Strange isomorphism: $R/(A + B) \cong (R/B)/\bar{A}$. $R$ is a ring and $A, B$ are ideals of $R.$ 
I was playing around with some stuff and I was wondering if $R/(A + B) \cong (R/B)/\bar{A}.$ for $\bar{A}$ being the image of $A$ under $R \rightarrow R/B.$ I'm pretty sure I have a proof for it. I will post a 'proof' of it as an answer if the isomorphism is correct. I simply don't know if it is correct because I have never seen this before. Or...is it just plain wrong? 

We have a map from $R \rightarrow (R/B)/\bar{A}$ which is a homomorphism as this is the composition of the natural maps $R \rightarrow R/B$ and $R/B \rightarrow (R/B)/\bar{A}.$ It is also surjective as it is the composition of surjective maps. It is not difficult to see that the kernel is $A + B.$ However, @JayTuma's answer seems to be a more direct approach using the second isomorphism theorem.
