Your question was:
2x-4y+3z=17 ........... [1]
x+2y-z=0 .............. [2]
4x-4y-z=6 ............. [3]
subtracting equation 3 from 2 we get,
1x + 2y - z = 0
4x - 4y - z = 6
- + + -
---------------
-3x + 6y = -6
Taking 3 common,
- x + 2y = -2
y= (-2 + x)/2
y = (x - 2)/2 ......... [4]
x + 2y - z = 0
x + 2(x-2)/2 - z = 0
x + x - 2 - z = 0
2x - 2 - z = 0
x = ( z + 2 ) / 2 ............[5]
Substituting 5 in 4,
Now we have x and y in terms of z,
x = (z+2)/2
y = (z-2)/4
Putting these relative values of x and y in equation 2 we get,
x + 2y - z = 0
Therefore, z = 0 ................[6]
Substituting this value of z in relative values of x and y we get,
x = (z+2)/2 = (0+2)/2 = 2/2 = 1
y = (z-2)/4 = (0-2)/4 = -2/4 = -1/2
Thus, x=1,y=-1/2,z=0
Cross check:
Hence,these values of x,y and z are correct.
Hope this helps,
Best of luck.