Solving the system $(18xy^2+x^3, 27x^2y+54y^3)=(12, 38)$

While answering this question, I got myself stumped with this crazy system with an evil graph: $$\begin{cases}
18xy^2+x^3=12 \\
\end{cases}$$ and I wonder whether there is some slick method to find the only real root $(x, y)=(2, 1/3)$ without relying on Cardano’s formula, ideally giving some intuition. This closely reassembles some kind of elliptic curves, so I’m tagging it as such, please remove if wrong. Number theoretic approaches are welcome.

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It makes sense to change variables, letting $v=3y$ and $x=2u$, so that what we’re looking for is $(u,v)=(1,1)$. The equations become


Multiplying the first by $19$ and the second by $3$ and subtracting gives


The quadratic term is easily seen to be non-negative:


Hence we must have $u=v$. Plugging this into either equation gives the desired result, $u=v=1$.

Added later: Just to elaborate, the given equations change into something of the form


This leads to


and the quadratic contributes nothing to the (real) solution if $(1-r-s)^2\lt4rs$.

Indeed, all rational solutions are given by $(x,y)=(2,1/3)$. We can see this without any theory on elliptic curves, but just with a computation:
Substitute $y^2=(12-x^3)/(18x)$
from the first equation into the second to obtain $y=19x/(6(2x^3+3)$. This in turn gives the following result:
(x-2)(x^2 + 2x + 4)(2x^2 + 4x + 3)(4x^4 – 8x^3 + 10x^2 – 12x + 9)=0
Obviously, only $x=2$ is a rational solution for $x$. Also we see, that it is the only real solution.

I see that you used Mathematica to find your “evil graph”. For a result with no clutter, you can also use the command,

 Factor[Resultant[18y^2x+x^3-12, 27x^2y+54y^3-38,y]]

which will eliminate the variable y and give you Burde’s neat result.