Algebra and Trigonometry: Structure and Method, Book 2

2000th Edition

ISBN: 9780395977255

Author: MCDOUGAL LITTEL

Publisher: McDougal Littell

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Question

Chapter 9.9, Problem 19WE

(a)

To determine

To find: The general expression for all its solutions.

(a)

Expert Solution

Answer to Problem 19WE

The general expressions for the solutions of the given system is {(5+z,2+2z,z)}

Explanation of Solution

Given:

x−2y+3z=1

x+y−3z=7

3x−4y+5z=7

Calculation:

Let us first transform the given system of equations into a triangular system.

Multiply first equation by −1 and add the result to the second equation.

x+y−3z=7−1(x−2y+3z=1)3y−6z=6x−2y+3z=13y−6z=63x−4y+5z=7

Then, multiply the first equation by −3 and add the result to the third equation.

−3(x−2y+3z=1)3x−4y+5z=72y−4z=4Resulting systemx−2y+3z=13y−6z=62y−4z=4

(1)

In (1) , divide the second equation by 3 and the third equation by −2 and add the results.

13(3y−6z=6)1−2(2y−4z=4)0=0Resulting systemx−2y+3z=13y−6z=60=0

(2)

The third equation in the resulting system shows that the given system is dependent and has infinitely many solutions. The solutions for x and y can be determined in terms of z .

In (2) solve 3y−6z=6 for y .

3y−6z=6y−2z=2y=2+2z

Substitute the value of y in the first equation and solve for x .

x−2(2+2z)+3z=1x−4−4z+3z=1x=1+4+z=5+z

Therefore, the general expressions for the solutions of the given system is {(5+z,2+2z,z)} .

Conclusion:

The general expressions for the solutions of the given system is {(5+z,2+2z,z)} .

(b)

To determine

To find: The three particular solutions.

(b)

Expert Solution

Answer to Problem 19WE

The three solutions of the given systems are (5,2,0),(8,8,3),(3,−2,−2)

Explanation of Solution

Given:

x−2y+3z=1

x+y−3z=7

3x−4y+5z=7

Calculation:

In order to find solutions of the system, substitute random values of z in the general expression.

Substitute 0,3 and −2 for z in the general expression to obtain three particular solutions.

z=0(5+0,2+2⋅0,0)=(5,2,0)z=3(5+3,2+2⋅3,3)=(8,8,3)z=−2(5−2,2−2(−2),−2)=(3,−2,−2)

Therefore, the three solutions of the given systems are (5,2,0),(8,8,3),(3,−2,−2)

Conclusion:

The three solutions of the given systems are (5,2,0),(8,8,3),(3,−2,−2)

Chapter 9.9, Problem 18WE

Chapter 9.9, Problem 20WE

Chapter 9 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 9.1 - Prob. 1OE Ch. 9.1 - Prob. 2OE Ch. 9.1 - Prob. 3OE Ch. 9.1 - Prob. 4OE Ch. 9.1 - Prob. 5OE Ch. 9.1 - Prob. 6OE Ch. 9.1 - Prob. 7OE Ch. 9.1 - Prob. 8OE Ch. 9.1 - Prob. 9OE Ch. 9.1 - Prob. 10OE

Ch. 9.1 - Prob. 11OE Ch. 9.1 - Prob. 12OE Ch. 9.1 - Prob. 13OE Ch. 9.1 - Prob. 14OE Ch. 9.1 - Prob. 15OE Ch. 9.1 - Prob. 1WE Ch. 9.1 - Prob. 2WE Ch. 9.1 - Prob. 3WE Ch. 9.1 - Prob. 4WE Ch. 9.1 - Prob. 5WE Ch. 9.1 - Prob. 6WE Ch. 9.1 - Prob. 7WE Ch. 9.1 - Prob. 8WE Ch. 9.1 - Prob. 9WE Ch. 9.1 - Prob. 10WE Ch. 9.1 - Prob. 11WE Ch. 9.1 - Prob. 12WE Ch. 9.1 - Prob. 13WE Ch. 9.1 - Prob. 14WE Ch. 9.1 - Prob. 15WE Ch. 9.1 - Prob. 16WE Ch. 9.1 - Prob. 17WE Ch. 9.1 - Prob. 18WE Ch. 9.1 - Prob. 19WE Ch. 9.1 - Prob. 20WE Ch. 9.1 - Prob. 21WE Ch. 9.1 - Prob. 22WE Ch. 9.1 - Prob. 23WE Ch. 9.1 - Prob. 24WE Ch. 9.1 - Prob. 25WE Ch. 9.1 - Prob. 26WE Ch. 9.1 - Prob. 27WE Ch. 9.1 - Prob. 28WE Ch. 9.1 - Prob. 29WE Ch. 9.1 - Prob. 30WE Ch. 9.1 - Prob. 31WE Ch. 9.1 - Prob. 32WE Ch. 9.1 - Prob. 1P Ch. 9.1 - Prob. 2P Ch. 9.1 - Prob. 3P Ch. 9.1 - Prob. 4P Ch. 9.1 - Prob. 5P Ch. 9.1 - Prob. 6P Ch. 9.1 - Prob. 7P Ch. 9.1 - Prob. 8P Ch. 9.1 - Prob. 9P Ch. 9.1 - Prob. 10P Ch. 9.1 - Prob. 11P Ch. 9.1 - Prob. 12P Ch. 9.1 - 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