It is quite often that we can prove that if you have a set of axioms $\mathbb A$ and some additional axiom $A$, that if $\mathbb A+ A$ is consistent, then $\mathbb A + \lnot A$ is also consistent.
For example, in plane geometry, if $\mathbb A$ is all the usual axioms minus the parallel postulate, and $A$ is the parallel postulate, then in $\mathbb A+A$ we can construct internally a model for $\mathbb A+\lnot A$, implying that $\mathbb A+\lnot A$ is also consistent if $\mathbb A+A$ is consistent.
Clearly, $A$ and $\lnot A$ are incompatible - they are the extreme case of incompatibility.
Perhaps a simpler example would be to start with the bare bones Peano axioms, with just the single "successor" operator, $S(n)$, defined for every natural number $n$. Let $\mathbb A$ be the axioms other than the induction axioms, and let $A$ be the induction axiom scheme. ($A$ is not one axiom, really, but a collection of axioms...) Then if $\mathbb A+A$ is consistent, we can show that $\mathbb A+\lnot A$ is consistent. In this case, since $A$ is not an axiom, but a list of axioms, we'd just mean by $\lnot A$ that some element of the set of axioms $A$ is provably not true.
We do this by defining the operator $T(n)=S(S(n))$ and showing that $T$ satisfies all the axioms other than the induction axioms.
Basically, this means showing:
$$\forall n: S(S(n))\neq 0$$
$$\forall n,m: S(S(n))=S(S(m))\implies n=m$$
Then you have to show that there is some theorem that is true by induction but is not true if $S$ is replaced everywhere by $T$. A simple such proposition is:
$$\forall n: n=0 \lor \exists m: S(m)=n$$
This can be proven in $\mathbb A+A$, but is not true when $S(m)$ is replaced by $S(S(m))$, since $S(0)$ is a natural number, and $S(0)\neq 0$ and $S(0)\neq S(S(m)))$ for any $m$.