Solve doesn't produce a solution, although one exists

Question

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?

Answer 1

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]}} *)

Answer 2

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}]

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]

And now

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

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