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?

Why not do it the easy way?
Since both equations involve the same modulus, this is no different from solving an ordinary pair of linear equations.  As a previous poster noted, substitution is still a valid technique.
Since $3x + y ≡ 2 \pmod 9$, therefore y ≡ 2 - 3x \pmod 9.
Substituting into the other equation,
$$x + 2(2 - 3x) ≡ 3 \pmod 9$$
$$x + 4 - 6x ≡ 3 \pmod 9$$
$$4 - 5x ≡ 3 \pmod 9$$
$$5x - 4 ≡ -3 \pmod 9$$
$$5x ≡ 1 \pmod 9$$
$$5x ≡ 10 \pmod 9$$
$$x ≡ 2 \pmod 9$$
Substituting back into the previous equation used,
$$y ≡ 2 - 3(2) \pmod 9$$
$$y ≡ -4 \pmod 9$$
$$y ≡ 5 \pmod 9$$
Thus $$x ≡ 2, y ≡ 5   \pmod 9$$
