Solving equations with mod So, I'm trying to solve the following equation using regular algebra, and I don't think I'm doing it right: $3x+5 = 1\pmod {11}$
I know the result is $x = 6$, but when I do regular algebra like the following, I do not get 6:
$3x=1  - 5\pmod{11}$
$x = \dfrac{(1 \pmod{11} - 5)} 3$
So, I figured that since $1 \pmod{11} = 1$ the equation becomes
$x = \dfrac{-4} 3$
Which is not 6! I am totally lost here and would appreciate any help......

The division in $\mathbb{F_{11}}$ is not like the division in $\mathbb{Q}$ !!! If you want to find $3(mod \; 11)^{-1}$ think of which element of $\mathbb{Z_{11}}$ multiplied by $3$ gives $1 \; (mod \; 11)$? It's $4$. So $$3x \equiv -4 \; (mod \; 11) \Leftrightarrow x \equiv -4*4 \; (mod \; 11)$$
And $-16 \equiv 6 \; (mod \; 11)$, because $11$ divides $-16 - 6 = -22$.
