How to solve the system of non-linear equations?

Question

$$\begin{cases} 3x^2 -3y =0 \\ 3y^2 - 3x = 0 \end{cases}$$

I really forgot how to solve this since my childhood, apperently it is not linear equations and Gauss' method cant'be applied

Answer 1

from the first equation we get $$y=x^2$$ plugging this in the second equation we obtain $$x^4-x=0$$ and from here we obtain $$x(x^3-1)=0$$ can you finish?

Answer 2

Divide by 3 and rearrange to see that the system is equivalent to the easier-looking: \begin{cases} y = x^2 \\ x = y^2 \end{cases} Now substituting $x=y^2$ into $y=x^2$ gives you: $$y=y^4 \iff y = 0 \; \vee y = 1$$ The corresponding $x$-values are identical.

If you draw the graphs of the two (easy) parabolas, you can easily find (or confirm) the solutions graphically.

Answer 3

If $x,y \in \mathbb R$

From the equation, we know $x >=0$, and $y>=0$ $$y=x^2,x=y^2$$ $$x^4-x=x(x^3-1)=0$$

Thus $(x = 0, y=0)$, or $(x = 1, y=1)$

If we allow complex numbers: $$y=x^2,x=y^2$$ $$x^4-x=x(x-1)(x^2+x+1)=x(x-1)(x-\frac{-1+\sqrt{3}i}{2})(x-\frac{-1-\sqrt{3}i}{2})=0$$

Thus $(x=0,y=0), (x=1, y=1), (x=\frac{-1+\sqrt{3}i}{2}, y=\frac{-1-\sqrt{3}i}{2}),(x=\frac{-1-\sqrt{3}i}{2},y=\frac{-1+\sqrt{3}i}{2})$

Answer 4

In this case, you can also see the solutions geometrically.

The graph of the first equation (which is equivalent to $y=x^2$) is a parabola with vertex at the origin and symmetric about the positive $y$-axis.

The graph of the second equation (which is equivalent to $x=y^2$) is a parabola with vertex at the origin and symmetric about the positive $x$-axis.

These graphs intersect exactly at $(0,0)$ and $(1,1)$, so these points are the only solutions to the system of equations.

Answer 5

We have $x^2-y^2+x-y=0$ or $(x-y)(x+y+1)=0$,

which gives $x=y$ and $\{(0,0),(1,1)\}$

or $x+y+1=0$, which with $x^2-y=0$ gives $x^2+x+1=0$, which is impossible.