Solve the equation $3x+7=6$ in $\mathbb{Z}_{13}$ The title is exercise 2.2 in The Fundamental Theorem of Algebra.  

The hint for the problem is: Find the value of $\frac{1}{3}$ in $\mathbb{Z}_{13}$ 


  (please realize that my knowledge of the subject is what I read in Chapter 2)


I have gotten that $x = -\frac{1}{3}$.  

I know that $\mathbb{Z}_{13}$ is the integers modulo 13.  Thus $x \equiv n \pmod{13}$? for some integer n,  $0\le n &lt; 13$.  

Thus for $x – n = km$ for some integer $k$, with $m \neq 0$ and an integer.  

How does m divide $(x-n)$, when $(x-n)$ is not an integer, if $x = -\frac{1}{3}$????
The notation can indeed be confusing the first few times one encounters in. When one says $x=-1/3$ in the context of $\mathbb{Z}_{13}$, it is simply short-hand for saying that $x$ is some number from $0$ to $12$ such that $-3x \equiv 1 \pmod{13}.$ So remember that $x$ is not the usual real number $-1/3$, but we use this notation because it behaves like $-1/3$ would, in that $-3x=1$ in $\mathbb{Z}_{13}.$
Hint:  what is $-1 \pmod {13}$ in the normal set of representatives $0 \le n \lt 13$?