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

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We have to solve the system of equations:

3x + 2y + 4z = 20…(1)

13x + 12y + z = 4…(2)

x+ y + z =9…(3)

Now (2) – (3)

=> 13x + 12y + z – (x+ y + z) = 4 – 9

=> 13x + 12y + z – x - y - z = -5

=> 12x + 11y= -5

4*(3) – (1)

=> 4x + 4y + 4z – 3x – 2y – 4z = 36 – 20

=> x + 2y = 16

=> x = 16 – 2y

substituting this in 12x – 11y= -5

=> 12*( 16 – 2y) + 11 y = -5

192 – 24 y + 11 y = -5

=> 13y = 197 => y = 197 / 13

x = 16 – 2y

=> x = 16 – 2*(197/13)

=> x = -186/13

x + y + z = 9

=> -186/13 + 197/13 + z = 9

=> z = 9 + 186/13 - 197/13

=> 106/13

**Therefore x = -186/13, y = 197 / 13 and z = 106/13**

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**

To solve:

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

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

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

4(3) - (1) gives: x+2y = 36-20 = 16

x+2y = 16....(4)

(2)-(3) gives: 12x+11y = 4-9 = -5.

12x+11y = -5....(5).

(5)- 12*(4) gives : 11y -24y = -5-12(16) = -197

Therefore -13y = -197.

y = 197/13.

From (4) , we get x = 16-2(197/13) = -186/3.

Therefore from (3), we get: z= 9-(x+y) = 9 - (197-186)/13) = 106/13.

Therefore x= -186/13 , y = 197/13 and z = 106/13.