0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jan 23 '15 at 16:32
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Nice. That's the correct answer I was trying to get $(3,0,0)(0,3,0)(0,0,3)$. Thank you. $\endgroup$ – João Pedro Jan 23 '15 at 16:39
1
$\begingroup$

(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

$\endgroup$
0
$\begingroup$

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
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.