Solve system of 2 equations in python

Question

I have a set of two equations with three unknowns that has some conditions. x, y and z must all be larger than zero. How can I solve this? There is only one solution and I already know it, but I want to know how to get to it correctly.

This is the equations:

100 = x + y + z

100 = 10x +2.5y + 0.5z

Need to find x, y and z. They are integers and positive.

This is the code I have that does not work:

from sympy import symbols, Eq, solve
x, y, z = symbols('x y z')
eq1 = Eq(x + y + z, 100)
eq2 = Eq(x*10 + y*2.5 + z*0.5, 100)
#eq3 = x, y, z must all be larger than zero and integers
solution = solve((eq1,eq2), (x,y,z))
solution

Answer 1

In sympy if you want to find integer solutions to equations then you should use diophantine . It doesn't handle systems of equations but you can put the solution from one equation into the other and call diophantine again:

In [69]: eq1 = x + y + z - 100

In [70]: eq2 = 10*x + 5*y/2 + z/2 - 100

In [71]: sol = diophantine(eq1, t, syms=[x, y, z])

In [72]: sol
Out[72]: {(t₀, t₀ + t₁, -2⋅t₀ - t₁ + 100)}

In [73]: [xt, yt, zt], = sol

In [74]: eq3 = eq2.subs({x:xt, y:yt, z:zt})

In [75]: eq3
Out[75]: 
23⋅t₀            
───── + 2⋅t₁ - 50
  2              

In [76]: t1, t2 = eq3.free_symbols

In [77]: [t1s, t2s], = diophantine(eq3, z, syms=[t1, t2])

In [78]: rep = {t1:t1s, t2:t2s}

In [79]: (xt.subs(rep), yt.subs(rep), zt.subs(rep))
Out[79]: (4⋅z₀ - 100, 500 - 19⋅z₀, 15⋅z₀ - 300)

The solution here is in terms of an integer parameter z0. This gives the set of solutions to the two equations but you also have the requirement that x, y, z are positive which constrains the possible values of z0:

In [80]: ineqs = [s.subs(rep) > 0 for s in [xt, yt, zt]]

In [81]: ineqs
Out[81]: [4⋅z₀ - 100 > 0, 500 - 19⋅z₀ > 0, 15⋅z₀ - 300 > 0]

In [82]: solve(ineqs)
Out[82]: 
               500
25 < z₀ ∧ z₀ < ───
                19

In [83]: 500/19
Out[83]: 26.31578947368421

We see that z needs to be 26 which gives a unique solution for x , y and z :

In [84]: z, = ineqs[0].free_symbols

In [85]: (xt.subs(rep).subs(z, 26), yt.subs(rep).subs(z, 26), zt.subs(rep).subs(z, 26))
Out[85]: (4, 6, 90)

Answer 2

You did not explicitly state it but according to your comment x,y and z should be integers. This complicates matters a bit. This is now an example of a mixed integer programming (MIP) problem. You could take a look at the following package for solving this in python: mip

The downside of solving MIP's is that they are NP hard. But for this small example this should not matter.

Answer 3

This type of problem can be solved by Z3py , a SAT/SMT solver:

from z3 import Ints, solve

x, y, z = Ints('x y z')
sol = solve(x + y + z == 100, x * 100 + y * 25 + z * 5 == 1000, x > 0, y > 0, z > 0)
print(sol)

Output: [z = 90, y = 6, x = 4] .

Note that in general, Z3 only looks for one solution. To find subsequent solutions, a clause needs to be added to prohibit the already found solutions. (In this case there seems to be only one solution.)