The difference contains $x^2-y^2$ and terms involving $x-y$. Factor out $x-y$ to create cases.

For each case, substitute into one original equation and use the discriminant to determine how many real values are possible. Then check whether the two cases share a solution.

Algebra is faster because subtracting the equations immediately produces two factorable cases. Use Desmos to verify the number of intersections.

Enter $x^2-3y=12$ and $y^2-3x=12$ as two equations. Desmos will graph each relation in the coordinate plane.

Click each intersection of the two graphs. Each intersection represents one ordered pair satisfying both equations; the graph confirms the algebraic count.

Because both expressions equal $12$, subtract the second equation from the first:

Therefore, either $x=y$ or $x+y=-3$.

Substitute $y=x$ into $x^2-3y=12$:

Its discriminant is $(-3)^2-4(1)(-12)=57$, which is positive. Thus, this case gives two real values of $x$ and therefore two ordered pairs of the form $(x,x)$.

Write $y=-3-x$ and substitute into the first equation:

The discriminant is $3^2-4(1)(-3)=21$, which is positive. This case also gives two real values of $x$, producing two ordered pairs.

A pair in both cases would have $x=y=-\frac{3}{2}$. But $x^2-3x=\frac{27}{4}$ when $x=-\frac{3}{2}$, not $12$, so this pair is not a solution. The two cases give $2+2=4$ distinct ordered pairs.