If $A$ is a subring of $R$ and $I$ is and ideal of $R$, let $A+I=\{a+i:a\in A,i\in I\}$, then prove $(A + I) / I \cong A/(A \cap I)$.
We need the second theorem of homomorphisms but I need to find the surjective homomorphism $\phi :(A+I)\to A/(A \cap I)$ and show $I$ is the kernel. What I understand is that $A/(A \cap I) = \{a+y:a \in A,y\in A\cap I\}$ is this right so far? What's next to find $\phi$ with kernel $I$?
Try instead $\phi:A \to (A+I)/I$ where $\phi(a)=a+I$.
Your idea can be made to work but there are some technical hiccups. See here