## solution to a tricky non-linear system

Posted: February 5, 2016 in Mathematics

Problem:

$x^2 + xy + y^2 = 3$

$y^2 + yz + z^2 = 4$

$z^2 + zx + x^2 = 5$

$x + y + z = \; ?$

Solution:

Although there is a 3-4-5 ratio, it has nothing to do with Pythagoras Theorem. First let’s try to play around with the variables.

$(z-x)(x+y+z)=z^2+yz-x^2-xy=4-3=1$

$(x-y)(x+y+z)=x^2+zx-y^2-yz=5-4=1$

Obviously $x+y+z \neq 0$, hence

$z-x=\dfrac{1}{x+y+z}=x-y$

$\therefore y+z=2x$

$\therefore x+y+z=3x$

Then we have the following,

$z-x = \dfrac{1}{3x} = x-y$

$\therefore y=x-\dfrac{1}{3x} \quad \text{and} \quad z=x+\dfrac{1}{3x}$

Substitute these equations into $y^2+yz+z^2=4$,

$\left(x-\dfrac{1}{3x}\right)^2 + \left(x-\dfrac{1}{3x}\right)\left(x+\dfrac{1}{3x}\right) + \left(x+\dfrac{1}{3x}\right)^2 = 4$

$2x^2 + \dfrac{2}{9x^2} + x^2 - \dfrac{1}{9x^2} = 4$

$3x^2 + \dfrac{1}{9x^2} - 4 = 0$

$27x^4-36x^2+1=0$

$x^2=\dfrac{36\pm\sqrt{36^2-4\cdot27}}{2\cdot27}$

$x^2 = \dfrac{36\pm\sqrt{36^2-4\cdot9\cdot3}}{2\cdot3\cdot9}$

$x^2=\dfrac{6\pm\sqrt{33}}{9}$

$\therefore x=\dfrac{\pm\sqrt{6\pm\sqrt{33}}}{3}$

Therefore,

$x+y+z=3x=\pm\sqrt{6\pm\sqrt{33}}$

Four solution for $x+y+z$, they are

$\sqrt{6+\sqrt{33}}\; ,\; \sqrt{6-\sqrt{33}}\; , \;-\sqrt{6+\sqrt{33}}\; ,\; -\sqrt{6-\sqrt{33}}$