How do I solve this system of equations?

Question

x-2y+z=-4
y+2z=-2
x+y+3z=-6

Answer 1

(i) x-2y+z=-4
(ii) y+2z=-2
(iii) x+y+3z=-6

substitution.
(i) says that x = 2y-z-4
so. using that, we get
y + 2z = -2
2y-z-4+y+3z = -6
so, that means we have
y + 2z = -2
3y + 2z = -2
Now, since y = -2z-2, we have
3(-2z-2) + 2z = -2
-4z = 4
z = -1
and y = -2z-2 = -2(-1)-2 = 0
x = 2y-z-4 = 0-(-1)-4 = -3
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elimination
subtract (i) from (iii) to get
x - 2y + z = -4
......y + 2z = -2
....3y + 3z = -6
now add 2*2nd to 1st, and subtract the 2nd from the 3rd/3, and you have
x + ... + 5z = -8
.......y + 2z = -2
...............z = -1
Now subtract multiples of 3rd from 1st and 2nd to get
x = -3
y = 0
z = -1
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Cramer's Rule
D =
|1 -2 1|
| 0 1 2|
| 1 1 3| = -4

Dx =
|-4 -2 1|
|-2 1 2|
|-6 1 3| = 12

Dy =
|1 -4 1|
|0 -2 2|
|1 -6 3| = 0

Dz =
|1 -2 -4|
|0 1 -2|
|1 1 -6| = 4
so, now e have
x = Dx/D = -3
y = Dy/D = 0
z = Dz/D = -1