Solving system of Congruences with Two Variables (x and y) I know a bit about the Chinese Remainder theorem but what do I do in the case I was asked to solve a system of congruences such as this with two variables:
$3x + y = 7$ (mod 8)
$4x + 3y = 1$ (mod 8)

The first equation is in a form suggesting substitution method as you have $y$ with coefficient $1$:


*

*$y \equiv 7-3x \pmod 8$
Plug this into the second equation and solve for $x$:
$$4x + 3(7-3x) \equiv 4x +21 -9x \stackrel{21=16+5}{\equiv} -5x +5 \equiv 1 \pmod 8$$
Now, note that the $5^2 \equiv (24+1) \equiv 1 \pmod 8$. Using this you get
$$\Leftrightarrow 5x \equiv 4 \pmod 8 \stackrel{\cdot 5}{\Leftrightarrow} \boxed{x =} 20 \equiv \boxed{4 \pmod 8}$$
Plugging this back into the equation $y \equiv 7-3x\pmod 8$ gives
$$\boxed{y=} 7-3\cdot 4 \equiv -5 \equiv \boxed{3 \pmod 8}$$
