Substitution Methods of Solving Linear System of Equations

Substitution

Example:
Solve the 2x2 system below by substitution:
2 x -  1 y = 2                        (1)
8 x + 2 y = 4                        (2)

Solution:
Solve equation (1) or (2) for one variable (say x) in terms of the other variable (say y):
from (1), solve for y:
y = 2 x - 2                             (3)

Substitute this into unused equation ie. into (2)
--> 8 x + 2 (2x - 2) = 4
--> 12x = 8
--> x = 8/12 = 2/3 » 0.67

Back substitute to find value of other variable.
Sub. into (3) that x = 2/3
\ y = 2 (2/3) - 2
\ y = 4/3 - 6/3 = -2/3
\ y = -2/3 » -0.07
solution is: x = 2/3 ; y = -2/3 or (2/3 , -2/3)

Example:
Solve the 3x3 system below by substitution:
2 x + 4 y + 6 z = 4                                  (1)
   x + 5 y + 9 z = 2                                  (2)
2 x +   y + 3 z = 7                                   (3)

Solution:
idea:
2x2 (sub.) ---> 1x1 ---> solve
3x3 (sub.) ---> 2x2 (sub.) ---> 1x1 ---> solve

From (2): x = 2 - 5 y - 9 z
sub. this into (1) and (3) (ie. sub. into all other unused equations)
(1) --> 2 (2 - 5 y - 9 z) + 4 y + 6 z = 4
(3) --> 2 (2 - 5 y - 9 z) +   y + 3 z = 7
simplifying...
-6 y - 12 z = 0      (4)
-9 y - 15 z = 3      (5)

Now 2x2!!
From (4): 6 y = -12 z --> y = -2 z   (6)
sub. (6) into (5):
-9 (-2 z) - 15 z = 3 ; 18 z - 15 z = 3

Now 1x1!!
From (6) we find: y = -2 z = -2 (1)
\ y = -2
\ x = 2 - 5 (-2) - 9 (1)
\ x = 3

\ x = 3 ; y = -2 ; z = 1 or (3 , -2 , 1)

Note: If had 4x4 system...
4x4 (sub. one variable) ---> 3x3 (sub. one variable) ---> 2x2 (sub.one variable) --> 1x1 ---> solve!!