Question 152205
<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  1 -2 | -3|
|0  0  1 -1 | -2|
|0  0  0  1 |  3|

<font size = 4 color = "indigo"><b>
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:

{{{system(w+2x-y=2,x+y-2z=-3,y-z=-2,z= 3)}}}

Now we do back substitution, starting at
the bottom with {{{z=3}}} and substitute 
that in {{{y-z=-2}}} 

{{{y-z=-2}}}
{{{y-(3)=-2}}}
{{{y-3=-2}}}
{{{y=3-2}}}
{{{y=1}}}

Now we substitute {{{y=1}}} and {{{z=3}}} in
{{{x+y-2z=-3}}}
{{{x+(1)-2(3)=-3}}}
{{{x+1-6=-3}}}
{{{x-5=-3}}}
{{{x=5-3}}}
{{{x=2}}}

Finally we substitute {{{x=2}}} and {{{y=1}}} in
{{{w+2x-y=2}}}
{{{w+2(2)-(1)=2}}}
{{{w+4-1=2}}}
{{{w+3=2}}}
{{{w=-1}}}

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

Edwin</pre>