Simple Modulo Question $6x = 9 \pmod{11}$ I am trying to solve $6x = 9\pmod{11}$ 
The solution suggests notice that $2*6 = 12 = 1 \pmod {11}$
$$6x = 9 \pmod {11}\\12x = 18 \pmod {11}\\x = 7 \pmod{11}$$
I don't get the final step from $12x = 18\pmod{11}$ to the solution.
 A: The fact that $2\cdot 6 \equiv 12 \equiv 1 \pmod {11}$ tells you that $2$ is the multiplicative inverse of $6 \pmod{11}$, i.e., $\;2\cdot 6 = 12 \equiv 1 \pmod{11}$
$$2\cdot 6 x\equiv 2\cdot 9 \pmod{11} \implies 1\cdot x \equiv 18 \pmod{11}$$ $$\iff \underbrace{x\equiv 18 - 11 \pmod{11} \iff x\equiv 7 \pmod{11}}_{\large *}$$
$(*)$ In short, we see that $18 \equiv 7 \pmod{11},\quad \text{since} 11 \mid (18 - 7) \;\text{i.e.} \;\; 11\mid 11$

Further notes:
It's also true that while $x = 7$ is a solution, $7$ here represents the residue class modulo $11$, which we can denote by $[7]$. $$x = [7] = \{11n + 7\mid n \in \mathbb Z\}$$
There are 11 residue classes modulo $11$:
$\{[0], [1], [2], \cdots , [10]\}$.
For $[k]$ modulo $11$, $0 \leq k \leq 10$, the elements of the class include all integers satisfying: $[k] = \{11n + k\mid n \in \mathbb Z\}$.
A: You should note that $12\equiv1\pmod{11}$ and $18\equiv7\pmod{11}$. That is why the last step holds.
A: $6x \equiv 9 \mod 11 \implies 11k+9=6x \implies 22k+18=12x$ call it ($i$)
$12x \equiv x \mod 11 \implies 11m+x=12x$ and call this ($ii$)
Equate $(i)$ and $(ii)$.
$22k+18 =11m+x \implies 11(2k-m+1)+7=x \implies 11j+7$
A: $6x \equiv 9 \pmod{11} \Rightarrow 2x\equiv 3 \pmod{11}  (\because (3,11)=1) \Rightarrow 2x\equiv 14 \pmod{11} (\because 14\equiv 3 \pmod{11}) \Rightarrow x\equiv 7 \pmod{11}.$  
