Solving Systems of Linear Equations in Three Variables

Report
Solving Systems of Linear
Equations in Three Variables
AII, 2.0: Students solve systems of linear equations and
inequalities (in two or three variables) by substitution, with
graphs, or with matrices.
LA, 6.0: Students demonstrate an understanding that linear
systems are inconsistent (have no solutions), have exactly
one solution, or have infinitely many solutions
Before we begin
QUIZ ON SOLVING A SYSTEM OF
TWO EQUATIONS (30 MINUTES)
Solving Systems of Linear Equations in
Three Variables
Objectives
1. Solve systems of linear
equations in three
variables
a)
Use the linear combination
method
b) Solve a system with NO
solution
c) Solve a system with Infinitely
many solutions
2. Solve systems of three
variables by substitution
Key Words
• Ordered triple
• Linear combination
• System of equations
The Linear Combination
Method (3-Variable
System)
1)
a)
WRITE THIS DOWN!!!
The steps to solving a system of linear
equations in three variables.
Example:
REWRITE the linear system in three
variables as a linear system in two
variables using the linear combination
method.
b)
2)
SOLVE the new linear system for both of
its variables.
a)
2 −  + 6 = −4
6 + 4 − 5 = −7
−4 − 2 + 5 = 9
b)
c)
d)
3)
−3,4,1
4)
Combine two equations and eliminate one
variable
Combine a different pair of equations to
eliminate the same variable
Combine the equations that resulted from
steps 1 and 2 to eliminate one of the two
variables that are left
Solve for the remaining variable
Substitute that value into one of the
equations with two variables
Solve for the second variable
SUBSTITUTE the values found in Step 2
into on of the original equations and solve
for the remaining variable.
CHECK the solution in each of the original
equations
Example 1
Use the Linear Combination Method
Solve the system.
3x + 2y + 4z = 11
Equation 1
2x – y + 3z = 4
Equation 2
5x – 3y + 5z = – 1
Equation 3
SOLUTION
STEP 1 Rewrite the system as a system in two variables.
First, add 2 times Equation 2 to Equation 1 to
eliminate y.
3x + 2y + 4z = 11
3x + 2y + 4z = 11
2x – y + 3z = 4
4x – 2y + 6z = 8
7x
+ 10z
= 19
New Equation 1
Example 1
Use the Linear Combination Method
Now add –3 times Equation 2 to Equation 3 to eliminate y.
5x – 3y + 5z = – 1
5x – 3y + 5z = – 1
2x – y + 3z = 4
– 6x + 3y – 9z = – 12
– x
– 4z = – 13 New Equation 2
STEP 2 Solve the new system of linear equations in two
variables. First, add 7 times new Equation 2 to new
Equation 1 to eliminate x.
7x + 10z
= 19
– x – 4z = – 13
7x + 10z
= 19
– 7x – 28z = – 91
– 18z = – 72
Example 1
Use the Linear Combination Method
z = 4
Solve for z.
Substitute 4 for z in new Equation 1 or 2 and solve for x to
=3. –
get x
STEP 3 Substitute –3 for x and 4 for z in one of the original
equations and solve for y.
2x – y + 3z = 4
2( – 3 ) – y + 3 ( 4 ) = 4
– 6 – y + 12 = 4
–y + 6 = 4
y = 2
Equation 2
Substitute –3 for x and 4 for z.
Multiply.
Combine like terms.
Solve for y.
Example 1
Use the Linear Combination Method
STEP 4 Check by substituting 3–for x, 2 for y, and 4 for z in
each of the original equations.
ANSWER
The solution is x
triple ( –3, 2, 4).
–
= 3,
y
=2,
and z
=4,
or the ordered
Example 2
Solve a System with No Solution
x + y + z = 2
Solve the system.
Equation 1
3x + 3y + 3z = 14
Equation 2
x – 2y + z = 4
Equation 3
SOLUTION
Multiply Equation 1 by 3– and add the result to Equation 2.
– 3x – 3y – 3z = – 6
3x + 3y + 3z = 14
0 = 8
Add –3 times Equation 1
to Equation 2.
False statement
Example 2
Solve a System with No Solution
ANSWER
Because solving the system resulted in the false statement
=
0 8, the original
system of equations has no solution.
Example 3
Solve a System with Infinitely Many Solutions
Solve the system.
x + y + z = 4
Equation 1
x + y – z = 4
Equation 2
3x + 3y + z = 12
Equation 3
SOLUTION
STEP 1 Rewrite the system as a system in two variables.
x + y + z = 4
Add Equation 1
x + y – z = 4
to Equation 2.
2x + 2y = 8
New Equation 1
Example 3
Solve a System with Infinitely Many Solutions
x + y – z = 4
3x + 3y + z = 12
4x + 4y = 16
Add Equation 2
to Equation 3.
New Equation 2
STEP 2 Solve the new system of linear equations in two
variables.
– 4x – 4y = – 16
4x + 4y = 16
0 = 0
Add –2 times new Equation 1
to new Equation 2.
Example 3
Solve a System with Infinitely Many Solutions
ANSWER
Because solving the system resulted in the true statement
=
0 0, the original
system of equations has infinitely many
solutions. The three planes intersect in a line.
Checkpoint
Solve Systems
Tell how many solutions the linear system has. If the system
has one solution, solve the system. Then check your
solution.
1.
x + y + 3z = 1
x + y – z = 1
ANSWER
1; (1, 0, 0)
ANSWER
no solution
– x – 3y + 4z = – 1
2.
x – y + z = 4
3x + y – 2z = – 2
2x – 2y + 2z = 5
Checkpoint
Solve Systems
Tell how many solutions the linear system has. If the system
has one solution, solve the system. Then check your
solution.
3.
x + y + 2z = 10
– x + 2y + z = 5
– x + 4y + 3z = 15
ANSWER
infinitely many solutions
The Substitution Method
(3-Variable System)
WRITE THIS DOWN!!!
The steps to solving a system of linear
equations in three variables.
Example:
−− =3
− + 2 + 5 = −1
 +  + 4 = 4
2, −2,1
1) SOLVE for one of the variables
in any of the equations
2) SUBSTITUTE the variable found
in Step 1 into another
equation.
3) SUBSTITUTE the variable found
in Step 1 into the last equation.
4) SOLVE the new linear system of
two equations in two variables
for both of its variables. NOTE:
use preferred method.
5) SUBSTITUTE the values found
in Step 4 into original equation
to find the last variable.
6) CHECK the solution in each of
the original equations
Conclusion
Summary
• How do you solve a system
of linear equations in three
variables?
– Rewrite the system as a linear
system in two variables by
eliminating one variable.
Solve that system for each
variable. Substitute those two
values into an equation in the
original system to get the
values of the third variable.
Assignment
• To be completed in class
– Pg156 #(14, 18, 22, 26, 31,
32)
• Extra Practice Assignment
to be completed by the end
of the week.
– Systems of two equations

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