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......

When we are working with congruence, $\text{mod } n$, for some $n$ (here, 11), we are interested in integral relations, i.e., relationships between integer values, and instead of division, we use the fact, in this case, that $3x +5= 11k + 1$, where $k$ is any integer (the quotient), and $1$ is the remainder when $3x +5$ is divided by $11$
$$3x + 5 \equiv 1\pmod {11} \implies  3x+5 -1\equiv 1-1 = 0 \pmod{11}\iff 3x+4 \equiv 0 \pmod{11}$$
So we are looking for $x$ such that $3x+4 = 11k+ 0 = 11k$ for some integer $k$. That means we need for $\dfrac{3x+4}{11}$ to equal the integer $k$. Note that the given solution is one such solution: When $x=6$, then $3x+4 = 18+4 = 22 = 2(11)$. So when $x = 6$, $11$ divides $3x+4$ evenly, with no remainder. 
There are infinitely many solutions $x$ that will meet this requirement, (test out $x = 6-11, x = 6+ 11= 17, x=6+ 2(11)$, etc.) So sometimes people use a representative congruence class $\overline x$, where $x$ is the least non-negative integer solution, and $$\overline x = \{6\pm 11k\mid k\in \mathbb Z\}$$
