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Solving Linear Systems

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Balcony. Mezzanine. Orchestra. Holt Algebra 2. 3-6. Solving Linear Systems. in ... the price of a mezzanine seat, and z represent the present of a balcony seat. ... – PowerPoint PPT presentation

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Title: Solving Linear Systems


1
Solving Linear Systems in Three Variables
3-6
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 2
2
Warm Up Solve each system of equations
algebraically. Classify each system and
determine the number of solutions.
(2, 2)
1.
2.
(1,3)
4.
3.
inconsistent none
consistent, independent one
3
Objectives
Represent solutions to systems of equations in
three dimensions graphically. Solve systems of
equations in three dimensions algebraically.
4
Systems of three equations with three variables
are often called 3-by-3 systems. In general, to
find a single solution to any system of
equations, you need as many equations as you have
variables.
5
Recall from Lesson 3-5 that the graph of a linear
equation in three variables is a plane. When you
graph a system of three linear equations in three
dimensions, the result is three planes that may
or may not intersect. The solution to the system
is the set of points where all three planes
intersect. These systems may have one, infinitely
many, or no solution.
6
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7
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8
Identifying the exact solution from a graph of a
3-by-3 system can be very difficult. However, you
can use the methods of elimination and
substitution to reduce a 3-by-3 system to a
2-by-2 system and then use the methods that you
learned in Lesson 3-2.
9
Example 1 Solving a Linear System in Three
Variables
Use elimination to solve the system of equations.
5x 2y 3z 7
2x 3y z 16
3x 4y 2z 7
Step 1 Eliminate one variable.
In this system, z is a reasonable choice to
eliminate first because the coefficient of z in
the second equation is 1 and z is easy to
eliminate from the other equations.
10
Example 1 Continued
5x 2y 3z 7
5x 2y 3z 7
3(2x 3y z 16)
6x 9y 3z 48
11x 11y 55
1
3x 4y 2z 7
3x 4y 2z 7
4x 6y 2z 32
2(2x 3y z 16)
7x 2y 25
11
Example 1 Continued
11x 11y 55
You now have a 2-by-2 system.
7x 2y 25
12
Example 1 Continued
Step 2 Eliminate another variable. Then solve
for the remaining variable. You can eliminate y
by using methods from Lesson 3-2.
2(11x 11y 55)
22x 22y 110
1
11(7x 2y 25)
77x 22y 275
55x 165
1
x 3
Solve for x.
13
Example 1 Continued
Step 3 Use one of the equations in your 2-by-2
system to solve for y.
11x 11y 55
1
11(3) 11y 55
Substitute 3 for x.
1
Solve for y.
y 2
14
Example 1 Continued
Step 4 Substitute for x and y in one of the
original equations to solve
for z.
2x 3y z 16
Substitute 3 for x and 2 for y.
1
2(3) 3(2) z 16
z 4
Solve for y.
1
The solution is (3, 2, 4).
15
Check It Out! Example 1
Use elimination to solve the system of equations.
x y 2z 7
2x 3y z 1
3x 4y z 4
Step 1 Eliminate one variable.
In this system, z is a reasonable choice to
eliminate first because the coefficient of z in
the second equation is 1.
16
Check It Out! Example 1 Continued
x y 2z 7
x y 2z 7
2(2x 3y z 1)
4x 6y 2z 2
5x 5y 5
1
x y 2z 7
x y 2z 7
6x 8y 2z 8
2(3x 4y z 4)
5x 9y 1
17
Check It Out! Example 1 Continued
5x 5y 5
You now have a 2-by-2 system.
5x 9y 1
18
Check It Out! Example 1 Continued
Step 2 Eliminate another variable. Then solve
for the remaining variable. You can eliminate x
by using methods from Lesson 3-2.
5x 5y 5
5x 9y 1
Solve for y.
4y 4
1
y 1
19
Check It Out! Example 1
Step 3 Use one of the equations in your 2-by-2
system to solve for x.
5x 5y 5
Substitute 1 for y.
5x 5(1) 5
1
Solve for x.
5x 5 5
1
5x 10
x 2
20
Check It Out! Example 1
Step 4 Substitute for x and y in one of the
original equations to solve
for z.
Substitute 2 for x and 1 for y.
2(2) 3(1) z 1
Solve for z.
1
4 3 z 1
z 2
1
The solution is (2, 1, 2).
21
You can also use substitution to solve a 3-by-3
system. Again, the first step is to reduce the
3-by-3 system to a 2-by-2 system.
22
Example 2 Business Application
The table shows the number of each type of ticket
sold and the total sales amount for each night of
the school play. Find the price of each type of
ticket.
23
Example 2 Continued
Step 1 Let x represent the price of an orchestra
seat, y represent the price of a mezzanine seat,
and z represent the present of a balcony
seat. Write a system of equations to represent
the data in the table.
200x 30y 40z 1470
Fridays sales.
250x 60y 50z 1950
Saturdays sales.
150x 30y 1050
Sundays sales.
A variable is missing in the last equation
however, the same solution methods apply.
Elimination is a good choice because eliminating
z is straightforward.
24
Example 2 Continued
Step 2 Eliminate z.
1000x 150y 200z 7350
5(200x 30y 40z 1470)
1000x 240y 200z 7800
4(250x 60y 50z 1950)
y 5
By eliminating z, due to the coefficients of x,
you also eliminated x providing a solution for y.
25
Example 2 Continued
150x 30y 1050
Substitute 5 for y.
150x 30(5) 1050
Solve for x.
x 6
26
Example 2 Continued
1
200x 30y 40z 1470
Substitute 6 for x and 5 for y.
Solve for x.
200(6) 30(5) 40z 1470
z 3
The solution to the system is (6, 5, 3). So, the
cost of an orchestra seat is 6, the cost of a
mezzanine seat is 5, and the cost of a balcony
seat is 3.
27
Check It Out! Example 2
Jadas chili won first place at the winter fair.
The table shows the results of the voting. How
many points are first-, second-, and third-place
votes worth?
28
Check It Out! Example 2 Continued
Step 1 Let x represent first-place points, y
represent second-place points, and z
represent third- place points. Write a
system of equations to represent the data in the
table.
3x y 4z 15
Jadas points.
2x 4y 14
Marias points.
2x 2y 3z 13
Als points.
A variable is missing in one equation however,
the same solution methods apply. Elimination is a
good choice because eliminating z is
straightforward.
29
Check It Out! Example 2 Continued
Step 2 Eliminate z.
9x 3y 12z 45
3(3x y 4z 15)
8x 8y 12z 52
4(2x 2y 3z 13)
x 5y 7
2(x 5y 7)
2x 10y 14
2x 4y 14
2x 4y 14
y 2
Solve for y.
30
Check It Out! Example 2 Continued
2x 4y 14
Substitute 2 for y.
2x 4(2) 14
Solve for x.
x 3
31
Check It Out! Example 2 Continued
Step 4 Substitute for x and y in one of the
original equations to solve for z.
2(3) 2(2) 3z 13
6 4 3z 13
Solve for z.
z 1
The solution to the system is (3, 2, 1). The
points for first-place is 3, the points for
second-place is 2, and 1 point for third-place.
32
The systems in Examples 1 and 2 have unique
solutions. However, 3-by-3 systems may have no
solution or an infinite number of solutions.
33
Example 3 Classifying Systems with Infinite Many
Solutions or No Solutions
Classify the system as consistent or
inconsistent, and determine the number of
solutions.
2x 6y 4z 2
3x 9y 6z 3
5x 15y 10z 5
34
Example 3 Continued
The elimination method is convenient because the
numbers you need to multiply the equations are
small.
First, eliminate x.
6x 18y 12z 6
3(2x 6y 4z 2)
2(3x 9y 6z 3)
6x 18y 12z 6
0 0
?
35
Example 3 Continued
10x 30y 20z 10
5(2x 6y 4z 2)
2(5x 15y 10z 5)
10x 30y 20z 10
?
0 0
Because 0 is always equal to 0, the equation is
an identity. Therefore, the system is
consistent, dependent and has an infinite number
of solutions.
36
Check It Out! Example 3a
Classify the system, and determine the number of
solutions.
3x y 2z 4
2x y 3z 7
9x 3y 6z 12
37
Check It Out! Example 3a Continued
The elimination method is convenient because the
numbers you need to multiply the equations by are
small.
First, eliminate y.
3x y 2z 4
3x y 2z 4
1(2x y 3z 7)
2x y 3z 7
x z 3
38
Check It Out! Example 3a Continued
6x 3y 9z 21
3(2x y 3z 7)
9x 3y 6z 12
9x 3y 6z 12
3x 3z 9
Now you have a 2-by-2 system.
x z 3
3x 3z 9
39
Check It Out! Example 3a Continued
Eliminate x.
3(x z 3)
3x 3z 9
3x 3z 9
3x 3z 9
?
0 0
Because 0 is always equal to 0, the equation is
an identity. Therefore, the system is
consistent, dependent, and has an infinite number
of solutions.
40
Check It Out! Example 3b
Classify the system, and determine the number of
solutions.
2x y 3z 6
2x 4y 6z 10
y z 2
41
Check It Out! Example 3b Continued
Use the substitution method. Solve for y in
equation 3.
y z 2
Solve for y.
y z 2
2x y 3z 6
2x (z 2) 3z 6
2x z 2 3z 6
2x 2z 4
42
Check It Out! Example 3b Continued
2x 4y 6z 10
2x 4(z 2) 6z 10
2x 4z 8 6z 10
2x 2z 2
Now you have a 2-by-2 system.
2x 2z 4
2x 2z 2
43
Check It Out! Example 3b Continued
Eliminate z.
2x 2z 4
1(2x 2z 2)
0 ? 2 ?
Because 0 is never equal to 2, the equation is a
contradiction. Therefore, the system is
inconsistent and has no solutions.
44
Lesson Quiz Part I
At the library book sale, each type of book is
priced differently. The table shows the number of
books Joy and her friends each bought, and the
amount each person spent. Find the price of each
type of book.
1.
hardcover 3
paperback 1
audio books 4
45
Lesson Quiz Part II
Classify each system and determine the number of
solutions.
2x y 2z 5
2.
3x y z 1
inconsistent none
x y 3z 2
9x 3y 6z 3
3.
12x 4y 8z 4
consistent dependent infinite
6x 2y 4z 5
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