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I am trying to solve the following system of equations:

$c (\alpha+\beta \gamma)=\alpha x, c(\gamma\alpha+\beta)=-\beta x$

with respect to $x$ ($\alpha, \beta,\gamma,c \in \mathrm{R}$)

The solution is easy to work by hand and is $x=\pm c \sqrt{1-\gamma^2}$

Now I try:

Solve[c (α + β γ) == α x && 
c (γ α + β) == -β x, x]
(* output *)
{}

no output, so I explicitly state that $\alpha, \beta,\gamma,c \in \mathrm{R}$

Solve[c (α + β γ) == α x && 
c (γ α + β) == -β x, 
x /; Element[α | β | γ | c | x, Reals]]
(* output *)
{}

again no output. As a final try I give the constraint $\gamma^2 < 1$ (although that shouldn't be necessary):

Solve[c (α + β γ) == α x && 
c (γ α + β) == -β x, x /; γ^2 < 1]
(* output *)
{}

no luck either. What am I missing?

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    $\begingroup$ Tow equations for one variable do not have a general solution. There is a solution only for specific values of the parameters. You may try "Reduce" to find these special solutions. $\endgroup$ Jun 7, 2023 at 10:39
  • $\begingroup$ You provided two equations and ask for one unknown, so your unknown is overdetermined. Solving the first equation gives $x = c(1+\beta \gamma / \alpha)$ while solving the second equation gives $x = - c(1 + \gamma \alpha / \beta)$. These are not identical (unless you've got more information than you've shared) so the solution set is empty. But given what you claim the answer is I suspect you've got more information. For example, try Solve[c (\[Alpha] + \[Beta] \[Gamma]) == \[Alpha] x && c (\[Gamma] \[Alpha] + \[Beta]) == -\[Beta] x, {x, \[Gamma]}] $\endgroup$
    – evanb
    Jun 7, 2023 at 10:41
  • $\begingroup$ @evanb This is an intermediate step of a larger calculation, actually $\alpha, \beta$ are also to be determined, after $x$ has been found $\endgroup$
    – geom
    Jun 7, 2023 at 10:46
  • $\begingroup$ @DanielHuber thanks for the comment, I also tried Reduce but didn't get the desired output $\endgroup$
    – geom
    Jun 7, 2023 at 10:51
  • $\begingroup$ @geom the point is: now is the time to determine one of alpha, beta, or gamma. $\endgroup$
    – evanb
    Jun 7, 2023 at 20:30

2 Answers 2

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You can also succeed by solving the two equations one after the other,for instance begin with the second and solve for a or b like

Solve[c (a + b g) == a x /. (Solve[{c (b + a g) == -b x}, a]) // 
Evaluate, x]
(* {{x -> -Sqrt[c^2 - c^2 g^2]}, {x -> Sqrt[c^2 - c^2 g^2]}} *)
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  • $\begingroup$ I was just thinking that :). Thanks $\endgroup$
    – geom
    Jun 7, 2023 at 11:22
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when you have less equations than the number of variables and want to solve for one variable, then if you gives all the variables to solve, it will do it and give all possible solutions

eq1 = c (α + β γ) == α x;
eq2 = c (γ α + β) == -β x;

sol = Solve[{eq1, eq2}, {x, α, c , β , γ, α}]

Mathematica graphics

If you pick the 3rd solution and do couple of manipulations, you will get what you want with assumption that α > 0

 solB = Solve[γ == -((    2 α β)/(α^2 + β^2)), β]

Mathematica graphics

And now

Solve[FullSimplify[x == (c (α^2 - β^2))/(α^2 + β^2) /. #, 
    Assumptions -> α > 0], x] & /@ solB

Mathematica graphics

I do not know if there is a more direct way to do this., May be someone knows.

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