Let n e Z. Prove that n² + 4 is divisible by 8 if and only if 3n + 6 is divisible by 12. (Hint: the statements n | m, 3k e Z: m = nk, m = 0 (mod n) are all equivalent)

In the following question: " Let a, b ∈ Z. Show that if a ≡ 5 (mod 6) and b ≡ 3 (mod 4), then 4a + 6b ≡ 6 (mod 8)." why do we ignore the -6 when substituting the variables of a and b, if we expand "4a + 6b ≡ 6 (mod 8)" we get (4a+6b)-6=8y, when we substitute we have to end up with the 8y right? Im just confused why the -6 is ignored?

Please give detailed steps. I’m really trying to understand this problem. Much appreciated!

(a) Find the order of a modulo 7 for a = 1, 2, 3, 4, 5, 6.