SOLUTION: Here is one that I am having some problems figuring out how to solve. It is one of the end of chapter 2 work exercises and it would help me understand this greatly. The solution

Question 84286This question is from textbook finite mathmatics
: Here is one that I am having some problems figuring out how to solve. It is one of the end of chapter 2 work exercises and it would help me understand this greatly.
The solution to the system of equations below is
X + 2Y - Z = -3
2X - Y + 3Z = 14
X + 4Y - 2Z = -8
1. (1, 1, 6)
2. (1, 3, 5)
3. (4, -1, 4)
4. (2, -1, 3)
PLEASE SHOW MO HOW TO GET THE CORRECT ANSWER.... I am going bonkers!!!
KeithThis question is from textbook finite mathmatics

Answer by PS(13) (Show Source):
You can put this solution on YOUR website!
Keith,
Here is how you should go about this.
EQ 1 : X + 2Y - Z = -3
EQ 2 : 2X - Y + 3Z = 14
EQ 3 : X + 4Y - 2Z = -8
Subtract from EQ 3 from EQ 1
EQ 1 : X + 2Y - Z = -3
- EQ 3 : X + 4Y - 2Z = -8
-2Y +Z = 5 --------------I
Add I to EQ 1
X + 2Y - Z = -3
-2Y +Z = 5
_________________________
X = 2
Substute this in EQ 2
EQ 2 : 2X - Y + 3Z = 14
4 - Y + 3Z = 14
- Y + 3Z = 10 ------II
Multiply II by 2
-2Y + 6Y = 20
Subtract I from this
-2Y + 6Y = 20
-2Y +Z = 5
____________________
5Z = 15
Z = 3
Substitute in EQ 1
X + 2Y - Z = -3
4 + 2Y - 3 = -3
2Y = -2
Y = -1
(X,Y,Z) is ( 2, -1, 3)
So the answer is option 4