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\begin{align*} x\lambda_{1}&=-\frac{1}{2}\\ y\lambda_{2}&=\frac{1}{2}\\ z\lambda_{1}+z\lambda_{2}&=1\\ x^{2}+z^{2}&=1\\ y^{2}+z^{2}&=1\\ \end{align*}

I wonder if there's any simple method of solving such systems of equations...is there?

I've been always doing this just intuitively, but there are too much equations here, I get stuck at $x^{2}=y^{2}$...

khernik
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    There is a general method if all equations are of the form $polynomial = 0$. However, I suggest you keep using your "intuitive" method because this general method is not very suitable for human calculation (unless you're desperate). – Tunococ Aug 03 '13 at 21:36
  • khernik: I saw your (now deleted) question. Have you checked out Paul's Online Notes? I linked you to "triple integrals" but there's also links, at that page, to many other topics in Calc II, Calc III: notes, step-by-step examples, exercises with solutions, etc. Also check out the khanacademy.com. – amWhy Aug 05 '13 at 13:04

1 Answers1

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For a system of quadratic equations, as pointed out by Tunococ there is no general method you'll want to carry out by hand. Mathematica and other software can be used to easily find the solutions, and if any old numerical solution will do (you don't need to find all of the solutions, and a very good approximation is OK i.e. you don't need the solution in closed form) a simple implementation of Newton's method often works.

Otherwise, you'll want to take advantage of special structure of the problem. For instance, you can see immediately from your last two equations that $x = \pm y$.

If $x=0$ the first equation is impossible. So $x\neq 0$. Similarly, let's take the plus case first. Then $\lambda_1 = -\lambda_2$ and the third equation is impossible. So we must have $x = -y$ and $\lambda_1 = \lambda_2$.

Eliminating lambda gives us

\begin{align*} \frac{2z}{-2x} &= 1\\ x^2+z^2 &= 1, \end{align*} in other words $x=z = \pm\frac{1}{\sqrt{2}}.$

user7530
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