# How to solve this system of equations (Lagrange Multipliers)

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:

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

• I have not checked your work up to the equations. Assuming they are right, $y$ is a factor of the second, and then we have $y=0$ or $-y-2x+3=0$. Substitute in first. – André Nicolas Jan 23 '15 at 16:32

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.

• Nice. That's the correct answer I was trying to get $(3,0,0)(0,3,0)(0,0,3)$. Thank you. – João Pedro Jan 23 '15 at 16:39

(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

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