SOLUTION: I do not understand how to solve systems of linear equations in three variables. Could someone please explain how I would solve this problem x+y-z=-1 -4x+-y+2z=-7

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Question 153681: I do not understand how to solve systems of linear equations in three variables.
Could someone please explain how I would solve this problem
x+y-z=-1
-4x+-y+2z=-7
2x+-2y+-5z=7

Found 2 solutions by Fombitz, ktsau:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
There are many ways to solve.
We'll use elimination.
Combine the equations to get rid of one of the variables.
1.x%2By-z=-1
2.-4x%2B%28-y%29%2B2z=-7
3.2x%2B%28-2y%29%2B%28-5z%29=7
Add eq.1 and eq.2, to get rid of y.
%28x%2By-z%29%2B%28-4x-y%2B2z%29=-1-7
4.-3x%2Bz=-8
Multiply eq. 2 by (-2) and add it to 3
-2%28-4x-y%2B2z%29%2B%282x-2y-5z%29=-2%28-7%29%2B7
8x%2B2y-4z%2B2x-2y-5z=14%2B7
5.10x-9z=21
Now you have two equations 4 and 5 in x and z.
Keep going.
Multiply eq. 4 by (9) and add to eq. 5 to get rid of z.
9%28-3x%2Bz%29%2B%2810x-9z%29=9%28-8%29%2B21
-27x%2B9z%2B10x-9z=-72%2B21
-17x=-51
highlight%28x=3%29
Now that you have x you can work back to get z and then y.
5.10x-9z=21
10%283%29-9z=21
-9z=21-30
-9z=-9
highlight%28z=1%29
Finally,
1.x%2By-z=-1
3%2By-1=-1
2%2By=-1
highlight%28y=-3%29
This method will work with as many number of equations as you have although it does get more tedious and prone to making mistakes when the number of equations gets too large.

Answer by ktsau(1) About Me  (Show Source):
You can put this solution on YOUR website!
Hi! We usually use simultaneous equations to solve this kind of question.
Firstly, name the formulas:
x+y-z=-1 -- (1)
-4x+-y+2z=-7 -- (2)
2x+-2y+-5z=7 -- (3)
the next thing we do is to prevent those three variables appear in a same formula, so, we need to remove one of them by subtracting or adding the formulas together. At the same time, we need to express a variable in terms of another variable. If you look careful enough, you will find 'y' can be removed by adding (1) and (2) together:
(1)+(2): x+y-z+(-4x-y+2z) = -1+(-7)
x-4x+y-y-z+2z = -8
-3x+z = -8
z = 3x-8 -- (4)
As shown above, i've expressed z in terms of x, and a new formula has produced. What i usually do is to give a name for new formulas: (4). Okay, let's ignore (4) for a second. And take a look at (2) and (3), i'm going to remove 'y' by substracting. But before that, we need to multiply (2) by 2, otherwise, the 'y' won't go.
(2)x2: (-4x+-y+2z)x2=-7x2
-8x-2y+4z =-14 -- (5)
Now, we can remove 'y' by substracting:
(5)-(3): -8x-2y+4z-(2x-2y-5z) = -14-(7)
-8x-2x-2y+2y+4z+5z = -21
-10x+9z = -21
At this moment, we substitute (4) into above formula.
-10x+9(3x-8) = -21
-10x+27x-72 = -21
17x = -21+72
17x = 51
x = 3
Now we've got one variable solved. Let's use 'x' to solve the others!
In here, we substitute x=3 into (4):
z=3(3)-8
z=1
Here's the last step, substitute all the knowns into one formula, i'll use the easiest one.
Sub. x=3, z=1 into (1)
(3)+y-(1) =-1
y =-1-3+1
y =-3
so, x=3, y=-3 and z=1. You may also check the answers by substituting them into other formulas. ( From (2): -4(3)-(-3)+2(1) = -7 )
Hope it helps!
see you later!