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I am trying to solve this system:

$$ \begin{aligned} x (2 + \lambda) &= 0 \\ 2 y (\lambda - 1) &= 0 \\ \frac12 x^2 + y^2 - 1 &= 0 \end{aligned} $$

I am not sure how do I need to consider the different possible cases. For example, from the first equation we have that $x=0$ or $\lambda=-2$ then what I exactly have to do with this? Is it just to substitute $x=0$ in the third equation and so get $y$? And same do foe $\lambda=-2$, if I subtitute it in the second equation I get $y=0$ so get $x$ from the third equation?

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2 Answers 2

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There can be EXACTLY two cases: either $\lambda=-2$ or $\lambda=1$. Otherwise, if $\lambda\notin\{1,-2\},$ then we will have from the first two equations $x=y=0$ which won’t satisfy the third equation.

Case I: $\textbf{$\lambda=-2$}$

Then, $-6y=0$ so that $y=0$. Thus, $x=\pm\sqrt 2$.

Case II: $\lambda=1$

$3x=0\implies x=0$ and $y=\pm 1$.

So the solutions are $$(x,y,\lambda)=(\pm\sqrt2,0,-2), (0,\pm1,1).$$

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One can start at any place here with a case distinction. The first equation, for example, leads to the following two cases:

Case 1: $x=0$. Then the third equation gives $y=\pm 1\neq 0$, so that the second equation gives $\lambda=1$.

Case 2: $x\neq 0$. Then $\lambda=-2$, so that the second equation gives $y=0$ and the third equation gives $x^2=2$.

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