SOLUTION: <pre> Write the system of linear equations, use x,y,z and if necessary w,x,y,z. Once system is written, use back-substitution to find its solution: |1 2 -1 0 | 2| |0 1

Write the system of linear equations, use x,y,z and if necessary w,x,y,z. 
Once system is written, use back-substitution to find its solution: 

|1  2 -1  0 |  2|
|0  1  1 -2 | -3|
|0  0  1 -1 | -2|
|0  0  0  1 |  3|


Put w's after the numbers in the first column:
Put x's after the numbers in the second column:
Put y's after the numbers in the third column:
Put z's after the numbers in the fourth column:

|1w 2x-1y 0z|  2|
|0w 1x 1y-2z| -3|
|0w 0x 1y-1z| -2|
|0w 0x 0y 1z|  3|

Put in + signs between terms if there's not a 
sign between them

|1w+2x-1y+0z|  2|
|0w+1x+1y-2z| -3|
|0w+0x+1y-1z| -2|
|0w+0x+0y+1z|  3|

Replace all the "|" after the z's with equal signs:

|1w+2x-1y+0z=  2|
|0w+1x+1y-2z= -3|
|0w+0x+1y-1z= -2|
|0w+0x+0y+1z=  3|

Erase all the terms with a 0 coefficient:

|1w+2x-1y   =  2|
|   1x+1y-2z= -3|
|      1y-1z= -2|
|         1z=  3|

Erase all the 1 coefficients:

| w+2x- y   =  2|
|    x+ y-2z= -3|
|       y- z= -2|
|          z=  3|

Write as a system of equations:



Now we do back substitution, starting at
the bottom with  and substitute 
that in  







Now we substitute  and  in







Finally we substitute  and  in






So the solution is (w,x,y,z) = (-1,2,1,3)

Edwin