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I am trying to solve a system of equation which seems to be simple but much as I tried I couldn't solve it in Mathematica and it says the coefficients need to be exact. Now, I was wondering if someone walk me through the step by step procedure of solving such equations.

I have a variable, phi, which is a function of P. P is a real number and 0 < P < 1.

phi = 45 beta^P; 

where beta is a constant.

Now the system of equations:

x^(-1/2) y^(-P) - x^(-1/2) y^P phi^2 = 3 phi
2 x y^(-P) - x y^(1 - P) = 4/phi
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    $\begingroup$ In Mathematica, equations are expressed with == not =, which is used for assignment. $\endgroup$
    – m_goldberg
    Jun 6, 2014 at 23:32

1 Answer 1

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This might get you started?

eqns = {x^(-1/2) y^(-P) - x^(-1/2) y^P ϕ^2 == 3 ϕ,
  2 x y^(-P) - x y^(1 - P) == 4/ϕ}


(*
==> {y^-P/Sqrt[x] - (ϕ^2 y^P)/Sqrt[x] == 3 ϕ, 
 2 x y^-P - x y^(1 - P) == 4/ϕ}
*)

Now first solve for x

eqns2 = eqns[[1]] /. First@Solve[eqns[[2]], {x}]

(*
==> y^-P/(2 Sqrt[-(y^P/((y - 2) ϕ))]) - (ϕ^2 y^P)/(
  2 Sqrt[-(y^P/((y - 2) ϕ))]) == 3 ϕ
*)

Then take a numerical example:

eqnn = eqns2 /. ϕ -> Pi/4 /. P -> 3 // Simplify

(*
==> -((Pi^2 y^6 - 16)/(
  64 y^3 Sqrt[y^3/(2 Pi - Pi y)])) == (3 Pi)/4
*)

For which we can solve for y

Solve[eqnn, y]

(*
==> {{y -> 
   Root[Pi^4 #1^13 - 2 Pi^4 #1^12 + 2304 Pi #1^9 - 
      32 Pi^2 #1^7 + 64 Pi^2 #1^6 + 256 #1 - 512 &, 1]}, {y -> 
   Root[Pi^4 #1^13 - 2 Pi^4 #1^12 + 2304 Pi #1^9 - 
      32 Pi^2 #1^7 + 64 Pi^2 #1^6 + 256 #1 - 512 &, 6]}, {y -> 
   Root[Pi^4 #1^13 - 2 Pi^4 #1^12 + 2304 Pi #1^9 - 
      32 Pi^2 #1^7 + 64 Pi^2 #1^6 + 256 #1 - 512 &, 7]}, {y -> 
   Root[Pi^4 #1^13 - 2 Pi^4 #1^12 + 2304 Pi #1^9 - 
      32 Pi^2 #1^7 + 64 Pi^2 #1^6 + 256 #1 - 512 &, 12]}, {y -> 
   Root[Pi^4 #1^13 - 2 Pi^4 #1^12 + 2304 Pi #1^9 - 
      32 Pi^2 #1^7 + 64 Pi^2 #1^6 + 256 #1 - 512 &, 13]}}
*)
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