A couple of questions about a system of equations with modulus 
Solve the system of equations below:
$$
\left\{ 
\begin{array}{c}
x \equiv 3 \pmod {11} \\ 
x \equiv 4 \pmod {17} \\ 
\end{array}
\right. 
$$

$11u + 17v = 1$
$u = -3$ and $v = 2$
$t = 4 \times 11 \times -3 + 3 \times 17 \times 2 = -30.$
The answer is supposed to be $x = -30 \pmod {187}$. Where does the $\pmod {187}$ part come from? 
$ax + by \neq 1$ doesn't imply $\gcd(a, b),$ correct? But $ax + by = 1 \rightarrow \gcd(a, b),$ right? So, when solving the system above we could use the Euclidean Algorithm to find $u, v.$ Is that correct?
 A: I don't know if I understand your whole question, especially the last two paragraphs, correctly. But I see that you are using the constructive method of Chinese remainder theorem.
The $187$ is of course $11\times 17$. It is easy to check that, if $x$ is a solution to the system of modular equations, then $(x+11\times 17n)$ is also a solution for integers $n$.
And let there are two solutions $x,y$ to the system of equations. Consider $x-y$. Then
$$\begin{align*}
x-y&\equiv 3-3 = 0 \pmod{11}\\
x-y&\equiv 4-4 = 0 \pmod{17}\\
\end{align*}$$
and $x-y$ is a common multiplier of $11$ and $17$. Since $11$ and $17$ are coprime, the smallest positive difference of any two solutions is $187$.
A: Let one of the roots of this equation be $x_0$ and let another root of this equation be $x_1$.
$x_0 = 3($mod $11) \Rightarrow x_0=11p+3$ for $p \in Z $
and $x_1 = 3($mod $11) \Rightarrow x_1=11q+3$.
Thus $x_1-x_0 = 11(q-p)$ - a multiple of $11$.
Also $x_0 = 4($mod $17) \Rightarrow x_0=17r+4$ 
and $x_1 = 4($mod $17) \Rightarrow x_1=17s+4$.
Thus $x_1-x_0 = 17(s-r)$ - a multiple of $17$.
Combined $x_1-x_0$ must be a multiple of $17 \times 11=187$.
$x_1-x_0 = 187k \Rightarrow x_1=187k+x_0 \Rightarrow x_1 = x_0($mod $187)$
