In the book I'm reading it says that there exist elements with multiplicative inverses in a integer ring if $gcd(a,m)=1$ where $a$ is the element and $m$ is the modulo.
The inverse $a^{-1}$ is defined such that $a \times a^{-1} \equiv 1$ mod $m$
Also it says that if an inverse exist for $a$ we can divide this element since $b/a \equiv b \times a^{-1}$ mod $m$
I'm confused as to what $b$ is in this case? My guess right now is that it is another member in one of the equivalence classes for the modulos $m$.
Yes. For example, let $m=7$ and $a=[2]$. Then $a^{-1}=[4],$ because $2\times4\equiv1\pmod7$.
For another example, if $m=7$ and $a=[3]$ then $a^{-1}=[5],$ because $3\times5\equiv1\pmod7$.
$b$ is any other element. Say you want to solve for $x$ in $ax\equiv b\pmod m$.
The answer is $x\equiv b\times a^{-1}\pmod m$.
For example, the solution to $2x\equiv5\pmod7$ is $x\equiv5\times2^{-1}\equiv5\times4=20\equiv6\pmod7$.