Example 3 - Elimination Methods of Solving Linear System of Equations

Example 3
Solve 3x3 system of equations by elimination methods
2 x + 4 y + 6 z = 4
1 x + 5 y + 9 z = 2
2 x + 1 y + 3 z = 7

Solution
Note:
3x3 system (eliminate a variable) --> 2x2 system (eliminate a variable) -->
1x1 system (solve for a variable) --> back substitution ...

2 x + 4 y + 6 z = 4                                                                                         (1)
2.(1 x + 5 y + 9 z = 2)                                                                                   (2)
1.(2 x + 1 y + 3 z = 7)                                                                                   (3)

use equation (1) to eliminate x in (2) e (3)
(could have also chosen to eliminate y or z)
eqn. (1) - eqn. (2):

     2 x + 4 y + 6 z      = 4
-
     2 x + 10 y + 18 z  = 4
________________    ___
     0 x - 6 y - 12 z      = 0                                                                               (4)

eqn. (1) - eqn. (3):

    2 x + 4 y + 6 z = 4
-
    2 x + 1 y + 3 z = 7
______________    __
    0 x + 3 y + 3 z = -3                                                                                   (5)


-6 y  - 12 z = 0                                                                                               (4)
2.(3 y + 3 z = -3)                                                                                           (5)

Now eliminate y or z between eqns. (4) & (5)
eliminate y: eqn. (4) + 2 eqn. (5)

   -6 y - 12 z = 0
+
   6 y + 6 z = -6
 ________    __
   0 y - 6 z = -6
\ z = 1

Back sub. to get y in eqn. (4) or eqn. (5).
(5): 3 y + 3 . 1 = -3  ,  (z = 1)
       3 y = -6  
\ y = -2


Now back sub. y = -2 and z = 1 in eqn. (1), (2) or (3).
(2): x + 5 (-2) + 9 (1) = 2
\ x = 3

\ solution is
[x, y, z]
[3 , -2 , 1]
NB: Satisfies equations 1-3
Check