Solve a System of Equations Algebraically

Use SymPy to algebraically solve a system of equations, whether linear or nonlinear. For example, solving \(x^2 + y = 2z, y = -4z\) for x and y (assuming z is a constant or parameter) yields \(\{(x = -\sqrt{6z}, y = -4z),\) \({(x = \sqrt{6z}, y = -4z)\}}\).

Alternatives to Consider

Examples of Solving a System of Equations Algebraically

Whether your equations are linear or nonlinear, you can use solve():

Solve a System of Linear Equations Algebraically

>>> from sympy import solve
>>> from sympy.abc import x, y, z
>>> solve([x + y - 2*z, y + 4*z], [x, y], dict=True)
[{x: 6*z, y: -4*z}]

Solve a System of Nonlinear Equations Algebraically

>>> from sympy import solve
>>> from sympy.abc import x, y, z
>>> solve([x**2 + y - 2*z, y + 4*z], x, y, dict=True)
[{x: -sqrt(6)*sqrt(z), y: -4*z}, {x: sqrt(6)*sqrt(z), y: -4*z}]

Guidance

Refer to Include the Variable to be Solved for in the Function Call and Ensure Consistent Formatting From solve().

There are two methods below for containing solution results: dictionary or set. A dictionary is easier to interrogate programmatically, so if you need to extract solutions using code, we recommend the dictionary approach.

Solve and Use Results in a Dictionary

Solve Into a Solution Given as a Dictionary

You can solve a system of equations for some variables (for example, \(x\) and \(y\)) leaving another symbol as a constant or parameter (for example, \(z\)). You can specify the variables to solve for as multiple separate arguments, or as a list (or tuple):

>>> from sympy import solve
>>> from sympy.abc import x, y, z
>>> equations = [x**2 + y - 2*z, y + 4*z]
>>> solutions = solve(equations, x, y, dict=True)
>>> solutions
[{x: -sqrt(6)*sqrt(z), y: -4*z}, {x: sqrt(6)*sqrt(z), y: -4*z}]

Use a Solution Given as a Dictionary

You can then extract solutions by indexing (specifying in brackets) the solution number, and then the symbol. For example solutions[0][x] gives the result for x in the first solution:

>>> solutions[0][x]
-sqrt(6)*sqrt(z)
>>> solutions[0][y]
-4*z

Solve Results in a Set

To get a list of symbols and set of solutions, use set=True instead of dict=True:

from sympy import solve
from sympy.abc import x, y, z
solve([x**2 + y - 2*z, y + 4*z], [x, y], set=True)
([x, y], {(-sqrt(6)*sqrt(z), -4*z), (sqrt(6)*sqrt(z), -4*z)})

Options That Can Speed up solve()

Refer to Options That Can Speed up solve().

Not All Systems of Equations Can be Solved

Systems of Equations With no Solution

Some systems of equations have no solution. For example, the following two systems have no solution because they reduce to 1 == 0, so SymPy returns an empty list:

>>> from sympy import solve
>>> from sympy.abc import x, y
>>> solve([x + y - 1, x + y], [x, y], dict=True)
[]

from sympy import solve
from sympy.abc import x, y, z
solve([x + y - (z + 1), x + y - z)], [x, y], dict=True)
[]

The following system reduces to \(z = 2z\), so it has no general solution, but it could be satisfied if \(z=0\). Note that solve() will not assume that \(z=0\), even though that is the only value of \(z\) that makes the system of equations consistent, because \(z\) is a parameter rather than an unknown. That is, solve() does not treat \(z\) as an unknown because it is not in the list of symbols specified as unknowns ([x, y]) and all such symbols are treated like parameters with arbitrary value. Whether a symbol is treated as a variable or a parameter is determined only by whether it is specified as a symbol to solve for in solve(). There is no such distinction made when creating the symbol using symbols() (or importing from abc).

>>> from sympy import solve
>>> from sympy.abc import x, y, z
>>> solve([x + y - z, x + y - 2*z], [x, y], dict=True)
[]

The following system is overconstrained, meaning there are more equations (three) than unknowns to be solved for (two, namely \(x\) and \(y\)). It has no solution:

>>> from sympy import solve
>>> from sympy.abc import x, y, z
>>> solve([x + y - z, x - (z + 1), 2*x - y], [x, y], dict=True)
[]

Note that some overconstrained systems do have solutions (for example, if an equation is a linear combination of the others), in which case SymPy can solve the overconstrained system.

Systems of Equations With no Closed-Form Solution

Some systems of equations cannot be solved algebraically, for example those containing transcendental equations:

>>> from sympy import cos, solve
>>> from sympy.abc import x, y, z
>>> solve([x - y, cos(x) - y], [x, y], dict=True)
Traceback (most recent call last):
    ...
NotImplementedError: could not solve -y + cos(y)

So you can use nsolve() to find a numerical solution:

>>> from sympy import cos, nsolve
>>> from sympy.abc import x, y, z
>>> nsolve([x - y, cos(x) - y], [x, y], [1,1])
    Matrix([
    [0.739085133215161],
    [0.739085133215161]])

Equations Which Have a Closed-Form Solution, and SymPy Cannot Solve

It is also possible that there is an algebraic solution to your equation, and SymPy has not implemented an appropriate algorithm. If SymPy returns an empty set or list when you know there is a closed-form solution (indicating a bug in SymPy), please post it on the mailing list, or open an issue on SymPy’s GitHub page. Until the issue is resolved, you can use a different method listed in Alternatives to Consider.

Report a Bug

If you find a bug with solve(), please post the problem on the SymPy mailing list. Until the issue is resolved, you can use a different method listed in Alternatives to Consider.