Question 588053
Here is one exactly like it

{{{system(2x + y - z = 3,x - y + z = 0
,y + 2z = 1)}}}


{{{(matrix(3, 5, 2, 1,-1, "|", 3,
                 1,-1, 1, "|", 0,
                 0, 1, 2, "|", 1))}}}
<pre><font size color = "indigo"><b>
Swap row 1 and 2 to get a 1 in the upper left corner

{{{(matrix(3, 5, 1,-1, 1, "|", 0, 
                 2, 1,-1, "|", 3,
                 0, 1, 2, "|", 1))}}}

Multiply Row 1 temporarily by -2, getting 

{{{(matrix(1, 5, -2,2, -2, "|", 0))}}}
and add that to row 2:
 {{{(matrix(3, 5, 1,-1, 1, "|", 0, 
                  0, 3,-3, "|", 3,
                  0, 1, 2, "|", 1))}}} 
Swap rows 2 and 3:

 {{{(matrix(3, 5, 1,-1, 1, "|", 0,
                  0, 1, 2, "|", 1, 
                  0, 3,-3, "|", 3))}}}

Multiply Row 2 temporarily by -3, getting 

{{{(matrix(1, 5, 0, -3, -6, "|", -3))}}}
and adding that to row 3:

 {{{(matrix(3, 5, 1,-1, 1, "|", 0,
                  0, 1, 2, "|", 1, 
                  0, 0,-9, "|", 0))}}}

Mutiply row 3 by {{{-1/9}}}

 {{{(matrix(3, 5, 1,-1, 1, "|", 0,
                  0, 1, 2, "|", 1, 
                  0, 0, 1, "|", 0))}}}

Convert back into a system of equations:

{{{system(x-y+z=0, 0x+1y+2z=1,0x+0y+1z=0)}}}

or just

{{{system(x-y+z=0, y+2z=1,z=0)}}}

Substitute {{{z=0}}} from the bottom equation
into the middle equation {{{y+2z=1}}}

{{{y+2(0)=1}}}
{{{y+0=1}}}
{{{y=1}}}

Substitute {{{z=0}}} and {{{y=1}}}
into the top equation {{{x-y+z=0}}}

{{{x-(1)+(0)=0}}}
{{{x-1=0}}}
{{{x=1}}}

So the solution is {{{"(x,y,z)"="(1,1,0)"}}}

Edwin</pre>