If $x, y$ and $z$ are real numbers with \begin{align*} x(x+y+z)&=2-yz\\ y(x+y+z)&=4-zx\\ z(x+y+z)&=8-xy \end{align*} what is the value of $|x+y+z|?$
(A) $2\quad$ (B) $2.5\quad$ (C) $3\quad$ (D) $3.5\quad$ (E) $4$
This is a problem from University of Waterloo's 2009 small C math contest. The correct answer is D, but there aren't any solutions on their website. I've found a system of equations that I solved for $3(x + y + z)^2 = x^2 + y^2 + z^2 + 28,$ but from there I'm not sure what to do. If anyone could show me how they did it, I'd appreciate it.
from our System we get $$(x+y)(x+z)=2$$ $$(x+y)(y+z)=4$$ and $$(x+z)(y+z)=8$$ so we have by dividing $$y=2x+z$$ $$z=x+2y$$ $$z=4x+3y$$ from here we obtain $$y=-3x$$ $$z=-5x$$ plugging this in the first equation of our System $$x(x-3x-5x)=2-15x^2$$ can you solve this? for your control: from here we get $$(x,y,z)=(\frac{1}{2};\frac{3}{2};\frac{5}{2})$$ or $$(x,y,z)=(-\frac{1}{2};-\frac{3}{2};-\frac{5}{2})$$