SOLUTION: 2x-3y+4z=1
x+3y-2z=4
3x-y+z=2
So far i have tried this..
2x-3y+4z=-1
x+3y-2x=4 = 3x+0+2z=3
and then i tried another problem to solve that one..
x+3y-2z=4
Algebra ->
Rational-functions
-> SOLUTION: 2x-3y+4z=1
x+3y-2z=4
3x-y+z=2
So far i have tried this..
2x-3y+4z=-1
x+3y-2x=4 = 3x+0+2z=3
and then i tried another problem to solve that one..
x+3y-2z=4
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Question 349108: 2x-3y+4z=1
x+3y-2z=4
3x-y+z=2
So far i have tried this..
2x-3y+4z=-1
x+3y-2x=4 = 3x+0+2z=3
and then i tried another problem to solve that one..
x+3y-2z=4 x+3y-2z=4
3(3x-y+z=2) = 9x-3y+3z=6
and my answer to tis solution was.. 10x-0+z=10
in my opinion it isn't right. Found 2 solutions by Fombitz, earmstrong620:Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! 1.
2.
3.
You can add eq. 1 and eq. 2 to eliminate .
4.
Multiply eq. 3 by 3 and add to eq. 2,
5.
.
.
.
Multiply eq. 5 by (-2) and add to eq. 4 to eliminate .
Now work backwards to solve for and
From eq. 5,
Then from eq. 3,
You can put this solution on YOUR website! 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