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I have the system of equations: $$2x(1+\lambda)=0$$$$2y(1+\lambda)=0$$$$2z(1-\lambda)=0$$$$x^2+y^2-(z^2+1)=0$$ It's easy to plug in a few values and see that the solution is $x^2+y^2=1$, $z=0$, and $\lambda=-1$, but is there a way to solve algebraically or some general method that doesn't include guessing and checking values?

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  • $\begingroup$ Just consider the tautology $\lambda =-1$ or $\lambda \neq -1$. The second case easily yields contradictions. $\endgroup$
    – Git Gud
    Apr 13, 2016 at 19:10

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This can be solved completely by algebraic methods. Suppose first that $\lambda=1$. Then $(x,y,z)=(0,0,\pm i)$. Now suppose that $\lambda=-1$. Then $(x,y,z)=(\pm \sqrt{1-y^2},y,0)$ for all $y$. Finally, let $\lambda\neq \pm 1$. Then there is no solution, because $x=y=z=0$ from the first three equations, but then we obtain $0=1$ in the last equation.

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  • $\begingroup$ So there is not a way to solve without breaking it into different cases of $\lambda$? $\endgroup$
    – Tburke2
    Apr 13, 2016 at 19:22
  • $\begingroup$ It depends what you mean by this. Suppose you have to solve the nonlinear system of one equation $\lambda x^2=0$. You must argue with $\lambda=0$ or $x=0$. What else do you want to do ? $\endgroup$ Apr 13, 2016 at 19:29
  • $\begingroup$ I guess I just assumed it could be solved like a linear system by using substitutions. $\endgroup$
    – Tburke2
    Apr 13, 2016 at 19:50

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