This seems like an incredibly simple and silly question to ask, but everything I've found about it has been too complex for me to understand.
I have these two very basic simultaneous equations:
X = 2x + 2z
Y = z - x
Given that I know both X and Y, how would I go about finding x and z? It's very easy to do it by hand, but I have no idea how this would be done in code.
This seems like an incredibly simple and silly question to ask
Not at all. This is a very good question, and it has unfortunately a complex answer. Let's solve
a * x + b * y = u
c * x + d * y = v
I stick to the 2x2 case here. More complex cases will require you to use a library.
The first thing to note is that Cramer formulas are not good to use. When you compute the determinant
a * d - b * c
as soon as you have a * d ~ b * c, then you have catastrophic cancellation. This case is typical, and you must guard against it.
The best tradeoff between simplicity / stability is partial pivoting. Suppose that |a| > |c|. Then the system is equivalent to
a * c/a * x + bc/a * y = uc/a
c * x + d * y = v
which is
cx + bc/a * y = uc/a
cx + dy = v
and now, substracting the first to the second yields
cx + bc/a * y = uc/a
(d - bc/a) * y = v - uc/a
which is now straightforward to solve: y = (v - uc/a) / (d - bc/a) and x = (uc/a - bc/a * y) / c. Computing d - bc/a is stabler than ad - bc, because we divide by the biggest number (it is not very obvious, but it holds -- do the computation with very close coefficients, you'll see why it works).
Now, if |c| > |a|, you just swap the rows and proceed similarly.
In code (please check the Python syntax):
def solve(a, b, c, d, u, v):
if abs(a) > abs(c):
f = u * c / a
g = b * c / a
y = (v - f) / (d - g)
return ((f - g * y) / c, y)
else:
f = v * a / c
g = d * a / c
x = (u - f) / (b - g)
return (x, (f - g * x) / a)
You can use full pivoting (requires you to swap x and y so that the first division is always by the largest coefficient), but this is more cumbersome to write, and almost never required for the 2x2 case.
For the n x n case, all the pivoting stuff is encapsulated into the LU decomposition, and you should use a library for this.