<s>[INST] system of modular equations. $x\equiv 2\pmod3$
$x\equiv 3\pmod 5$
$x\equiv 7 \pmod{11}$
How can I solve this system for $x$? I've tried all kinds of things using divisibility but no success. Any hints of solutions are greatly appreciated. What value of $x$ satisfies these three equations?
 [/INST] When I create new variables, they're integers.
$x\equiv 2\,\pmod{\! 3}\iff x=3k+2$.
$x\equiv 3\,\pmod{\! 5}\iff 3k+2\equiv 3\,\pmod{\! 5}$
$\iff 3k\equiv 1\equiv 6\stackrel{:3}\iff k\equiv 2\,\pmod{\! 5}$.
I.e. $k=5m+2$, then $x=3(5m+2)+2=15m+8$.
Here we could've solved $3k\equiv 1\pmod{\! 5}$ by using Extended Euclidean algorithm (as below) to find $k_1,b\in\Bbb Z$ such that $3k_1+5b=1$, in which case $3k_1\equiv 1\pmod{\! 5}$. 
EEA can be used in a table (as explained here), where e.g. $5=5(1)+3(0)$. We subtract consecutive rows:
$$\begin{array}{l|c|r}5& 5(1)& 3(0)\\\hline 3& 5(0)& 3(1)\\\hline 2& 5(1)& 3(-1)\\\hline 1& 5(-1)& 3(2)\end{array}$$
$5(-1)+3(2)=1$ implies $3(2)\equiv 1\pmod{\! 5}$, i.e. $k\equiv 2\pmod{\! 5}$.
Gauss' algorithm can also be used:
$3k\equiv 1\stackrel{\cdot 2}\iff 6k\equiv 2\iff k\equiv 2\,\pmod{\! 5}$
$x\equiv 7\,\pmod{\! 11}\iff 15m+8\equiv 7\,\pmod{\! 11}$
$\iff 15m\equiv 4m\equiv -1\equiv -12\stackrel{:4}\iff m\equiv -3\equiv 8\,\pmod{\! 11}$
I.e. $m=11a+8$, then $x=15(11a+8)+8=165a+128$.
In the same way we could've used Extended Euclidean or Gauss' algorithm to solve $4m\equiv -1\,\pmod{\! 11}$, like above.
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