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

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)}.