Consider the following system of linear equations. 3x +2y = 7 -2y -z = -7 3x + 4y +3z = 8 Solve the system by completing the steps below to produce a reduced row-echelon form. R₁, R₂, and R3 denote the first, second, and third rows, respectively. The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right. Here is the augmented matrix: 3 2 0 0-2 -1 7 olo X 08 Español

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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Chapter9: Quadratic Functions And Equations
Section9.7: Solving Systems Of Linear And Quadratic Equations
Problem 29HP
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Consider the following system of linear equations.
3x +2y = 7
-2y -z = -7
3x + 4y +3z = 8
Solve the system by completing the steps below to produce a reduced row-echelon form.
R₁, R₂, and R3 denote the first, second, and third rows, respectively.
The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the
right.
Here is the augmented matrix:
Enter the missing coefficients for the row operations.
(1)
(2)
(3)
(4)
(-/-)
R₁ R₁:
R₁ + R₂ R3:
R₂ → R₂:
•R₂ + R₁ R₁:
+
3
0
R₂ + R3 R3:
1
1
2 0
2 -1 1
4
3
0 -2 -1
4
3
1
W|N
0
N N W/N
0 2
0-2 -1
W|N
0 1
2
10
0
01
00
0
3
-la
-16
3
1
2
2
7
-7
8
لاس
34
-7
باس را
B
0
7
2
-6
Español
?
Transcribed Image Text:Consider the following system of linear equations. 3x +2y = 7 -2y -z = -7 3x + 4y +3z = 8 Solve the system by completing the steps below to produce a reduced row-echelon form. R₁, R₂, and R3 denote the first, second, and third rows, respectively. The arrow notation (→) stands for "replaces," where the expression on the left of the arrow replaces the expression on the right. Here is the augmented matrix: Enter the missing coefficients for the row operations. (1) (2) (3) (4) (-/-) R₁ R₁: R₁ + R₂ R3: R₂ → R₂: •R₂ + R₁ R₁: + 3 0 R₂ + R3 R3: 1 1 2 0 2 -1 1 4 3 0 -2 -1 4 3 1 W|N 0 N N W/N 0 2 0-2 -1 W|N 0 1 2 10 0 01 00 0 3 -la -16 3 1 2 2 7 -7 8 لاس 34 -7 باس را B 0 7 2 -6 Español ?
(2)
(3)
(4)
(5)
(6)
R₁ + R₂ → R3:
Solution:
R₂- → R₂:
(-3) =
·R₂ + R₁ R₁:
R₂ + R3 → R3:
R3 → R3:
(1) R₁ + R₂₁
·R3 + R₁ → R₁ :
R₂ + R₂ → R₂:
1
y =
3
0 -2 -1
0
2
3
1
0 1
0
als
2
10
0 1
1 0
0 1
00
00 2
1 0 0
0 1 0
00 1
z =
0
-0
0
2
3
wit
1
1
-|~
3
1
2
1
Enter the missing coefficient for the row operation, fill in the missing matrix
entries, and give the solution.
NM N
7
3
2
1
3
-7
72
0
I 7
2
-3
Transcribed Image Text:(2) (3) (4) (5) (6) R₁ + R₂ → R3: Solution: R₂- → R₂: (-3) = ·R₂ + R₁ R₁: R₂ + R3 → R3: R3 → R3: (1) R₁ + R₂₁ ·R3 + R₁ → R₁ : R₂ + R₂ → R₂: 1 y = 3 0 -2 -1 0 2 3 1 0 1 0 als 2 10 0 1 1 0 0 1 00 00 2 1 0 0 0 1 0 00 1 z = 0 -0 0 2 3 wit 1 1 -|~ 3 1 2 1 Enter the missing coefficient for the row operation, fill in the missing matrix entries, and give the solution. NM N 7 3 2 1 3 -7 72 0 I 7 2 -3
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