Prime number divisibility The following line is in a proof I'm reading, and I don't understand the logic:

Let $\frac{a}{b}$ be an arbitrary element ($a$ and $b$ both integers).  Since $p$ is a
  prime, and $p$ doesn't divide $b$, the congruence $bx ≡ a$ $(mod$ $p)$ has  a solution.

So, we know $b$ is not divisible by prime $p$, $a$ is just some integer (possibly divisible by $p$).  Therefore $bx - a = py$ for some integer $x$ and $y$?  Why is this the case?
 A: For all prime numbers $p$, the ring $\mathbb{Z}_p$ of integers modulo $p$ is a field, meaning that every element of $\mathbb{Z}_p$ other than $0$ has a multiplicative inverse. Let $[b]$ be the element of $\mathbb{Z}_p$ corresponding to $b$. Since $p$ doesn't divide $b$, we know that $[b] \neq 0$ in $\mathbb{Z}_p$, and so $[b]$ has a multiplicative inverse $[b]^{-1}$ in $\mathbb{Z}_p$.
Let $c$ be some integer whose corresponding element in $\mathbb{Z}_p$ is $[b]^{-1}$. Since $[b] \cdot [b]^{-1} = 1$ in $\mathbb{Z}_p$, $b c \equiv 1 \pmod p$. Let $x = c a$; then $b x = b c a$, and so $b x \equiv a \pmod p$.
A: Recall the following equivalent translations between equalities in quotient rings, congruences, and linear Diophantine equations:
$$\begin{eqnarray}\rm (b\!+\!n\Bbb Z)(x\!+\!n\Bbb Z) &=\,&\rm\: a\!+\!n\Bbb Z \,\in\, \Bbb Z/n\Bbb Z\\
\rm [b]\,[x] &=\,&\rm [a]\,\in\,\Bbb Z/n\Bbb Z\\
\rm b\,x\, &\equiv\,&\rm\: a\ \ \,(mod\ n)\\
\rm b\,x\, &=&\rm\: a + n\,y,\ \ for\  \ some\ \ y \in \Bbb Z\end{eqnarray}$$
The prior equation has a  solution $\rm\ (x,y)\in\Bbb Z^2\!\iff gcd(b,n) \mid a,\ $ by Euclid/Bezout. $ $ The solution is unique $\rm\,mod\ n\,$ iff $\rm\,gcd(b,n)= 1;\:$ if so the fraction $\rm\, x \equiv a/b\,$ is well-defined $\rm\,mod\ n.$  
Generally normal fractional arithmetic makes sense $\rm\:mod\ n\:$ as long as one restricts to fractions with denominator coprime to $\rm\,n\:$ (else the fraction may not uniquely exist, $ $  i.e. the equation $\rm\: ax\equiv b\,\ (mod\ n)\:$ might have no solutions, or more than one solution). The reason why such fraction arithmetic works here (and in analogous contexts) will become clearer when one learns about the universal properties of fraction rings (localizations). 
A: Since $p\nmid b$ it follows that if $$R=\{r:r \text{ is the remainder of $s$ over $p$ for } s\in\{0,b,2b,3b,\ldots,(p-1)b\}\}$$
then $R=\{0,1,2,\ldots,p-1\}$ (if $xb$ and $yb$ leave the same remainder when divided by $p$ then $|x-y|b$ is divisible by $p\Rightarrow x=y$. Therefore $R$ has $p$ elements.)
We conclude that there is an $x\in\{0,1,\ldots,p-1\}$ such that $xb$ and $a$ leave the same remainder when divided by $p$. Hence $p\mid xb-a.$
