Turn fractions into $\mathbb Z_7$ elements I had to perform a division between two polinomials $2x^2+3x+4$ and $3x+4$, my book suggests to do this operation without worrying about the modulo. So my result is $(3x+4)(\frac{2}{3}x+\frac{1}{9})+\frac{32}{9}$. Unfortunately my book fails to explain how should I perform the conversion from fractions to $\mathbb Z_7$ elements, like it's kind of obvious, it only reports $\frac{1}{3}=5$, $\frac{2}{3} = 10 = 3$, $\frac{32}{9}=2$. Will you please break it down for me?

Deal first with $1/3$. In the modulo $7$ world, this is a number $x$ such that $3x\equiv 1\pmod{7}$.
How do we find such an $x$? Well $7$ is small, so try all the possibilities for $x$.  These are $1$ (well, not really!), $2$, $3$, $4$, $5$, and $6$. Try $x=2$. Does it work? Let's check. We have $(3)(2)\equiv 6\pmod{7}$, doesn't work. 
Try $x=3$. We have $(3)(3)=9\equiv 2\pmod{7}$, doesn't work. Try $x=4$. We have $(3)(4)=12\equiv 5\pmod{7}$, doesn't work. Try $x=5$. We have $(3)(5)=15\equiv 1\pmod{7}$. Bingo! 
By the way, our first trial almost worked. We have $(3)(2)\equiv 6\equiv -1\pmod{7}$. So if we replace $2$ by $-2$, the product will be congruent to $1$. Thus $x\equiv -2\equiv 5\pmod{7}$.
Now for the $2/3$, this is easy, it is twice $1/3$. Modulo $7$, we get $10$, which is congruent to $3$.
We can get to $32/9$ in many ways. For example, to get $1/9$, square our $1/3$ found above. We get $25$, which is congruent to $4$ modulo $7$. But $32$ is congruent to $4$, so for $32/9$ we will want $(4)(4)$, which is congruent to $2$ modulo $7$.
There are much easier ways. For example, $32$ is congruent to $4$, and $9$ is congruent to $2$, so the fraction should be congruent to $\dfrac{4}{2}$, which is $1$.
Remark: The "guess and test" approach described above is only practical for relatively small primes. For larger primes, one needs a more efficient approach. You will probably learn one soon, the Extended Euclidean Algorithm.
