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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
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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