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 Solve system of equations {$$x+yz=2,\; y+zx=2, \; z+xy=2$$}
02-17-2010, 12:36 PM
Post: #1
 elim Moderator Posts: 581 Joined: Feb 2010 Reputation: 0
Solve system of equations {$$x+yz=2,\; y+zx=2, \; z+xy=2$$}
Clearly we have $$(x-y)(1-z)=0$$ from the 1st two equations.
If $$x=y$$ then the last two equations become {$$x(1+z)=2, \; z+x^2=2$$}, so $$2 = x(1+z)=x(1+(2-x^2))$$
i.e. $$(x-1)^2(x+2)=0$$ thus $$(x=1)∨(x=-2)$$. Therefore the solutions are
$$x=y=z=1$$ or $$x=y=z=-2$$.
If $$z=1$$, then the original equations become {$$x+y=2, \; xy=1$$} so $$x(2-x)=1$$ i.e. $$(x-1)^2=0$$.
Again we get $$x=y=z=1$$
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