How do you solve linear congruences with two variables. For example: Solving for $x$ and $y$ given the following linear congruences.
$x + 2y \equiv 3 \pmod9\,$, $3x + y \equiv 2 \pmod9$
So far, I've tried taking the difference of the two congruences.
Since $x + 2y \equiv 3 \pmod9 \Rightarrow x + 2y = 3 + 9k\,$, and $3x + y \equiv 2 \pmod9 \Rightarrow 3x + y = 2 + 9l$.
$$(x + 2y = 3 + 9k) - (-3x + y = 2 + 9l) \Rightarrow -2x + y = 1 + 9(k - l)$$.
So now, are you supposed to solve this like a normal linear Diophantine equation?

Here's a method with only logical reasoning, and not the CRT. 
Considering each equation modulo $3$, we get $x \equiv y \mod 3$ and $y \equiv 2 \mod 3$. So any solution must satisfy $x \equiv y \equiv 2 \mod 3$. But if $x \equiv 2 \mod 3$, then $3x \equiv 6 \mod 9$, so in the second equation we get $y \equiv 5 \mod 9$. Plugging this into the first equation, this leads to the only solution: $(x, y) \equiv (2, 5) \mod 9$.
