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!
 A: Show that $3$ divides $a+2$ and $b+4$, and deduce that $a+b+6$ (and a fortiori $a+b$) is divisible by $3$.
A: $\rm\begin{eqnarray}\rm {\bf Hint}\ \ \ \dfrac{4\!+\!2a}{6} = &&\rm\dfrac{2\!+\!a}{3},\ \ \ \dfrac{4\!+\!b}{3} = \dfrac{12\!+\!3b}{9}\ \  \ both \in\Bbb Z \\  
\\
\rm therefore\   &&\rm\dfrac{2\!+\!a}{3}\, +\, \dfrac{4\!+\!b}{3} \,-\, 2\,=\,\dfrac{a\!+\!b}{3}\,\in\,\Bbb Z
\end{eqnarray}$
A: $$6 \mid (2a+4) \implies 3 \mid a+2 \,\, \text{(Why?)}$$
$$9 \mid (3b+12) \implies 3 \mid b+4 \,\, \text{(Why?)}$$
Hence, $$3 \mid (a+2+b+4) \,\, \text{(Why?)} \implies 3 \mid (a+b+6) \,\, \text{(Why?)} \implies 3 \mid (a+b) \,\, \text{(Why?)}$$
A: $6|(2a+4)$ implies that $2a+4=6k_1$ for some integer $k_1$ and $9|(12+3b)$ implies that $3b+12=9k_2$ for some integer $k_2$.
So $a+2=3k_1$ and $4+b=3k_2$.
Thus $a+2+4+b=3(k_1+k_2)$.
The rest is easy.
A: 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!
