Find an integer $x$ satisfying the congruence: $$x \equiv \ 1 \pmod3$$
$$x \equiv \ 2 \pmod5$$
$$x \equiv \ 8 \pmod{11}$$
From the first, I have $x=3k+1$, $x=5j+2$ from the second and $x=11l+8$ from the third.
Subbing the third into the second I get  
$11l+8 \equiv 2 \pmod5$
$l \equiv -6\pmod5$
$l \equiv -1\pmod5$  
So $l=5m-1$
Subbing back into the third equation I had, I get
$x=11(5m-1)+8=55m-3$
Thus $x \equiv -3\pmod{55}$.
I was only asked to find any integer x which satisfies the system of congruences, so any of x of that form should work; i.e. $x=52$.  
I want to know if this is the smallest such $x$ which would work, or could I have approached the question in a different way and arrived at another answer?    
If I'd been asked to find the smallest $x$, or all $x$, I'm not sure if this approach would have worked. Is there a way to confirm that this is the smallest such $x$, or that all $x$ of the form $x \equiv -3\pmod{55}$ are the only $x$ which satisfy the system?

For small moduli, I came up with this method. Make the following table.
\begin{array}{r|ccc}
       & \mod 3 & \mod 5 & \mod{11}\\
    \hline
    55 & 1 & 0 & 0\\
    33 & 0 & 3 & 0\\
    15 & 0 & 0 & 4\\
   \hline
   103 & 1 & 3 & 4
\end{array}
Let $N = 3 \times 5 \times 11 = 165$. Then 
$55=\dfrac N3,\; 
33= \dfrac N5;$ and 
$15 = \dfrac N{11}$.
 The table consists of those three numbers modulo $3, 5, $ and $11.$ We need for the diagonal, $[1\; 3\; 4],$ to become $[1\; 2\; 8],$
We can convert $3 \pmod 5$ to $2 \pmod 5$ by multiplying $33$ by $-1$.
We can convert $4 \pmod{11}$ to $8 \pmod{11}$ by multiplying $15$ by $2$.
\begin{array}{r|ccc}
       & \mod 3 & \mod 5 & \mod{11}\\
    \hline
     55 & 1 & 0 & 0\\
    -33 & 0 & 2 & 0\\
     30 & 0 & 0 & 8\\
   \hline
   52 & 1 & 2 & 8
\end{array}
The number we seek is
$55-33+30 \pmod{3 \times 5 \times 11} = 52 \pmod{165}$
By the CRT, $52$ is the smallest such non negative number.
The smallest such negative number would therefore be $-113$.
