Solving a quadratic system

Question

Just confused on method here, I have the following system

$3x^2-y=0\\-x-3y^2=0$

How would I go about solving this? I am aware of what the four solutions are but I can't see how I could personally find them. Any help is appreciated!

Answer 1

From first equation, $\;y=3x^2\;$ , and now substitute in the second one:

$$-x-3(3x^2)^2=0\implies x(1+27x^3)=0\implies x= 0,\,-\frac13$$

Take it from here...(if you want also complex roots add them in the above line already)

Answer 2

From the first equation we have that $y=3x^2$. Substituting this into the second equation we obtain, $x+3(3x^2)^2 = 0$ which simplifies to $x + 27x^4 = 0$, or equivalently, $x(1+27x^3) = 0$. One solution is thus $(x,y) = (0,3\cdot 0^2) = (0,0)$. Can you find the other three?

Answer 3

Substitute $y=3x^2$ in the second equation to get $$-x-27x^4=0.$$

Then

$$x=0\text{ or }x^3=-\frac1{3^3}.$$