Proving equivalence relation for 7 | (3a + 4b)

Question

I know this might be quite trivial, but I just can't seem to figure out how to prove $$R = \{(a,b) \in \mathbb{Z} \times \mathbb{Z} : 3a + 4b \text{ is divisible by } 7\}$$ is a symmetric relation, i.e., if $aRb$, then $bRa$, where $\mathbb{Z}$ is the set of integers

Answer 1

Hint:

$(3a+4b)+(4a+3b)=7(a+b)$.

Answer 2

HINT

Assume $aRb$. Then $7 \mid (3a+4b)$. What can you say about $7\mid(4a+3b)$ instead?

Answer 3

Hint: $3a+4b=7c\implies3b+4a=7(a+b-c)$