<s>[INST] How to solve a system of linear equations modulo n? For example, 
$4x - 10y \equiv 8\pmod {20}$
$7x + 2y \equiv 5\pmod {20}$
It resembles linear diophantine equations and the Chinese Remainder Theorem, but I don't know how to actually solve it.. 
 [/INST] First you can multiply the system by any number that has an inverse, that is $\gcd(x,20)=1$.
So in particular you cannot multiply or divide by $2,4,5,10$ as you would not multiply or divide by $0$ in a normal non-modular system. Well you can do it, but you'll loose equivalence in the way.
For instance here we would like to have $4x-3x$. We notice that $7^{-1}=3\pmod {20}$ so let's multiply the second line by $9$.
$\begin{cases} 4x-10y\equiv 8\pmod{20}\\ 7x+2y\equiv 5\pmod{20}\end{cases}\iff\begin{cases} 4x-10y\equiv 8\pmod{20}\\ 3x+18y\equiv 5\pmod{20}\end{cases}$
Now we can subtract two lines like we do in normal systems
$\begin{cases} 4x-10y\equiv 8\pmod{20}\\ x-28y\equiv 3\pmod{20}\end{cases}\iff\begin{cases} 4x-10y\equiv 8\pmod{20}\\ x\equiv 8y+3\pmod{20}\end{cases}$
We now have $x$ and can report in the first equation
$\begin{cases} 32y+12-10y\equiv 8\pmod{20}\\ x\equiv 8y+3\pmod{20}\end{cases}\iff\begin{cases} 2y\equiv 16\pmod{20}\\ x\equiv 8y+3\pmod{20}\end{cases}$
$2$ is not invertible modulo $20$ but since all coefficients are even, we can divide everything by $2$ including the modulo.
$\begin{cases} y\equiv 8\pmod{10}\\ x\equiv 8y+3\pmod{20}\end{cases}$
Finally report in second equation to get $x$, by introducing a dummy variable $k$ for expressing $y$
$\begin{cases} y\equiv 8\pmod{10}\\ x\equiv 8(10k+8)+3\pmod{20}\equiv 64+3\equiv 7 \pmod{20}\end{cases}$
I have detailed a lot, since this is apparently your first time solving this.

Answering question in comment

$\begin{cases} 2x-6y\equiv 11\pmod{20}\\ 4x+8y\equiv 9\pmod{20}\end{cases}$
Rem: for parity reasons this has no solution, but let's do like we ignore that.
For this one $9$ and $11$ are invertible (and their own inverse), so let's multiply the lines by $9$ and $11$.
You get 
$\begin{cases} 2x+14y\equiv 1\pmod{20}\\ 16x+12y\equiv 1\pmod{20}\end{cases}\iff \begin{cases} 2x+14y\equiv 1\pmod{20}\\ 14x-2y\equiv 0\pmod{20}\end{cases}$
Which reduce to $y\equiv 7x\pmod{10}$, and this system has no solution ($2x+14y$ is divisible by $20$).
</s>