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# solve x+y+z=1 2xy-z^2=1 for real x,y,z

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solve x+y+z=1 2xy-z^2=1 for real x,y,z

dhantheswar  Sep 14, 2014

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Since both equations = 1, we have

x + y + z = 2xy - z^2    or, alternatively

x + y - 2xy + z^2 + z = 0

As Melody indicates, because we have more variables than equations, there are infinitely many solutions.....

The graph of this is a "two-sheeted" hyperboloid....... here it is displayed in WolframAlpha

CPhill  Sep 14, 2014
Sort:

#2
+80928
+10

Since both equations = 1, we have

x + y + z = 2xy - z^2    or, alternatively

x + y - 2xy + z^2 + z = 0

As Melody indicates, because we have more variables than equations, there are infinitely many solutions.....

The graph of this is a "two-sheeted" hyperboloid....... here it is displayed in WolframAlpha

CPhill  Sep 14, 2014

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