Solve a+3b+2c= 6, a-b+4c= 2 and c-b-a =2

a + 3b + 2c = 6
.........(1)

a - b + 4c =
2.............(2)

-a - b + c = 2
.............(3)

To solve the system, we are going to use
th elimination methos.

First we will add (2) and
(3):

==> -2b + 5c = 4
...........(4)

Now subtract (2) from
(1)

==> 4b - 2c = 4
..........(5)

Now multiply (4) by (2) and add tp
(5):

==> 8c =
12

==> c=
12/8

==> c =
3/2

Now to calculate b, we will substitute
in (4)

-2b + 5c = 4

==>
b= (5c - 4)/2 = 5*3/2 - 4 )/2 =
7/4

==> b =
7/4

Now to find a, we will substitute in
(2)

==> a -b + 4c =
2

==> a= b - 4c +
2

            = 7/4 - 4*3/2 +
2

              = (7- 24 + 8)/4 =
-9/4

==> a=
-9/4