Solve 5x + 6y + z = 9, 6x + 3 y = 6, 2x + 6y + z =4

5x+6y+z=9...........(1)

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

2x+6y+z=4
.................(3).

The 2nd equation is free from
z.

So we eleminate z between equations (1) and (3) by
Eq(1)-eq(3) which also eliminates y. So we solve x
:

(5x+6y+z) - (2x+6y+z) = 9 - 4 =
5

3x = 5.

x =
5/3.

Substitute x = 5/3 in
eq(2):

6x+3y = 6. Or 3y = 6-6x = 6-6(5/3) = 6-10 =
-4.

3y = -4.

y=
-4/3.

Substitute x = 5/3 and y =-4/3 in
(1):

5x+6y+z =
9

5(5/3)+6(-4/3)+z = 9

25/3
-24/3 +z = 9

1/3 +z = 9

z =
9-1/3 = 26/3

Therefore x = 5/3 , y =-4/3 and z
=28/3