<s>[INST] Solve the following system of equations: $x^2 - 3x\sqrt{y - 2} + 3y = 9$ and $x^3 + 9(3y - 7)\sqrt{y - 2} = 3x$. $(x, y \in \mathbb R)$ 
Solve the following system of equations $(x, y \in \mathbb R)$ $$\large \left\{ \begin{align} x^2 - 3x\sqrt{y - 2} + 3y = 9\\ x^3 + 9(3y - 7)\sqrt{y - 2} = 3x \end{align} \right.$$

I have provided a solution below where there is an obsolete step for replacing $ -3 = \dfrac{x}{z}$ but who cares.
I hope there are other solutions which are better than mine.
 [/INST] Let $\sqrt{y - 2} = z$ $(z \ge 0)$, we have that $\left\{ \begin{align} x^2 - 3zx + 3z^2 = 3\\ x^3 + 9z(3z^2 - 1) = 3x \end{align} \right.$
$$ \iff 3z^2 - 3zx + x^2 - 3 = 27z^3 + x^3 - 9z - 3x = 0$$
$$ \iff 3z^2 - 3zx + x^2 - 3 = (3z + x)(9z^2 - 3zx + x^2 - 3) = 0$$
$$ \iff 3z^2 - 3zx + x^2 - 3 = 6z^2(3z + x) = 0$$
$$ \iff 3z^2 - 3zx + x^2 - 3 = \left[ \begin{align} z\\ \dfrac{x}{z} + 3\\ \end{align} \right. = 0$$
If $z = 0 \iff y = 2$ and $x^2 - 3 = 0 \iff x = \pm \sqrt 3$
If $\dfrac{x}{z} = -3 \implies \left(-\dfrac{x}{z}\right) \cdot z^2 + \dfrac{x}{z} \cdot zx + x^2 + \dfrac{x}{z} = 0 \iff \dfrac{x}{z} \cdot (2zx - z^2 + 1) = 0$
$$ \iff 2zx - z^2 + 1 = 0 \iff 2z \cdot (-3z) - z^2 + 1 = 0 \iff z^2 = \dfrac{1}{7} \iff y = \dfrac{15}{7}$$
and $x = \dfrac{\mp 3}{\sqrt 7}$. We have that $(x, y) = \left\{(\pm \sqrt 3, 2), \left(\dfrac{\mp 3}{\sqrt 7}, \dfrac{15}{7}\right)\right\}$.
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