Simplifying Quotient of Tensor Products

Consider

$$(A\otimes C)/(B\otimes C)$$

where $B$ is a submodule of $A$. ($A,B,C$ are $R$-modules).

Is it true that $$(A\otimes C)/(B\otimes C)\cong(A/B)\otimes C$$?

Thanks. If no, are there any easy counter-examples?

This is sort of true.

There is a natural map from $B\otimes C$ to $A\otimes C$, but in general it is not injective, so that we cannot think of $B\otimes C$ as a submodule of $A\otimes C$. But the image $I$ of this map is a submodule of $A\otimes C$, and $(A\otimes C)/I$ is isomorphic to $(A/B)\otimes C$.

What is going on is that the tensor product is right exact. We have an exact sequence $$0\to B\to A\to A/B\to 0$$ and when we tensor with $C$ we get that $$B\otimes C\to A\otimes C\to (A/B)\otimes C\to 0$$

is exact.