Download Solving a System of Three Linear Equations in 3 Variables (3 by 3

Solving a System of Three Linear Equations in 3
Variables (3 by 3 linear systems)
A 3 by 3 linear system is a set of three equations in three variables x, y, and
z and looks like:
a1x + b1y + c1z = d1
a2 x + b2 y + c2 z = d 2 , where the a ' s , the b ' s , the c ' s , and the d ' s are all real numbers.
a3 x + b3 y + c3 z = d3
A solution will be a value of x and a value of y and a value of z that satisfy ALL
three equations.
The method of finding a solution consists of the following 5 steps:
1. Choose a pair of equations and eliminate one of the variables (your choice).
(At this stage you will have one linear equation with the other two variables.)
2. Choose a different pair of equations and eliminate the same variable as in
step 1.
(At this stage you will have another linear equation with the same two
variables as in step 1.)
3. The two equations from steps 1 and 2 now make up a 2 by 2 linear system
that can be solved using any of the three methods that you learned from
Class Practices on Systems of Equations #1 – #6.
4. Return to any of the original equations, and substitute the two numbers found
in step 3, and this will give the third number.
5. Check the three numbers in ALL three of the original equations.
EXAMPLE: Solve the system:
5 x − 2y − 4z = 3
(1)
Note that we number
3 x + 3 y + 2z = − 3
−2 x + 5 y + 3 z = 3
(2)
(3)
the equations for
reference purposes.
SOLUTION:
Step 1 Choosing equations (1) and (2), we can eliminate the variable z:
5 x − 2y − 4z = 3
5 x − 2y − 4z = 3
Leave as is -->
3 x + 3 y + 2z = − 3 Multiply by 2 --> 6 x + 6 y + 4z = − 6
11x + 4 y
= −3
Call this equation (4)
Step 2 Next, choose equations (2) and (3) and eliminate the variable z again:
3 x + 3 y + 2z = − 3 Multiply by − 3 -->
−2 x + 5 y + 3 z = 3
Multiply by 2
− 9 x − 9 y − 6z = 9
− 4 x + 10 y + 6z = 6
− 13 x + y
15
Call this equation (5)
-->
Step 3 Next, combine equations (4) and (5) to form a 2 by 2 linear system:
Equation (4)
11x + 4 y = − 3 Leave as is
--> 11x + 4 y = − 3
Equation (5) − 13 x + y = 15 Multiply by − 4 --> 52 x − 4 y = − 60
63 x
= − 63
x
= −1
Next, substitute x = –1 into either Equation (4) or (5). We will use
Equation (5).
−13 x
+ y = 15
−13( −1) + y = 15
13 + y = 15
y =2
Step 4 With x = –1 and y = 2, return to equation (1) to find the value of z:
5 x − 2y − 4z = 3
5( −1) − 2(2) − 4z = 3
− 5 − 4 − 4z = 3
− 9 − 4z = 3
− 4z = 12
z = −3
It seems as though we have the solution x = –1, y = 2, and z = –3 or (–1, 2, –3).
Step 5
5x
Check in (1)
− 2y − 4z
Check in (2)
=3
3x
+ 3 y + 2z
Check in (3)
=−3
− 2x
+ 5 y + 3z
=3
5( −1) − 2(2) − 4( −3) = 3
3( −1) + 3(2) + 2( −3) = −3 − 2( −1) + 5(2) + 3( −3) = 3
− 5 − 4 + 12
=3
−3 + 6
− 6 = −3
2 + 10 − 9
=3
=3
−3
= −3
=3
3
3
The solution x = –1, y = 2, and z = –3 or (–1, 2, –3) does check in ALL three
equations.