Say you have some m equations in n variables, x1, x2, ... xn

lhs1 = rhs1
lhs2 = rhs2
...
lhs_m = rhs_m

You can pick some variable, let's just say x1, and solve each equation for that variable and get an equivalent system:
x1 = rearranged1
x1 = rearranged2
...
x1 = rearranged_m

rearranged1...m only contain variables x2, ..., xn

You can "hold on" to the first equation, x1 = rearranged1, and then reduce this system to:

rearranged1 = rearranged2
...
rearranged1 = rearranged_m

This is now just a linear system w/ (m-1) equations and (n-1) variables. So you can just repeat the process until you run out of either equations or variables

If m = n, then you eventually get down to 1 equation in 1 variable, and you can solve for that variable like normal.

You can then substitute this back into each of the previous equations that you "held onto". This should reduce the previous equation you held onto to an equation in 1 variable, and you can then solve for that one, and repeat, all the way back up to x1 = rearranged1, at which point the system is solved.

Some different considerations need to be made if m > n, or m < n, that's what I'm working on now.

Example:
x + y + z = 5
x + 2y + 3z = 7
x + 3y + 4z = 9

Solve for x in each equation:
x = 5 - y - z
x = 7 - 2y - 3z
x = 9 - 3y - 4z

Hold onto x = 5 - y - z

Equate the RHSs, reduce to a system with 2 equations in just y and z:

5 - y - z = 7 - 2y - 3z
5 - y - z = 9 - 3y - 4z

Repeat:
y = 2 - 2z
y = 2 - (3/2)z

Hold onto y = 2 - 2z

Equate the RHSs, reduce to a system w/ 1 equation in just z:
2 - 2z = 2 - (3/2)z

Solve for z:
z = 0

Substitute z back into y = 2 - 2z:
y = 2

Substitute y and z back into x = 5 - y - z:
x = 3

Solution:
x = 3
y = 2
z = 0

Verify against Wolfram Alpha:

https://www.wolframalpha.com/input/?i=row+reduce+%7B%7B1%2C+1%2C+1%2C+5%7D%2C+%7B1%2C+2%2C+3%2C+7%7D%2C+%7B1%2C+3%2C+4%2C+9%7D%7D

* The right-most column in the "Result" matrix