<PRE><B>Question 349108</B>
Ok This is a simple problem as i have figured out over the last two minutes.

the problem is : 2x-3y+4z=-1
                 x+3y-2z=4
                 3x-y+z=2


now the solutions for solving this problem..

2x-3y+4z=-1
x+3y-2z=4
____________

3x+2z=3



and so you reduce another problem to get x and z alone like in the other problem..


x+3y-2z=4
3x-y+z=2  but you want to multiply the second equation by 3


x+3y-2z=4
3(3x-y+z=2)

and that gets you...


x+3y-2z=4
9x-3y+3z=6
___________ 
10x+z=10


ok so now you want to reduce from the first problem that we solved..

3x+2z=3
10x+z=10   but now you want to multiply z by -2 
_________

3x+2z=3           3x+2z=3 
-2(10x+z=10)   = -20x-2z=-20
_____________     ____________

now your answer to get x is..

-17x=-17
now divide by -17 and your answer to x is..

                         x=1

Now to solve z..

2(1)-3y+4z (we are now subbing out x)
1+3y-2z=4
___________
3+0+2z=3
_________ now subtract 3 from both side and you get 0..

2z=0

and so ..                        z=0


Now to solve for y..(now we are subbing x and z)

 1+3y-0=4
-1     -1
__     __
0       3   and now  3y= 3



and now divide by 3 and your answer is y=1



  So your solution is..  S= (1,1,0)
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