Put in missing variables with 0 coefficients
and give single variables 1 coefficents



Put the coefficients in a 4×5 augmented matrix and do row operations
to get 0's below the diagonals:



-1*R1+R4->R4   (Multiply row 1 by -1, add to row 4, put it in row 4.) 



R2+R4->R4    (Add row 2 and R4 and put it in row 4)  



-R3+R4->R4





Eliminate the 0s and 1 coefficients, which
means we eliminate the entire bottom equation. 



We let 



Solve the bottom equation of the system: 

Substitute in the middle equation and solve:



{{x[2]-5-s = 5}}}
{{x[2]= 10+s}}}

Substitute in the top equation












or in the form that you gave:

[x1] = [5] + [-1]s 
[x2] = [10] + [1]s
[x3] = [-5] + [-1]s 
[x4] = [0] + [1]s

Edwin