ and $C=B+D$ for the other we have
$$u=x-b\equiv
\begin{cases}
-d\mod(B-D)\\
0\mod B\\
d\mod(B+D)
\end{cases}$$
The central congruence means we have $u=kB$ for some $k$.  Noting that we can write $kB$ as $k(B-D)+kD$ and $k(B+D)-kD$, we find that there is a single congruence to solve:
$$kD\equiv-d\mod(B^2-D^2)$$
(To be more precise, the modulus is lcm$(B-D,B+D)$, but in the OP's problem $D=2$ and $B$ is odd.)  Applied to the OP's problem we have
$$2k\equiv-1\mod117$$
which has $k=58$ as one obvious solution, from which we find $u=11k=638$ and thus $x=u+2=640$.  
A: Follows immediately by CCRT for congruences in A.P. (Arithmetic Progression), i.e.
$$\begin{align}x \equiv 1\!+\!j\!\!\!\pmod{\color{#c00}{9\!+\!2j}}\! \iff &\, 2x \equiv 2\!+\!\color{#c00}{2j} \equiv 2\!\color{#c00}{-\!9} \equiv -7\!\!\!\pmod{\!1287 \!=\! 9(11)13}\\[.3em]
\iff &\, x \equiv \dfrac{-7}2\equiv\dfrac{1287\!-\!7}2\equiv 640 \end{align}$$
