Solve for x, y , z.

3x+4y+2z=10..........(1)

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

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

(1)- (3):

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

2y+z = 6............(4)

3(2)-(1):

3(x+2y+3z) - (3x+4y+2z) = 3*6-10

2y +7z = 8...........(5)

(5)-(4):

7z-z = 8-6 = 2

6z = 2

z = 2/6 = 1/3. Substituting z= 1/3 in (5), we get:

2y+7(1/3) = 8.

2y = 8-7/3 = 17/3.

Threfore y = (17/3)/2 = 17/6.

Substitute y = 17/6 and z = 1/3 in (1)

3x+4(17/6) + 2(1/3) = 10

3x = 10 -34/3 -2/3 = 10

3x = 10 -12

3x = -2

3x = -2/3

Therefore x = -2/3 , y = 17/6 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**

We'll note the equations as:

3x + 2y + z = 4 (1)

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

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

We'll subtract (2) from (1):

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

We'll combine and eliminate like terms:

2x - 2z = -2

We'll divide by 2:

x - z = -1 (4)

We'll multiply (1) by 2 and we'll get:

2*(3x + 2y + z) = 8

We'll remove the brackets:

6x + 4y + 2z = 8 (5)

We'll subtract (3) from (5):

6x + 4y + 2z -3x - 4y - 2z = 8 – 10

We'll combine and eliminate like terms:

3x = -2

We'll divide by 3:

**x = -2/3**

We'll substitute x in (4)

-2/3 - z = -1

We'll multiply by -1:

2/3 + z = 1

We'll subtract 2/3 both sides:

z = 1 - 2/3

**z = 1/3**

We'll substitute x = -2/3 and z = 1/3 in (2)

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

-2/3 + 2y + 1 = 6

2y = 5 + 2/3

2y = (15 + 2)/3

We'll divide by 2:

**y = 17/6**

**The solution of the system is: {(-2/3 , 17/6, 1/3)}.**