2014/010) Solve this system of congruences, 2x +3y ≡ 22 mod27, and x +4y ≡ 7 mod27

2x + 3y ≡ 22 mod 27  … (1)
x + 4y ≡ 7 mod 27 … (2)

multiply the second by 2
2x + 8y ≡ 14 mod 27  … (3)

subtract the first from the last one
5y ≡ -8 mod 27

as GCD(5,27) = 1, which divides any number, the congruence has one solution between 0 and 26, namely y = 20

substitute in the second equation
x + 80 ≡ 7 mod 27
x   ≡ - 73  mod 27 or 8

hence general solution is
(x = 8 + 27h, y = 20 + 27k)