Unit 5 - Objective 6 - Solving 3 Equations in 3 Unknowns Algebraically


Now solve a system of 3 equations in 3 unknowns and find its solution which is the set of values x, y, and z that satisfy all three equations. The general method is to pair 2 equations and eliminate one variable. Pair 2 other equations and eliminate the same variable. That reduces the system to 2 equations and 2 unknowns. Solve as in the last objective and back substitute to get value for x, y, and z. Sometimes you get lucky so watch for ways to make the work easier.

Examples: Solve

(1) 2x + y - z = 4
(2) 4x - 3y - 2z = - 2
(3) 8x - 2y - 3z = 3

 

- 4x - 2y + 2z = - 8 (1) multiplied by - 2
4x - 3y - 2z = - 2 (2)

0x - 5y +0z = - 10 adding
(4) y = 2

 

2x + (2) - z = 4 substituting (4) into (1)
(5) 2x - z = 2

 

8x - 2 (2) - 3z = 3 substituting (4) into (3)
8x - 4 - 3z = 3
(6) 8x - 3z = 7

 

2x - z = 2 equation 6
8x - 3z = 7 equation 7

 

(8) - 6x + 3z = - 6 (6) multiplied by -3
8x - 3z = 7 (7)

2x + 0z = 1 adding
(9) x = 1/2

 

2 (1/2) + (2) - z = 4 substituting (9) and (5)
1 + 2 - z = 4
3 - z = 4
- z = 1
z = -1

 

So the solution x = 1/2, y = 2, and z = -1 satisfied the 3 equations or (1/2, 2, -1)

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