# Trying to solve simultaneous equations based on format found on official website [closed]

$\def\1{x_1}\def\2{x_2}\def\3{x_3}\def\f{f(\1,\2,\3)}\def\bs{\bigskip}\def\b{\begin{pmatrix}}\def\e{ \end{pmatrix}}\def\le{\left}\def\ri{\right}\def\g{g(\1,\2,\3)}\def\l{\lambda}\def\n{\nabla}\def\x{\lambda_1}\def\y{\lambda_2}$ I am putting in the following input:

Solve[{2 + 2 a*d + 2 a*e == 0, 1 - 2 b + 2 d*b + 2 e*b == 0,
1 + 2 d*c + 2 e*c == 0, -2 + a^2 - 2 b + b^2 + c^2 ==
0, -2 + a^2 + b^2 + c^2 == 0}, {a, b, c, d, e}]


And it is returning:

(* {} *)


What am I doing wrong? Is this not computable?

\begin{align} 2+2\x\1+2\y\1 = 0 \\ %8 1+2\x\2+2\y\2-2\y = 0 \\ %9 1+2\x\3+2\y\3=0 \\ %10 \1^2 + \2^2 - 2\2+\3^2-2=0 \\ %11 \1^2 + \2^2 + \3^2 - 2 = 0 %12 \end{align}

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## closed as off-topic by Michael E2, Artes, RunnyKine, Daniel Lichtblau, m_goldbergAug 24 '14 at 18:24

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• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Michael E2, Artes, RunnyKine, Daniel Lichtblau, m_goldberg
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Solve gives solutions in terms of rules of the form: {} no solutions. –  Apple Aug 24 '14 at 12:48
try Reduce .. –  george2079 Aug 24 '14 at 12:49
@george2079 it returned FALSE, does that mean I am out of luck? –  Display Name Aug 24 '14 at 12:52
you should easily work through this by hand if you dont believe the mathematica result. –  george2079 Aug 24 '14 at 13:06
@george2079 by the below answer it is hand calculable. I obvious don't know how to utilize the Mathematica function solve or reduce. –  Display Name Aug 24 '14 at 13:11

Oh, you just incorrectly type the equation in Mathematica, your second one should be: 1 - 2 d + 2 d b + 2 e b == 0, check the 2 d term. It's not a 2 b.

Solve[{2 + 2 a d + 2 a e == 0, 1 - 2 d + 2 d b + 2 e b == 0,
1 + 2 d c + 2 e c == 0, a^2 + b^2 + c^2 - 2 == 2 b,
a^2 + b^2 + c^2 - 2 == 0}, {a, b, c, d, e}]
(* {{a -> -2 Sqrt[2/5], b -> 0, c -> -Sqrt[(2/5)], d -> 1/2,
e -> 1/8 (-4 + 2 Sqrt[10])}, {a -> 2 Sqrt[2/5], b -> 0,
c -> Sqrt[2/5], d -> 1/2, e -> 1/8 (-4 - 2 Sqrt[10])}} *)


\begin{align} \label{1}2+2\x\1+2\y\1 = 0\tag{1} \\ \label{2}1+2\x\2+2\y\2-2\y = 0\tag{2} \\ \label{3}1+2\x\3+2\y\3=0\tag{3} \\ \label{4}\1^2 + \2^2 - 2\2+\3^2-2=0\tag{4} \\ \label{5}\1^2 + \2^2 + \3^2 - 2 = 0\tag{5} \end{align}

It's easier to solve by hands, consider this heuristics:

Last two equations:

\begin{align} \1^2 + \2^2 + \3^2 - 2 & = 2\2 \\ \1^2 + \2^2 + \3^2 - 2 & = 0 \end{align}

Now it's clear that $\2 = 0$. Then you have a restriction: $\lambda_2 = 1/2$ (from $\eqref{2}$ equation).

Now you have only two equations (from $\eqref{1}$ and $\eqref{3}$):

\begin{align} 2 + 2\x\1 + \1 = 0 \\ 1 + 2\x\3 + \3 = 0 \end{align}

Than the solution, for $\forall\x$:

\begin{align} \1 & = -\frac{2}{1 + 2\x} \\ \2 & = 0 \\ \3 & = -\frac{1}{1 + 2\x} \\ \y & = \frac{1}{2} \end{align}

But: We have one equation left:

$$\1^2 + \3^2 - 2 = 0$$

Substituting gives:

$$\frac{2^2}{(1 + 2\x)^2} + \frac{1}{(1 + 2\x)^2} = 2$$

From where: $\x = \frac{1}{4}(-2 \pm\sqrt{10})$. And finally: $\1 = -2\sqrt\frac{2}{5}$ and $\3 = 2\sqrt\frac{2}{5}$. Or with interchanged signs.

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You are pretty amazing to be honest. –  Display Name Aug 24 '14 at 13:03
Thank you for getting the Mathematica to work, and I must note, overall your hand answer is much more helpful! Thank you so much! –  Display Name Aug 24 '14 at 13:19
@DisplayName I just started in assumption that Mathematica is right and tried to prove there is no solution, but when found one - try to find, why Mathematica didn't found any. –  m0nhawk Aug 24 '14 at 13:20

You have a typo in your second equation.

eqns = {2 + 2 a*d + 2 a*e == 0,
1 - 2 e + 2 d*b + 2 e*b == 0,
1 + 2 d*c + 2 e*c == 0,
-2 + a^2 - 2 b + b^2 + c^2 == 0,
-2 + a^2 + b^2 + c^2 == 0};

Solve[eqns, {a, b, c, d, e}] // Simplify


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