Prove that $ I + J = R \iff b + J $ is invertible in $ R / J$, where $ I = \langle b \rangle$

Question

Let $R$ be an integral domain and $I, J$ its ideals, where $I = \langle b \rangle$, for a $ b \in R$. Prove that $ I + J = R \iff b + J $ is invertible in $ R / J$

If it is a duplicate, I apologize, but I've already looked up for it. I was able to show this: $ I + J = R \iff (\forall r \in R) r = i + j $, for some $ i \in I, j \in J \iff (\forall r \in R) r = ba + j $, for some $ a \in R, j \in J$. Am I missing something or am I in a completely wrong direction? Thank you.

Answer 1

You were almost there. Note that $R$ is an integral domain, so it contains the unity. Thus, $I+J=R$ if and only if $1\in I+J$. That is, $I+J=R$ iff $ab+j=1$ for some $a\in R$ and $j\in J$. (That is, you only have to take $r$ to be $1$ in your attempt.) What next?