Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![Solve the system of linear equations and check any solution algebraically.
5x + 4y + 14z = 0
6x + 5y + 19z = 0
5x + 3y + 16z =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ba61890-9fab-4450-9896-917b96677c7a%2F57fa9deb-3f50-4b21-a701-d407a41a8d71%2F7ckvj7c_processed.png&w=3840&q=75)
Transcribed Image Text:Solve the system of linear equations and check any solution algebraically.
5x + 4y + 14z = 0
6x + 5y + 19z = 0
5x + 3y + 16z =
![The solution is the ordered triple (x, y, z) that satisfies all three equations. Convert the system to an
equivalent system that is in the row echelon form, by using row operations. In the row echelon form, the
equations have a stair-step pattern with leading coefficients of 1.
Write the system with the first two equations interchanged.
]× )v + (D
|× )z
= 0 Equation 1
+
= 0 Equation 2
= 0 Equation 3
5x
4y
14z
5x
3y
16z
Multiply Equation 2 by (-1) and add with Equation 1 to make the leading coefficient in the first row 1.
|× )x + 5y +
]× )v
]× ) .
6х +
+ 19z +
= 0 +
+ Eq 1 + (-1)Eq 2
+ Eq 2
+ Eq 3
5x
4y
14z
= 0
5x
3y
+
16z
= 0
Write the equivalent system with the resultant Equation 1.
|× )z
x + y +
= 0 + Eq 1
= 0 + Eq 2
= 0 + Eq 3
5х + 4y +
14z
5x + 3у +
16z](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ba61890-9fab-4450-9896-917b96677c7a%2F57fa9deb-3f50-4b21-a701-d407a41a8d71%2Fz3l6igu_processed.png&w=3840&q=75)
Transcribed Image Text:The solution is the ordered triple (x, y, z) that satisfies all three equations. Convert the system to an
equivalent system that is in the row echelon form, by using row operations. In the row echelon form, the
equations have a stair-step pattern with leading coefficients of 1.
Write the system with the first two equations interchanged.
]× )v + (D
|× )z
= 0 Equation 1
+
= 0 Equation 2
= 0 Equation 3
5x
4y
14z
5x
3y
16z
Multiply Equation 2 by (-1) and add with Equation 1 to make the leading coefficient in the first row 1.
|× )x + 5y +
]× )v
]× ) .
6х +
+ 19z +
= 0 +
+ Eq 1 + (-1)Eq 2
+ Eq 2
+ Eq 3
5x
4y
14z
= 0
5x
3y
+
16z
= 0
Write the equivalent system with the resultant Equation 1.
|× )z
x + y +
= 0 + Eq 1
= 0 + Eq 2
= 0 + Eq 3
5х + 4y +
14z
5x + 3у +
16z
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