Systems of equations

This single number D packages both the "how many solutions" question and the "what are they" answer: check one sign, and you know whether elimination will succeed before you even attempt it.

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 $ad-bc\neq0$.

D=0 means the two lines have the same slope: geometrically they are either perfectly parallel and never meet (no solution), or actually the same line drawn twice (every point is a solution) — never a single crossing point.

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.