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I am using findroot to solve 2 sets of transcendental equations which has a form

a/b = c/d and e/f = g/h. All these (a, b, c ,d , e, f, g, h) are function of one variable,x. Can findroot solve 2 sets of equations simultaneously for one variable. If it can can anyone tell me how to write it in mathematica?

n = 144/100;
c = 3*10^(8);
a = n/c;
b = 1/c;
f1[x_, r_] := 2^(1/3)/m^(1/3)*AiryAi[-2^((1/3))/m^(1/3)*((b*x*r) - m)];
f2[x_, r_] := 2^(1/3)/m^(1/3)*AiryAi[-2^((1/3))/m^(1/3)*((a*x*r) - m)];
f3[x_, r_] := -2^((1/3))/m^(1/3)*
   AiryBi[-2^((1/3))/m^(1/3)*((a*x*r) - m)];
f4[x_, r_] := 
 2^(1/3)/m^(1/3)*AiryAi[-2^((1/3))/m^(1/3)*((b*x*r) - m)] + 
  I *(-2^((1/3))/ (m^((1/3)))*AiryBi[-2^((1/3))/m^(1/3)*((b*x*r) - m)])
f5[x_, r_] := D[f2[x, r], r];
f6[x_, r_] := D[f3[x, r], r];
f7[x_, r_] := D[f4[x, r], r];
f8[x_, r_] := D[f1[x, r], r];
f9[x_] := (((f8[x, r]*f2[x, r]) - (f5[x, r]*f1[x, r]))/((f6[x, r]*
       f1[x, r]) - (f8[x, r]*f3[x, r]))) /. r -> 140*10^(-6); 
f10[x_] := (f8[x, r]/f1[x, r] - (f5[x, r] + (f9[x]*f6[x, r]))/(
     f2[x, r] + (f9[x]*f3[x, r]))) /. r -> 140*10^(-6);
f11[x_] := ((f5[x, r] + (f9[x]*f6[x, r]))/(
     f2[x, r] + (f9[x]*f3[x, r])) - f8[x, r]/f4[x, r]) /. 
   r -> 160*10^(-6);

I need to solve f10[x_] and f11[x_] simultaneously

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  • 1
    $\begingroup$ This seems the same question as your earlier one. My suggestion is to respond to the advice there, edit that question with the missing information, and see if anyone can help. What is added here about the structure ("a/b =...") provides no useful information than you have an overdetermined, (hopefully) consistent system. But there's still not enough information to formulate an answer to your question beyond, "Yes, sometimes." $\endgroup$ – Michael E2 Apr 13 '17 at 0:17
  • $\begingroup$ Possible duplicate of Finding roots simultaneously for transcendental equations $\endgroup$ – Michael E2 Apr 13 '17 at 0:17
  • $\begingroup$ @MichaelE2: I have edited the question. $\endgroup$ – Raj Apr 13 '17 at 0:30
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    $\begingroup$ @MichaelE2:m=1000 $\endgroup$ – Raj Apr 13 '17 at 0:39
  • $\begingroup$ Thanks, again. (Shouldn't that be put in the code). But note that f10 simplifies to zero: FullSimplify[f10[x]]. So you just have to solve f11[x] == 0, yes? $\endgroup$ – Michael E2 Apr 13 '17 at 0:46
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Like this?

FullSimplify[f10[x]]
(*  0  *)

FindRoot[f11[x] == 0, {x, -5*^12}, WorkingPrecision -> 20]
(*  {x -> -6.2886017387938934990*10^12}  *)
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