(Dis)prove $\,a\mid b\,\Rightarrow\, a+b\mid b+c$

Question

Can someone help me here? I'm completed stuck in this simple problem.

Let a, b, c $\in \mathbb{N} $, check if the statement below is true or false:
if   a|b then (a+b) | (b+c)

Any tips?
What i've done so far:

If a|b, then b=a.c
To prove that (a+b)|(b+c) there must be a j, such as (b+c) = (a+b)*j

1. (a+b) -> a+a.c

2. (b+c) -> a.c + c

3. (a+b)*j =?= (a.c+c)

Oh lord, just tried one case and found out that it is false.
Take a = 3, b = 6, c =2

1. a|b -> b =a.c -> 6=3*2
   
2. (a+b) | (b+c) -> (b+c) = (a+b)*j -> 6+2=(3+6)j -> 8=9j
   (there's no j in N that satifies this equation

Answer 1

I've just tried one case and found out that it is false. Take a = 3, b = 6, c =2

a|b -> b =a.c -> 6=3*2 (a+b) | (b+c) -> (b+c) = (a+b)*j -> 6+2=(3+6)j -> 8=9j (there's no j in N that satifies this equation