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