Solve x + 2y + 3z = 6, 3x + 2y + z = 4, 3x + 4y + 2z = 10

We have 3 equations to solve for x, y and
z

x + 2y + 3z = 6…(1)

3x + 2y
+ z = 4…(2)

3x + 4y + 2z =
10…(3)

(1) – (2)

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

=> -2x + 2z =
2

=> -x + z =
1…(4)

2*(2)- (3)

2*(3x + 2y +
z)- (3x + 4y + 2z) = 8 – 10

=> 6x + 4y + 2z – 3x -4y
-2z = -2

=> 3x =
-2

=> x = -2/3

Use this
in (4)

2/3 +z = 1

=> z
= 1- 2/3

=> z =
1/3

Substitute x = -2/3 and z = 1/3 in
(2)

=> -2 + 2y +1/3
=4

=> 2y = 4 +2
-1/3

=> 2y = 6 –
1/3

=> y = 3 –
1/6

=> y =
17/6

Therefore x= -2/3 , y = 17/6 and z =
1/3