<s>[INST] How can these two equations be solved by elimination? In this question, the following two equations were solved using elimination. With a google crash course I sort of get how elimination works, but it seems like these are much too complex to add the left sides together in a way that cancels out x or y.
$$
\frac{3x(3x^2+9)}{2y}
-
\frac{(3x^2+9)}{8y^3}^3
-
y
=
6 \pmod {23}
$$
$$
\frac{(3x^2+9)^2}{(2y)^2}
-
2x
=
12 \pmod {23}
$$
Ultimately I'm trying to figure out any way (the easiest preferably) to reduce these such that x=18, y=10 or x=19, y = 3 in order to convert it to a function in a scripting language.
If elimination is the way to go, what were the steps involved in arriving at the following equation?
$$x^4-48 x^3-18 x^2+13968 x=-86481 \pmod {23}$$
 [/INST] If you would like to see how to get the equation by hand:
Introduce a third variable $$z = \frac{3x^2+9}{2y}$$ We have three equations
$$2yz - 3x^2 = 9\\3xz - z^3 - y = 6\\z^2 - 2x = 12$$
Multiplying the second equation by $2z$ and adding the first eliminates $y$:
$$6xz^2 - 2z^4 - 3x^2 = 12z + 9$$
Substituting for $z^2$ from the final equation gives $$6x(2x + 12)  - 2(2x +12)^2- 3x^2 = 12z + 9\\x^2 - 24x - 297 = 12z$$
Squaring both sides, then substituting for $z^2$ again
$$(x^2 - 24x - 297)^2 = 144(2x + 12)\\x^4 - 48x^3 - 18x^2  +13968x + 86481 = 0$$
Modulo $23$, this reduces to
$$x^4 - 2x^3 + 5x^2 + 7x + 1 \equiv 0 \mod 23$$
</s>