I have 2 equations: $$x^2 + y^2 = 12 \tag{1}$$ $$x^2 + y^2 -6x -2 \times 3^{1/2}y = 0 \tag{2}$$
On substituting the value of (1) in (2) we get: $$12 - 6x -2 \sqrt[3]{y} = 0 \tag{3}.$$
If we express $x$ in terms of $y$ we get: $$x = 12 - 2 \sqrt[2]{y}$$
This satisfies Eqn (3) but not for (2).
How is this is a root for equation (3) but not for (2)?
Well, we have:
$$ \begin{cases} x^2+\text{y}^2=12\\ \\ x^2+\text{y}^2-6x-2\cdot\sqrt[3]{3}\cdot\text{y}=0 \end{cases}\tag1 $$
Notice that the second equation is a quadratic equation in $x$:
$$x^2+\text{y}^2-6x-2\cdot\sqrt[3]{3}\cdot\text{y}=x^2-6x-2\cdot\sqrt[3]{3}\cdot\text{y}+\text{y}^2=0\space\Longleftrightarrow\space$$ $$x=\frac{-\left(-6\right)\pm\sqrt{\left(-6\right)^2-4\cdot1\cdot\left(-2\cdot\sqrt[3]{3}\cdot\text{y}+\text{y}^2\right)}}{2\cdot1}=3\pm\sqrt{9+\text{y}\cdot\left(2\cdot\sqrt[3]{3}-\text{y}\right)}\tag2$$
So, we get:
$$x^2+\text{y}^2=\left(3\pm\sqrt{9+\text{y}\cdot\left(2\cdot\sqrt[3]{3}-\text{y}\right)}\right)^2+\text{y}^2=12\space\Longleftrightarrow\space$$ $$\text{y}\approx-1.88511\space\space\space\vee\space\space\space\text{y}\approx3.4471\tag3$$