Solution to ax+b=c in a Boolean algebra

Question

I have a question. In another forum, a user asked if there is a solution to ax+b=c in a Boolean algebra, where "ax+b=c" is "$(A \wedge X) \vee B = C$". The idea is that, in a Boolean ring, this equation admits the solution x=c/a-b/a; however, I'm not entirely sure if it's possible to find a similar solution in a Boolean algebra, as the inverse operations are not defined. Further, it seems to me that there's no general solution, as, if the domain is {0,1}, if we assign A=0, B=1, and C=0, there's no X satisfying the equation.

Any comments?

Answer 1

$c=0\iff b=0$ and $ax=0$ ; $c=1\iff b=1$ or $a=x=1$.

Answer 2

The function $f(X)=(A\wedge X)\vee B$ is monotonically increasing. Inasmuch as $f(0)=B$ and $f(1)=A\vee B$, the equation $f(X)=C$ has no solution unless $B\le C\le A\vee B$.
If $B\le C\le A\vee B$, then $X=C$ is a solution, and the set of all solutions is $\{X:C\wedge B'\le X\le C\vee A'\}$ where $\ '\ $ is complementation.