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I have the following system of equations:

(substituting any real values for a,b,c,r,Alpha,Beta will return the same answer for x)

eq1 = x == -(((a - b) (a + b + 2 r) + ((a - c) (b + r) (a + c + 2 r) Sin[\[Alpha]])/((c + r) Sin[\[Beta]]))/(2 (a + r + ((b + r) ((a + r) Sin[\[Alpha]] + (c + r) (Cos[\[Beta]] Sin[\[Alpha]] + Cos[\[Alpha]] Sin[\[Beta]])))/((c + r) Sin[\[Beta]]))))

eq2 = x == -(((a - c) (b + r) (a + c + 2 r) + ((a - b) (c + r) (a + b + 2 r) Sin[\[Beta]])/Sin[\[Alpha]])/(2 (a + r) (b + r) + (2 (c + r) ((a + r) Sin[\[Beta]] + (b + r) (Cos[\[Beta]] Sin[\[Alpha]] + Cos[\[Alpha]] Sin[\[Beta]])))/Sin[\[Alpha]]))

eq3 = x == -((((b + r) Sin[\[Alpha]] + (c + r) Sin[\[Beta]]) ((a - c) (a + c + 2 r) - ((b - c) (c + r) (b + c + 2 r) Sin[\[Beta]])/((b + r) Sin[\[Alpha]] + (c + r) Sin[\[Beta]])))/(2 ((a + r) (b + r) Sin[\[Alpha]] + (c + r) ((a + r) Sin[\[Beta]] + (b + r) (Cos[\[Beta]] Sin[\[Alpha]] + Cos[\[Alpha]] Sin[\[Beta]])))))

I require a single expression for X in terms of a,b,c,r. (Alpha and Beta should cancel). (3 equations, 3 unknowns)

I am trying to run the following code, but after many hours it still has not returned an answer:

Solve[Eliminate[{eq1, eq2 , eq3}, {\[Alpha] , \[Beta]}], x, Reals]

Is this the most efficent way to get the answer? I have tried solving for Alpha and Beta separatly but again... hours passed without an answer.

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eqn= {eq1, eq2, eq3} /. Equal -> Subtract

Assuming -Pi<\[Alpha],\[Beta]<Pi Weierstrasssubstituion

eqnu = eqn /. { \[Alpha] -> 2 ArcTan[u\[Alpha]], \[Beta] ->2 ArcTan[u\[Beta]]} // TrigExpand //FullSimplify 

evaluates three identical equations eqnu

Union[eqnu]
(*{(-b^2 (c + r) (1 + u\[Alpha]^2) u\[Beta] - 
c r (c + 2 r) u\[Alpha] (1 + u\[Beta]^2) + 
a^2 ((b + r) u\[Alpha] + (c + r) (1 + u\[Alpha]^2) u\[Beta] + (b +
       r) u\[Alpha] u\[Beta]^2) + 
2 r (2 r (u\[Alpha] + u\[Beta]) + 
   c (u\[Alpha] + 2 u\[Beta] - u\[Alpha] u\[Beta]^2)) x + 
2 a ((b + r) u\[Alpha] + (c + r) (1 + u\[Alpha]^2) u\[Beta] + (b +
       r) u\[Alpha] u\[Beta]^2) (r + x) + 
b (-c^2 u\[Alpha] (1 + u\[Beta]^2) - 
   2 c (u\[Alpha] + u\[Beta]) (r + r u\[Alpha] u\[Beta] - x + 
      u\[Alpha] u\[Beta] x) - 
   2 r (r (1 + u\[Alpha]^2) u\[Beta] - 
      2 u\[Alpha] x + (-1 + u\[Alpha]^2) u\[Beta] x)))/(2 ((b + 
     r) (a + c + 2 r) u\[Alpha] + (c + r) (a + b + 
     2 r + (a - b) u\[Alpha]^2) u\[Beta] + (a - c) (b + 
     r) u\[Alpha] u\[Beta]^2))}*)

That's why elimination of \[Alpha],\[Beta] is not possible!

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  • $\begingroup$ Nice trick. Just to be clear, you're saying that all three equations are actually identical? $\endgroup$ Jun 21 at 15:47
  • $\begingroup$ Yes, Union[eqnu] gives one equation! $\endgroup$ Jun 21 at 15:48
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    $\begingroup$ Or Simplify[eq1[[2]] - eq2[[2]]]. $\endgroup$
    – user64494
    Jun 21 at 15:50

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