Solving a congruence/modular equation : $(((ax) mod M) + b) mod M = (ax + b) mod M$

I've been trying to prove this equation for my homework.

$$(((ax) \bmod M) + b) \mod M = (ax + b) \bmod M$$

We have that $a,x,b,M > 0$, and $a ≡ b \pmod M$

Reading KhanAcademy website, I found the following properties that looked promising.

 - Multiplication property : 
\[(A * B) mod C = (A mod C * B mod C) mod C\]
 - Addition property :
\[(A + B) mod C = (A mod C + B mod C) mod C\]

I tried developping the left side of the Equation :

$(((ax) \bmod M) + b) \bmod M \rightarrow((a \bmod M \cdot x \bmod M) \bmod M + b) \bmod M$ (multiplication property)

And if I develop the right side of the Equation :

$$(ax + b) \bmod M = (ax \bmod M + b \bmod M) \mod M$$ (addition property)

Which gives this after applying the multiplication property :

$$(((a \bmod M \cdot x \bmod M)\bmod M) + b \bmod M) \bmod M$$

So I have

$$((a\bmod M\cdot x \bmod M)\bmod M+b) \bmod M = (((a \bmod M \cdot x \bmod M)\bmod M) + b \bmod M) \bmod M$$

It's almost the answer, but one side has $b \bmod M$ and the other only has $b.$ I've been looking for more congruence properties but I can't seem to find one that would allow me to solve my problem. Have I been tackling this problem from the correct angle? Or did I make a mistake from the beginning (by applying the addition and multiplication properties)?

Any help would be greatly appreciated! Thanks