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

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### 2 Answers

x+2y+3z=6.............(1)

3x+2y+z=4..............(2)

3x+4y+2z =10..........(3)

(1)-(2):

-2x+2z = 2

Divide by -2:

x-z = -1.........(4)

(3)-2*(2):

(3x-6x) = 10-2*4 = 2

-3x = 2

3x = -2. Substitute x = -2/3 in (4): (-2/3) -z = -1. So z = -2/3+1 = 1/3

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

x+2y+3z =6

-2/3+2y +3(1/3) = 6

2y =6 +2/3-1 = 17/3

y = 23/6 = 23/6

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

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**