Solve the system of equations: 3x + 2y + 4z = 20, 13x + 12y + z = 4 and x+ y + z =9

3x + 2y + 4z =
20..............(1)

13x + 12y + z =
4...............(2)

x + y + z =
9....................(3)

We have a system of three
equations and three variables. Then, we will use the substitution and elimination method
to solve the system.

Let us rewrite
(3).

x + y + z = 9

==>
x = (9 - y -z)

We will substitute in (1) and
(2).

3x + 2y + 4z =
20

==> 3(9-y-z) + 2y + 4z =
20

==> 27 - 3y - 3z + 2y + 4z =
20

==> 7 - y + z =
0

==> y - z = 7
....................(4)

13x + 12y + z =
4........(2)

==> 13( 9- y - z) + 12y + z =
4

==> 117 - 13y -13z + 12y + z =
4

==> 113 - y - 12z =
0

==> y + 12z = 113
..............(5)

Now we will subtract (4) from
(5):

==> 13z = 
106

==> z= 106/
13

==> y= 7+ z = 7 + 106/13 =
197/13

==> y=
197/13

==> x = 9 - y  z = 9 - 197/13
- 106/13

               = ( 117 - 197 - 106)/13 = -186/
13

==> x= -186/
13