When D=0: no solution or infinitely many

The elimination steps from the previous theorem did not require $ad-bc\neq0$: they show that any solution $(x,y)$ of $\begin{cases}ax+by=e\\cx+dy=f\end{cases}$ must satisfy $(ad-bc)x=ed-bf$, regardless of the value of $D$. When $ad-bc=0$, the left side $(ad-bc)x$ equals $0\cdot x=0$ for every $x$, so this necessary equation collapses to the numerical statement $0=ed-bf$.

**Case $ed-bf\neq0$ (no solution).** The statement $0=ed-bf$ is then simply false, independent of $x$ and $y$. Since every solution of the system would have to make this false statement true, no solution can exist. Geometrically, the two equations describe lines with the same slope (since $D=0$) but different intercepts, i.e. two parallel, non-intersecting lines.

**Case $ed-bf=0$ (infinitely many solutions).** Now the necessary equation is the true but empty statement $0=0$, so eliminating y gives no information at all — it is automatically satisfied by any $x$. Concretely, since $D=ad-bc=0$ and $(a,b)\neq(0,0)$, there is a constant $k$ with $c=ka$ and $d=kb$ (the second equation's x,y-coefficients are proportional to the first's); combined with $ed-bf=0$ one finds $f=ke$ as well, so the second equation $cx+dy=f$ is literally $k$ times the first equation $ax+by=e$. Every point $(x,y)$ satisfying the first equation automatically satisfies the second (multiply the first equation by $k$), so the whole line $ax+by=e$ — infinitely many points — solves the system.