# equation solving – how to solve system of polynomials restricted to positive reals? I’m new to Mathematica and would like to use it to solve a system of polynomials. In this system, every variable (and every solution I’m interested in) is a positive real number.

I tried this:

``````Solve(k1*A*N - k2*C == 0 && k3*B*N - k4*D == 0 &&
N0 - C - D - N == 0 && A0 - C - A == 0 && B0 - D - B == 0, {A, B, C,
D, N})
``````

which generates an enormous expression and isn’t constrained to positive reals. I am using Mathematica 11.3.0.0 which I believe doesn’t support `PositiveReals` as an argument to `Solve`. When I try to encode the constraints as additional equations, Mathematica hangs:

``````Solve(k1*A*N - k2*C == 0 &&
k3*B*N - k4*D == 0 &&
N0 - C - D - N == 0 &&
A0 - C - A == 0 &&
B0 - D - B == 0 &&
k1 >= 0 && k2 >= 0 && k3 >= 0 && k4 >= 0 && A >= 0 && B >= 0 && N >= 0 && C >= 0 && D >= 0 && A0 >= 0 && B0 >= 0 && N0 >= 0,
{A, B, C, D, N}, Reals)
``````

Trying to encode the constraints with `Assuming` also hangs:

``````Assuming(k1 >= 0 && k2 >= 0 && k3 >= 0 && k4 >= 0 && A >= 0 &&
B >= 0 && N >= 0 && C >= 0 && D >= 0 && A0 >= 0 && B0 >= 0 &&
N0 >= 0,
Solve(k1*A*N - k2*C == 0 &&
k3*B*N - k4*D == 0 &&
N0 - C - D - N == 0 &&
A0 - C - A == 0 &&
B0 - D - B == 0,
{A, B, C, D, N}, Reals))
``````

any ideas on how to solve this system? I don’t expect it to have any enormously complicated solutions. Posted on