Okay yeah, that was my main question as to whether Sympy had the capability 
of doing so. Thanks! 

On Thursday, February 11, 2021 at 6:14:34 PM UTC-5 Oscar wrote:

> On Thu, 11 Feb 2021 at 22:54, mpierro3 <[email protected]> wrote:
> >
> > Yes, thank you. The rest is as follows:
> >
> > from sympy import *
> > import numpy as np
> >
> > M_e = 1
> > gamma_1 = 1.667
> > gamma_4 = 1.667
> > MW_1 = 39.948
> > MW_4 = 4.0026
> > D_4 = 5 / 39.37
> > D_1 = 3 / 39.37
> > A_4 = (np.pi / 4) * Pow(D_4, 2)
> > A_1 = (np.pi / 4) * Pow(D_1, 2)
> > a_4 = (gamma_4 + 1) / (gamma_4 - 1)
> > a_1 = (gamma_1 + 1) / (gamma_1 - 1)
> > A_4_A_1 = A_4 / A_1
> >
> > M_3a = symbols('M_3a', positive=true, nonzero=true)
> > eqn = nonlinsolve([Eq((M_e / M_3a) * Pow((2 + (gamma_4 - 1) * Pow(M_3a, 
> 2)) / \
> > (2 + (gamma_4 - 1) * Pow(M_e, 2)), a_4 / 2), A_4_A_1), M_3a <= 1], M_3a)
> > print(eqn)
>
> Your equation to be solved is (truncating the numbers for clarity):
>
> Eq(0.56*(0.33*M_3a**2 + 1)**1.99/M_3a, 2.77)
>
> I think it is unlikely that you will get analytic expressions for the
> solutions to an equation like this.
>
> Sympy can solve it numerically though:
>
> In [31]: nsolve(eq[0], M_3a, 0.2)
> Out[31]: 0.208399106769846
>
>
> --
> Oscar
>

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