Download System of Equations: Row Reduction Method

Systems with Three Variables
Objective:
I can solve systems with three or more variables.
How much does
each box
weigh?
Explain your
reasoning.
Solve the system.
𝟐𝒙 − 𝒚 + 𝒛 = 𝟒
𝒙 + 𝟑𝒚 − 𝒛 = 𝟏𝟏
𝟒𝒙 + 𝒚 − 𝒛 = 𝟏𝟒
𝟐𝒙 − 𝒚 + 𝒛 = 𝟒
𝒙 + 𝟑𝒚 − 𝒛 = 𝟏𝟏
𝟑𝒙 + 𝟐𝒚
= 𝟏𝟓
𝟑𝒙 + 𝟐𝒚 = 𝟏𝟓
𝟑 𝟑 + 𝟐𝒚 = 𝟏𝟓
𝟗 + 𝟐𝒚 = 𝟏𝟓
𝟐𝒙 − 𝒚 + 𝒛 = 𝟒
𝟒𝒙 + 𝒚 − 𝒛 = 𝟏𝟒
𝟔𝒙
= 𝟏𝟖
𝒙 =𝟑
(𝟑, 𝟑, 𝟏)
𝟐𝒚 = 𝟔
𝒚=𝟑
𝟐𝒙 − 𝒚 + 𝒛 = 𝟒
𝟐(𝟑) − (𝟑) + 𝒛 = 𝟒
𝟑+𝒛=𝟒
𝒛=𝟏
Solve the system.
𝒙 + 𝒚 + 𝟐𝒛 = 𝟑
−𝒙 − 𝟐𝒚 + 𝒛 = 𝟏𝟎
−𝒚 + 𝟑𝒛 = 𝟏𝟑
−𝒚 + 𝟑(𝟑) = 𝟏𝟑
𝒙 + 𝒚 + 𝟐𝒛 = 𝟑
𝟐𝒙 + 𝒚 + 𝟑𝒛 = 𝟕
−𝒙 − 𝟐𝒚 + 𝒛 = 𝟏𝟎
𝟐𝒙 + 𝒚 + 𝟑𝒛 = 𝟕
−𝒙 − 𝟐𝒚 + 𝒛 = 𝟏𝟎
𝟐𝒙 + 𝒚 + 𝟑𝒛 = 𝟕
−𝟐𝒙 − 𝟒𝒚 + 𝟐𝒛 = 𝟐𝟎
−𝟑𝒚 + 𝟓𝒛 = 𝟐𝟕
𝒙 + 𝒚 + 𝟐𝒛 = 𝟑
𝒙 + (−𝟒) + 𝟐(𝟑) = 𝟑
−𝒚 + 𝟗 = 𝟏𝟑
−𝒚 = 𝟒
𝒚 = −𝟒
−𝒚 + 𝟑𝒛 = 𝟏𝟑
−𝟑𝒚 + 𝟓𝒛 = 𝟐𝟕
𝟑𝒚 − 𝟗𝒛 = −𝟑𝟗
−𝟑𝒚 + 𝟓𝒛 = 𝟐𝟕
−𝟒𝒛 = −𝟏𝟐
𝒛 =𝟑
𝒙−𝟐=𝟑
(𝟏, −𝟒, 𝟑)
𝒙=𝟏
You manage a clothing store and budget $6000 to restock 200 shirts.
You can buy T-shirts for $12 each, polo shirts $24 each and rugby shirts
for $36 each. You want to have twice as many rugby shirts as polo
shirts. How many of each type shirt should you buy?
𝑻 + 𝑷 + (𝟐𝑷) = 𝟐𝟎𝟎
𝑻 + 𝑷 + 𝑹 = 𝟐𝟎𝟎
𝟏𝟐𝑻 + 𝟐𝟒𝑷 + 𝟑𝟔(𝟐𝑷) = 𝟔𝟎𝟎𝟎
𝟏𝟐𝑻 + 𝟐𝟒𝑷 + 𝟑𝟔𝑹 = 𝟔𝟎𝟎𝟎
𝑻 + 𝟑𝑷 = 𝟐𝟎𝟎
𝑹 = 𝟐𝑷
𝑹 = 𝟐(𝟔𝟎) = 𝟏𝟐𝟎
𝑻 + (𝟔𝟎) + (𝟏𝟐𝟎) = 𝟐𝟎𝟎
𝑻 = 𝟐𝟎
Pg. 171
#9-11, 30
𝟏𝟐𝑻 + 𝟗𝟔𝑷 = 𝟔𝟎𝟎𝟎
−𝟏𝟐𝑻 − 𝟑𝟔𝑷 = −𝟐𝟒𝟎𝟎
𝟏𝟐𝑻 + 𝟗𝟔𝑷 = 𝟔𝟎𝟎𝟎
𝟔𝟎𝑷 = 𝟑𝟔𝟎𝟎
𝑷 = 𝟔𝟎
Systems with Three Variables
Objective:
I can use matrices to solve systems.
Use the rules below to change figure 1 into figure 2?
Matrix:
• A rectangular array of numbers.
•Dimensions: rows × columns
Matrix
Name
2 5 11
𝐴 =
3 −2 5
Systems to matrices
 A1 x  B1 y  C1

 A2 x  B2 y  C2
𝑥 + 4𝑦 = 5
2𝑥 + 5𝑦 = 4
A1 B1 C1 
matrix
A B2 C2 
 2

matrix
1 4
2 5
5
4
1 0 a 
0 1 b 


(−𝟑, 𝟐)
Reduced Row Echelon form (rref)
-4 × Row 2 + Row 1
Divide Row 2 by -3
-2 × Row 1 + Row 2
Put in Row 1
Put in Row 2
Put in Row 2
1
0
4
5
−3 −6
1 4
0 1
5
2
1 0
0 1
−3
2
Systems to matrices
 A1 x  B1 y  C1

 A2 x  B2 y  C2
2𝑥 + 𝑦 = 5
5𝑥 + 3𝑦 = 13
matrix
matrix
A1 B1 C1 
A B2 C2 
 2

2 1
5 3
5
13
1 0 a 
0 1 b 


rref
rref
1 0
0 1
[2nd],[x-1],[►] ,[►], [enter]
Reduced Row-Echelon Form
enter rows [enter]
[2nd], [x-1], [►], [alpha], [apps]
enter columns [enter]
or
enter matrix;
[2nd], [x-1], [►], [▼] to rref( , [enter]
[2nd], [mode]
[2nd], [ x-1], [enter], [enter]
2
1
(𝟐, 𝟏)
Solve each system using matrices
 2 x  5 y  11

 3 x  8 y  1
 2
 3

1
0

5
8
0
1
11 
 1

3
1

Solution:
( 3, 1 )
Solution:
( 1 ,  1, 0 )
 abc  0

 4a  2b  c  2
16a  4b  c  12

1
4


16
1
1
2
1
4
1
1
0


0
0
0
1
0
0
1
0
2

12

1 
 1

0 

Solve each system using matrices
 4 x  2 y  7

 3x  y  5
Solution:
( -1.5, -0.5)
 4 x  2 y  7

 3 x  y  5
 4
 3

1
0

2
1
0
1
7 

 5
 1.5 
 0.5

Solution:
( -2 , 3, 5)
Pg. 179
#15-20,
24-27,
32,33,36,37
x  3 y  z  2

 x  2z  8
 2y  z 1

1
1


0
1
0


0
3
1
0
2
2
1
0
0
1
0
0
1
2
8

1

 2
3 

5 
