Taking modulo of both sides On the following page at the bottom there is an algorithm for calculating modular inverses. In the proof I am confused with the line 'Taking both sides modulo $m$'. How does that work getting a congruence from the above equation?
 A: $j = k\,\Rightarrow\, \color{#0a0}{j\bmod m} = k\bmod m\,$ by $\,f(x) = x\bmod m\,$ is a function so $\,j=k\,\Rightarrow\, f(j) = f(k)$.
Your special case has $\, \color{#c00}0 \le \underbrace{m\bmod i}_{\Large\color{#0a0} j} < i < \color{#c00}m\,$ so $\,\color{#0a0}{j\bmod m = j}\,$ is already reduced $\!\bmod \color{#c00}m$
Remark $ $ Note that the modular inversion algorithm discussed there is the same as the algorithm I discussed in your prior question, i.e. Gauss's modular inversion algorithm (in non-fractional form). 
A: The congruence $a\cdot x\equiv c\mod m$ means $m$ divides $a\cdot x -c$ and is thus equivalent to $a\cdot x+b\cdot m = c$ for some integer $b$. In this way, one can get a congruence from an equation and vice versa.
A: If $x=y$ then $x \equiv y (mod m)$, so far, so good.
Now, let's say we have what I think you may be struggling with $ax + bm = y$
Then $ax + bm \equiv y\space (mod m)$, 
but $bm \equiv 0 \space (mod m)$, so that's why we get $ax \equiv y \space (mod m)$ To get there, you may want to subtract $bm$ from both sides
