Prove that if $6 \mid (2a + 4)$ and $9 \mid (12 + 3b)$ then $3 \mid (a + b)$ Could you help me with the problem below?

Prove that if $6 \mid (2a + 4)$ and $9 \mid (12 + 3b)$ then $3 \mid (a + b)$.

Thank you!

Since $6 \mid (2a+4)$, let $$(2a+4) = 6p.$$
$$a = (6p-4)/2.$$
$$a = (3p-2) \tag{1}$$
Again since $9 \mid (12+3b)$, let $$(12+3b) = 9q.$$
$$b = (9q-12)/3.$$
$$b = (3q-4)\tag{2}$$
(1) + (2) gives you: $$(a+b) = 3(p+q)-6 \quad \Rightarrow \quad (a+b) = 3(p+q-2).$$
Hence $3$ divides $(a+b)$ or $3 \mid(a+b)$.
Hope the answer is clear!
