Elementary Proof: If $x \in [a]$ and $y \in [b]$, then $x+y \in [a+b]$ where $a\sim b$ is an Equivalence relation if $5| (b-a)$

Question

If $x \in [a]$ and $y \in [b]$, then $x+y \in [a+b]$ where $a\sim b$ is an Equivalence relation on $\mathbb{Z}$ if $5| (b-a)$.

Honestly I don't know where to start on this one, I know the definition of an equivalence relation and some the key proposition used such as

Given $\sim$ equivalence relation on $A$ I know the equivalence class of $a \in A$ is

$$[a]:= b \in A:b\sim a$$

and by some proposition I know

(i) $a \in [a]$

(ii) $a \sim b$ if and only if $[a]=[b]$

Where do I go from here?

Edit:

I know that if $x \in [a]$ then $5| (x-a)$

and if $y \in [b]$ then $5|(y-b)$.

This means by definition of divisibility $(x-a) =5j$ where $j \in $ $\mathbb{Z}$ and $(y-a) =5r$ where $r \in $ $\mathbb{Z}$

So I want to show $x+y \in [a+b]$ which means $5| (x+y) -(a+b)$. By some basic math I see $5| (x-a)+(y-b)$ and by replacement...

$5| 5(j + r)$ which is clearly true.

Answer 1

If $x \in [a]$ then $5|x-a$.

Similarly $5|y-b $.

Now you need to show that $5|(x+y)-(a+b)$. Can you take it from here?