Solve system of equations {\(x+yz=2,\; y+zx=2, \; z+xy=2\)}

02172010, 12:36 PM
Post: #1




Solve system of equations {\(x+yz=2,\; y+zx=2, \; z+xy=2\)}
Clearly we have \((xy)(1z)=0\) from the 1st two equations.
If \(x=y\) then the last two equations become {\(x(1+z)=2, \; z+x^2=2\)}, so \(2 = x(1+z)=x(1+(2x^2))\) i.e. \((x1)^2(x+2)=0\) 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 \(x(2x)=1\) i.e. \((x1)^2=0\). Again we get \(x=y=z=1\) 

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