how many solutions to non-linear simultaneous equations I'm doing a Lagrange multiplier optimization problem, and I wound up with the following simultaneous equations:
$2x + 1 -2\lambda x = 0$
$4y-2 \lambda y = 0$
$6z-2 \lambda z = 0$
$-x^2 - y^2 - z^2 + 1 = 0$
I am asked to find ALL solutions, and I managed to find 2 solutions (had to assume that y and z were both zero so equation 2 and 3 would make sense) :
$\lambda_1 = \frac{1}{2}, x_1 = -1$
$\lambda_2 = \frac{3}{2}, x_2 = 1$
But how do I know that these two are ALL the solutions? Is there any way to tell how many solutions I should be able to get, or maybe suggest how to get more solutions?
Thanks.
 A: Since the 2nd and 3rd equation are contradictory when both y and z are not zero, I must try several scenarios where just y, just z, and both y and z are equal to 0. For the y=z=0 case:
Solve equation 1 for x:
$x = \frac{-1}{2-2\lambda}$
plug into equation 4, solve for lambda:
$\lambda = \frac{8 \pm \sqrt{64-48}}{8}$
For each of these two answers, plug back into the x equation for 2 solutions:
$
 x_1 = -1, 
 y_1 = 0, 
 z_1 = 0, 
 \lambda_1 = \frac{1}{2}$  
$
 x_2 = 1 ,
 y_2 = 0 ,
 z_2 = 0 ,
 \lambda_2 = \frac{3}{2}
$
Now take the case where z = 0,
Solve equation 2 for lambda:
$\lambda = 2$
Into equation 1:
$x = \frac{1}{2}$
both of these into equation 4:
$y = \pm \sqrt{\frac{3}{4}}$
Giving two more solutions of:
$x_3 = \frac{1}{2}, y_3 = \sqrt{\frac{3}{4}}, z_3 = 0, \lambda_3 = 2$
$x_4 = \frac{1}{2}, y_4 = -\sqrt{\frac{3}{4}}, z_4 = 0, \lambda_4 = 2$  
Taking the case where y = 0:
Solve equation 3 for lambda:
$\lambda = 3$
Into equation 1:
$x = \frac{1}{4}$
both into equation 4:
$z = \pm \sqrt{\frac{15}{16}}$
Giving the final two solutions of:
$x_5 = \frac{1}{4}, y_5 = 0, z_5 = \sqrt{\frac{15}{16}}, \lambda_5 = 3$
$x_6 = \frac{1}{4}, y_6 = 0, z_6 = -\sqrt{\frac{15}{16}}, \lambda_6 = 3$
I can't think of any other possibility that would be mathematically legal. If I assume x=0, the first equation becomes $1=0$ which can't happen. Similarly, if I assume lambda = 0, then the 4th equation will eventually become impossible.
A: Here is a somewhat different way to organize solving the equations. Observe that the three equations may be expressed as 
$$2x(\lambda-1)=1,\quad 2y(\lambda-2)=0,\quad 2z(\lambda-3)=0.$$ The last two equations imply three cases:


*

*$\boxed{\lambda\neq 2,3}\quad$ We require $y=z=0$, and then the constraint $x^2+y^2+z^2=1$ yields $\hspace{2.25cm}$ $x=\pm 1$. The first equation then gives $\lambda=1\mp\frac12=\frac12,\frac32$.

*$\boxed{\lambda=2}\quad$ Solving the first and third equations yields $(x,z)=
(\frac12,0)$. The constraint then $\hspace{1.8cm}$ gives $y=\pm \frac{\sqrt{3}}{2}$.

*$\boxed{\lambda=3}\quad$ Solving the first and second equations yields $(x,y)=
(\frac14,0)$. The constraint then $\hspace{1.8cm}$ gives $z=\pm \frac{\sqrt{15}}{4}$.
This gives a total of six solutions. As a check, the objective function for this problem is $F(x)=x^2+x+2y^2+3z^2=(x+\frac{1}{2})^2+2y^2+3z^2$ i.e. level sets of ellipses centered at $(-\frac12,0,0)$. It's worth visualizing that to confirm that six solutions are the expected number.
