### The Question

Let $R$ be a unital commutative ring, and let $a,b_1,b_2\in R$. The following is a basic commutative algebra exercise.

**Lemma.** If $Ra+Rb_1=R$ and $Ra+Rb_2=R$, then $Ra+Rb_1b_2=R$.

**Proof.** Let $P$ be any prime ideal containing $Ra+Rb_1b_2$. Since $b_1b_2\in P$, then $b_1\in P$ or $b_2\in P$. In either case, $P=R$. Therefore, no prime ideal contains $Ra+Rb_1b_2$, so it is all of $R$.

Its a slick proof, but its also very nonconstructive. My question is; given $f_1,f_2,g_1,g_2\in R$ such that $$ f_1a+g_1b_1=1= f_2a+g_2b_2$$ can you construct $f,g\in R$ such that $$ fa+gb_1b_2=1?$$ I'm willing to be fairly lax in my standards for a 'construction', in that it doesn't have to be a closed formula. I just don't want it to use an embedding in a hypothetical prime ideal.

### My Motivation

My practical interest in this comes from a non-commutative analog of this problem. I am considering non-commutative $R$ and *quasi-commuting* elements $a,b_1,b_2$. That is, $ab_1=\lambda b_1a$ for some unit $\lambda$ (and likewise for other pairs).

I would like to deduce that $$Ra+Rb_1=R \text{ and } Ra+Rb_2=R \text{ implies } Ra+Rb_1b_2=R$$ Quasi-commuting elements are close enough to commutative that many constructions still work. However, one tool which does not generalize is primary decomposition. Therefore, I would like a more explicit commutative proof, in the hopes that it will work in the quasi-commuting case also.