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Since you have linear equations, it'd be more straightforward and efficient to employ linear algebra rather than symbolic solve() function.

If I got you manipulations correctly, then instead of my_solve you can simply use

M = matrix(QQ, 3, [2,1,0,-1,0,   0,1,0,1,1,   1,0,-1,2,0])
v = vector(QQ, [4,4,0])
s = M.solve_right(v)
K = M.right_kernel().basis()
print(s)
print(K)

which prints

(0, 4, 0, 0, 0)
[
(1, 0, 5, 2, -2),
(0, 1, 2, 1, -2)
]

These are exactly the "free term" and the vectors (up to a factor) you multiply by w and u in the solution you search for.

Since you have linear equations, it'd be more straightforward and efficient to employ linear algebra rather than symbolic solve() function.

If I got you your manipulations correctly, then instead of my_solvemy_solve() you can simply use

M = matrix(QQ, 3, [2,1,0,-1,0,   0,1,0,1,1,   1,0,-1,2,0])
v = vector(QQ, [4,4,0])
s = M.solve_right(v)
K = M.right_kernel().basis()
print(s)
print(K)

which prints

(0, 4, 0, 0, 0)
[
(1, 0, 5, 2, -2),
(0, 1, 2, 1, -2)
]

These are exactly the "free term" and the vectors (up to a factor) you multiply by w and u in the solution you search for.

Since you have linear equations, it'd be more straightforward and efficient to employ linear algebra rather than symbolic solve() function.

If I got your manipulations correctly, then instead of my_solve() you can simply use

M = matrix(QQ, 3, [2,1,0,-1,0,   0,1,0,1,1,   1,0,-1,2,0])
v = vector(QQ, [4,4,0])
s = M.solve_right(v)
K = M.right_kernel().basis()
print(s)
print(K)

which prints

(0, 4, 0, 0, 0)
[
(1, 0, 5, 2, -2),
(0, 1, 2, 1, -2)
]

These are exactly the "free term" and the vectors (up to a factor) you multiply by real coefficients w and u in to obtain the solution you search for.