Question 897529: Hello. I hope I put this question in the correct topic. I am norwegian, and I don't know the English equivalent of my question. Anyway, here goes:
x+2y+3z=14 I
4x-y-z=-1 II
7x+5y+z=20 III
What is x, y and z? (If you can show it from where I last learned, and figured out, that would be great):
x = 14-2y-3z
-> 4(14-2y-3z)-y-z=-1
-> 7(14-2y-3z)+5y+z=20
PS: I got
-9y-13z=-57
-9y-20z=-78 What then? (...if correct up to here)
Thank you so much in advance!! :)
Found 2 solutions by richwmiller, MathTherapy:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! 1,2,3,14
4,-1,-1,-1
7,5,1,20
add down (-4) *row 1 to row 2
1,2,3,14
0,-9,-13,-57
7,5,1,20
add down (-7) *row 1 to row 3
1,2,3,14
0,-9,-13,-57
0,-9,-20,-78
divide row 2 by -9
1,2,3,14
0,1,-13/-9,-57/-9
0,-9,-20,-78
add down (9) *row 2 to row 3
1,2,3,14
0,1,13/9,19/3
0,0,-7,-21
divide row 3 by -7
1,2,3,14
0,1,13/9,19/3
0,0,1,3
We now have the value for the last variable.
We will work our way up and get the other solutions.
add up (-13/9) *row 3 to row 2
1,2,3,14
0,1,0,2
0,0,1,3
add up (-3) *row 3 to row 1
1,2,0,5
0,1,0,2
0,0,1,3
add up (-2) *row 2 to row 1
1,0,0,1
0,1,0,2
0,0,1,3
1,0,0,1
0,1,0,2
0,0,1,3
"1","2","3"
(1,2,3)
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website! Hello. I hope I put this question in the correct topic. I am norwegian, and I don't know the English equivalent of my question. Anyway, here goes:
x+2y+3z=14 I
4x-y-z=-1 II
7x+5y+z=20 III
What is x, y and z? (If you can show it from where I last learned, and figured out, that would be great):
x = 14-2y-3z
-> 4(14-2y-3z)-y-z=-1
-> 7(14-2y-3z)+5y+z=20
PS: I got
-9y-13z=-57
-9y-20z=-78 What then? (...if correct up to here)
Thank you so much in advance!! :)
x + 2y + 3z = 14 ------- eq (i)
4x - y - z = - 1 ------- eq (ii)
7x + 5y + z = 20 ------- eq (iii)
PS: I got
-9y-13z=-57
-9y-20z=-78 What then? (...if correct up to here)
Yes, you are CORRECT up to here. Good job!!
- 9y - 13z = - 57 -------- eq (iv)
- 9y - 20z = - 78 -------- eq (v)
7z = 21 -------- Subtracting eq (v) from eq (iv)
z =  , or 
**Notice that when eq (v) was subtracted from eq (iv), the y variable was ELIMINATED
Next, you substitute 3 for z in either eq (iv) or eq (v)
- 9y - 13(3) = - 57 -------- Substituting 3 for z in eq (iv)
- 9y - 39 = - 57
- 9y = - 57 + 39
- 9y = - 18
y =  , or 
x + 2(2) + 3(3) = 14 ------- Substituting 3 for z, and 2 for y in eq (i)
x + 4 + 9 = 14
x + 13 = 14
x = 14 - 13
Solution: 
You can do the check!!
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