###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.