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

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?

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    $\begingroup$ You can add $(x+y+z)^2=2+3+4=9$. $\endgroup$ – Bernard Nov 18 '20 at 18:38
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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 $$x=\pm\dfrac23,\qquad y=\pm 1,\qquad z=\pm\dfrac43.$$

Edit: Btw, here we hove only two solutions for the system.

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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 $$ x+y+z = 9t $$

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Solve first equation for $y$, substitute in the others, simplify...

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I solved it by solving for (x+y+z) in all three original equations. Then you can compare the results to each other.

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