How to solve this system of equations (Lagrange Multipliers)

Question

I was doing a question on Lagrange multipliers and stucked when trying to evaluate the point.

The system of equations that I can't solve is this:

$$y^2-x^2+3x-3y=0$$ $$-y^2-yx+3y-xy=0$$

I just can't find a way to isolate $x$ or $y$...

Just in case anyone wondering the original system was:

$$yz=\gamma$$ $$xz=\gamma$$ $$xy=\gamma$$ $$x+y+z=3$$

Please, help me.

Answer 1

I'm assuming x and y commute, right?

So, if you add the equations you end up with:

-x^2-2xy+3x=0

or

-x-2y+3=0

Answer 2

Note that $xy \cdot yz = \gamma^2$. If $\gamma \neq 0$ then dividing by $xz$ gives $y^2 = \gamma$. Similarly for $x,z$.

If $\gamma = 0$, then exactly two of $x,y,z$ are zero and the other is 3.

Answer 3

(0,0) satisfies both of your equations , Check its nature at that point .Also factoring first equation you get $x=y$ and $x+y=3$ .use in second equation