We have two equations: $$x^2+y^2=12\tag1$$ $$x^2+y^2-6x-2\sqrt3y=0\tag2$$ From these two equations we get a third equation $$12-6x-2\sqrt3y=0\tag3$$ Thus $$x=\frac{12-2\sqrt3y}6 ---(*) $$ Now $x$ is a root for $(3)$. Why isn't it for $(2)$ as well?
Reasoning: In $(1)$, $x^2+y^2=12$ so basically $(3)$ is $(2)$ with the value of $x^2+y^2$ substituted.
What I tried doing: When I actually put $x$ from $(3)$ in equation $(1,2)$ i actually get the same quadratic $y^2-\sqrt3y-6=0$
Note: I am currently comfortable with algebra-precalculus and single variable calculus.
You might note that geometrically, (1) and (2) are equations of circles. The two circles intersect at two points. Equation (3) represents a straight line that passes through both points. But those are the only points where the line intersects either circle.