Question 2831
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,

<pre>
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]
</pre>
<br>
x + 2y - z = 0
x + 2(x-2)/2 - z = 0
x + x - 2 - z = 0
2x - 2 - z = 0
<br>
x = ( z + 2 ) / 2 ............[5]
<br>
Substituting 5 in 4,
{{{ y = (x-2)/2 }}}
{{{ y = (((z+2)/2)-2)/2 }}}
{{{ y = (z-2)/4 }}}
<br>
Now we have x and y in terms of z,
x = (z+2)/2
y = (z-2)/4
<br>
Putting these relative values of x and y in equation 2 we get,
x + 2y - z = 0
<br>
{{{ (z+2)/2 + 2((z-2)/4) -z = 0 }}}
{{{ (z + 2 + z - 2)/2 - z = 0 }}}
{{{ 2z - z = 0 }}}
<br>
Therefore,<B> z = 0 </b> ................[6]
<br>
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
<br>
Thus, <B>x=1,y=-1/2,z=0</b>
<P>
Cross check:
{{{ x + 2y - z = 0 }}}
{{{ (1) + 2(-1/2) - (0) = 0 }}}
{{{ 1 - 1 - 0 = 0 }}}
{{{ 0 - 0 = 0 }}}
<B> Hence,these values of x,y and z are correct.</b>
<P>
Hope this helps,
Best of luck.