System of $4$ equations in three variables.

Question

I'm preparing for my optimization exam and I'm having a really hard time solving the following system of equations. Can anyone suggest to me an approach to solving this, because I'm desperate right now:

$x_{2}x_{3}-k(x_{2}+x_{3})=0$

$x_{1}x_{3}-k(x_{1}+x_{3})=0$

$x_{1}x_{2}-k(x_{1}+x_{2})=0$

$x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=6$

Answer 1

Hint: multiply the first equation by $x_1$, the second by $x_2$, the third by $x_3$ and sum them up. Use the fourth to get $4k=x_1x_2x_3$. Substitute this $k$, cancel and compare pairwise to get $xy=xz=yz=2$. Can you continue from here?

Answer 2

You can write $$x_2=\frac {kx_3}{x_3-k}$$ and similarly for $x_1$ from the second, which shows $x_1=x_2$. Similarly you can show $x_2=x_3$ The last gives that each is $\sqrt 2$ Then the first gives $$2-2k\sqrt 2=0\\k=\frac {\sqrt 2}2\\x_1=x_2=x_3=\sqrt 2$$

Answer 3

HINT.- Noting $(1),(2),(3)$ the three first equations you have $$(1)-(2) \iff (x_2-x_1)(x_3-k)=0\\(1)-(3)\iff(x_3-x_1)(x_2-k)=0\\(2)-(3)\iff(x_3-x_2)(x_1-k)=0$$ The result follows.