# Solving 3 non-linear equations (with 3 unknowns)

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?

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

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 $$x^2=2$$.