Question 664130
The system


x-y=2
y-z=0
x+z=6 


is the same as


1x-1y+0z=2
0x+1y-1z=0
1x+0y+1z=6 


and that converts to this matrix


<pre>
1	-1	0	2
0	1	-1	0
1	0	1	6
</pre>


Now perform row reduction 


<pre>
1	-1	0	2
0	1	-1	0
1	0	1	6


1	-1	0	2	
0	1	-1	0	
0	1	1	4	R3 + (-1)*R1


1	0	-1	2	R1 + (1)*R2
0	1	-1	0	
0	1	1	4	


1	0	-1	2	
0	1	-1	0	
0	0	2	4	R3 + (-1)*R2


1	0	-1	2	
0	1	-1	0	
0	0	1	2	0.5*R3


1	0	0	4	R1 + (1)*R3
0	1	-1	0	
0	0	1	2	


1	0	0	4	
0	1	0	2	R2 + (1)*R3
0	0	1	2	


</pre>


The right hand column has the values, 4, 2 and 2.


So x = 4, y = 2, and z = 2 are the three solutions


The solution as an ordered triple is therefore (4, 2, 2)