# Solve the given system of quadratic equations

$$x^2 + xy + xz = 2$$ $$y^2 + yz + xy = 3$$ $$z^2 + zx + yz = 4$$

I tried solving it but i just ended up with $$(x-y)(x+y+z) = -1$$ $$(y-z)(x+y+z) = -1$$ $$(x-z)(x+y+z) = -2$$

I'm not sure what to do. Any hints?

• You can add $(x+y+z)^2=2+3+4=9$. – Bernard Nov 18 '20 at 18:38

Your equations are equivalent to $$x(x+y+z)=2,\quad y(x+y+z)=3,\quad z(x+y+z)=4.$$ Set $$x+y+z=t,$$ then $$t^2=2+3+4=9$$ and we have $$t=\pm 3.$$ Hence $$x=\pm\dfrac23,\qquad y=\pm 1,\qquad z=\pm\dfrac43.$$
Take $$x = 2t \; , \; \; y = 3 t \; , \; \; z = 4t \; . \;$$ We know this is valid because $$x+y+z$$ must be nonzero, so too the individual letters, and $$\frac{y}{x} = \frac{3}{2} \; , \; \; \frac{z}{x} = \frac{4}{2} \; .$$Then $$x+y+z = 9t$$
Solve first equation for $$y$$, substitute in the others, simplify...