Unique solution exactly when D≠0

Deriving a necessary condition (eliminating y). Multiply the first equation of $\begin{cases}ax+by=e\\cx+dy=f\end{cases}$ by $d$ and the second by $b$: $adx+bdy=ed$ and $bcx+bdy=bf$. Subtracting the second from the first cancels the $bdy$ terms exactly, leaving $(ad-bc)x=ed-bf$. This equation is a necessary consequence of the system: any pair $(x,y)$ that solves both original equations must also satisfy it, since it was built only by adding/subtracting multiples of the two original equations.

Deriving the twin condition (eliminating x). Symmetrically, multiply the first equation by $c$ and the second by $a$, then subtract the first from the second to cancel the $x$ terms: this leaves $(ad-bc)y=af-ec$, again a necessary consequence of any solution.

Existence when D≠0. If $ad-bc\neq0$, the single-unknown equation $(ad-bc)x=ed-bf$ has, by the theorem on linear equations in one unknown, exactly the one solution $x=\dfrac{ed-bf}{ad-bc}$; likewise $(ad-bc)y=af-ec$ gives exactly $y=\dfrac{af-ec}{ad-bc}$. Substituting these two values into the original equations of $\begin{cases}ax+by=e\\cx+dy=f\end{cases}$ and simplifying algebraically confirms both are satisfied simultaneously (the reader can verify $a\cdot\dfrac{ed-bf}{ad-bc}+b\cdot\dfrac{af-ec}{ad-bc}=e$ by combining fractions over the common denominator $D$) — so a solution genuinely exists.

Uniqueness when D≠0. Since $(ad-bc)x=ed-bf$ and $(ad-bc)y=af-ec$ are necessary for every solution of $\begin{cases}ax+by=e\\cx+dy=f\end{cases}$ (not merely sufficient), any solution $(x,y)$ of the original system, whatever it may be, is forced to satisfy $x=\dfrac{ed-bf}{ad-bc}$ and $y=\dfrac{af-ec}{ad-bc}$ exactly, by the uniqueness part of the one-unknown theorem. So there cannot be two different solutions: combined with existence, the solution is unique precisely when $ad-bc\neq0$.