Solve the following system of equations

1. X+y-z= 6
2x-y +z=-9
X-2y+3z=1

2. x+z=-3
y+z=3
x+y=8

3 answers

1. 1st plus 2nd ... 3x = 3

twice the 1st plus 3rd ... 3x + z = 13

2. 1st minus 2nd plus 3rd ... 2x = 2
1.

x + y - z = 6
+
2 x - y + z = - 9
______________

3 x = - 3

Divide both sides by 3

x = - 1

2 x - y + z = - 9 Multiply both sides by - 3

- 6 x + 3 y - 3 z = 27

- 6 x + 3 y - 3 z = 27
+
x - 2 y + 3 z = 1
_________________

- 5 x + y = 28

- 5 ∙ ( - 1 ) + y = 28

5 + y = 28

Subtract 5 to both sides

5 + y - 5 = 28 - 5

y = 23

x + y - z = 6

- 1 + 23 - z = 6

22 - z = 6

Subtract 22 to both sides

22 - z - 22 = 6 - 22

- z = - 16

Multiply both sides by - 1

z = 16

The solutions are:

x = - 1 , y = 23 , z = 16

2.

x + z = - 3

Subtract z to both sides

x + z - z = - 3 - z

x = - 3 - z

x + y = 8

Subtract y to both sides

x + y - y = 8 - y

x = 8 - y

x = x

- 3 - z = 8 - y

Add 3 to b oth sides

- 3 - z + 3 = 8 - y + 3

- z = 11 - y

Multiply both sides by - 1

z = - 11 + y

z = y - 11

Now:

y + z = 3

y + y - 11 = 3

2 y - 11 = 3

Add 11 to both sides

2 y - 11 + 11 = 3 + 11

2 y = 14

Divide both sides by 2

y = 14 / 2

y = 7

x = 8 - y

x = 8 - 7

x = 1

z = y - 11

z = 7 - 11

z = - 4

The solutions are:

x = 1 , y = 7 , z = - 4
x+y-z= 6
2x-y +z=-9
X-2y+3z=1

using substitution, you can do:

from #3,
x = 2y-3z+1
Using that in #1 and #2, you get

(2y-3z+1)+y-z = 6
2(2y-3z+1)-y+z = -9

3y-4z = 5
3y-5z = -11

Now, using 3y=4z+5,
(4z+5)-5z = -11
-z = -16
z = 16
3y = 4*16+5 = 69, so y = 23
x = 2*23-3*16+1 = -1

and similarly for the 2nd problem