Precalculus: Mathematics for Calculus - 6th Edition
Precalculus: Mathematics for Calculus - 6th Edition
6th Edition
ISBN: 9780840068071
Author: Stewart, James, Redlin, Lothar, Watson, Saleem
Publisher: Cengage Learning
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Chapter 10, Problem 31RE
To determine

To find: the solution of the system

Expert Solution & Answer
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Answer to Problem 31RE

Therefore, the solution to the system of equations is (1−4t,−1−t,t) .

Given:

  {x−3y+z=44x−y+15z=5

Explanation:

Consider the system of equations

  x−3y+z=4      Equation 14x−y+15z=5     Equation 2

Recall the Gauss elimination method,

To solve the system, use the system equivalent operations and reduce the system to triangular form. Then use back substitution to solve for the variables.

And operations that yield equivalent system are

  1. Add a non-zero multiple of one equation to another
  2. Multiply an equation by a non-zero constant
  3. Interchange the positions of two equations.

Use these together to solve the system. Change the system to triangular form by eliminating the x−term from the equation 2 by adding equation 2 to −4 times equation 1 .

The work is shown as follows.

  −4x+12y−4z=−164x−y+15z=511y+11z=−11

So, the equivalent system one step closer to triangular form is

  x−3y+z=411y+11z=−11       

The system is now triangular form, but there are only two equations, use them to solve for x and y in terms of the variable z but z can take any value. So the system has infinite number of solutions.

To find the complete solution of x begin by solving for y in terms of z .

From the second equation,

  y+z=−1y=−1−z 

Then solve for x by substituting y and z in first equation.

  x−3(−1−z)+z=4x+3+3z+z=4x+4z+3=4x=4−4z−3=1−4z

To describe the complete solution, let t be any real number. The solution is

  x=1−4ty=−1−tz=t

Therefore, the solution to the system of equations is (1−4t,−1−t,t) .

Conclusion:

Therefore, the solution to the system of equations is (1−4t,−1−t,t) .

Explanation of Solution

Given:

  {x−3y+z=44x−y+15z=5

Consider the system of equations

  x−3y+z=4      Equation 14x−y+15z=5     Equation 2

Recall the Gauss elimination method,

To solve the system, use the system equivalent operations and reduce the system to triangular form. Then use back substitution to solve for the variables.

And operations that yield equivalent system are

  1. Add a non-zero multiple of one equation to another
  2. Multiply an equation by a non-zero constant
  3. Interchange the positions of two equations.

Use these together to solve the system. Change the system to triangular form by eliminating the x−term from the equation 2 by adding equation 2 to −4 times equation 1 .

The work is shown as follows.

  −4x+12y−4z=−164x−y+15z=511y+11z=−11

So, the equivalent system one step closer to triangular form is

  x−3y+z=411y+11z=−11       

The system is now triangular form, but there are only two equations, use them to solve for x and y in terms of the variable z but z can take any value. So the system has infinite number of solutions.

To find the complete solution of x begin by solving for y in terms of z .

From the second equation,

  y+z=−1y=−1−z 

Then solve for x by substituting y and z in first equation.

  x−3(−1−z)+z=4x+3+3z+z=4x+4z+3=4x=4−4z−3=1−4z

To describe the complete solution, let t be any real number. The solution is

  x=1−4ty=−1−tz=t

Therefore, the solution to the system of equations is (1−4t,−1−t,t) .

Conclusion:

Therefore, the solution to the system of equations is (1−4t,−1−t,t) .

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Chapter 10, Problem 30RE
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Chapter 10, Problem 32RE

Chapter 10 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

Ch. 10.1 - The system of equations {2x+3y=75xy=9 is a system...Ch. 10.1 - Prob. 2ECh. 10.1 - A system of two linear equations in two variables...Ch. 10.1 - The following is a system of two linear equations...Ch. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - Prob. 9ECh. 10.1 - Prob. 10E
Ch. 10.1 - Prob. 11ECh. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14ECh. 10.1 - Prob. 15ECh. 10.1 - Prob. 16ECh. 10.1 - Prob. 17ECh. 10.1 - Prob. 18ECh. 10.1 - Prob. 19ECh. 10.1 - Prob. 20ECh. 10.1 - Prob. 21ECh. 10.1 - Prob. 22ECh. 10.1 - Prob. 23ECh. 10.1 - Prob. 24ECh. 10.1 - Prob. 25ECh. 10.1 - Prob. 26ECh. 10.1 - Prob. 27ECh. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.1 - Prob. 32ECh. 10.1 - Prob. 33ECh. 10.1 - Prob. 34ECh. 10.1 - Prob. 35ECh. 10.1 - Prob. 36ECh. 10.1 - Prob. 37ECh. 10.1 - Prob. 38ECh. 10.1 - Prob. 39ECh. 10.1 - Prob. 40ECh. 10.1 - Prob. 41ECh. 10.1 - Prob. 42ECh. 10.1 - Prob. 43ECh. 10.1 - Prob. 44ECh. 10.1 - Prob. 45ECh. 10.1 - Prob. 46ECh. 10.1 - Prob. 47ECh. 10.1 - Prob. 48ECh. 10.1 - Prob. 49ECh. 10.1 - Prob. 50ECh. 10.1 - Prob. 51ECh. 10.1 - Prob. 52ECh. 10.1 - Prob. 53ECh. 10.1 - Prob. 54ECh. 10.1 - Prob. 55ECh. 10.1 - Prob. 56ECh. 10.1 - Prob. 57ECh. 10.1 - Prob. 58ECh. 10.1 - Value of Coins A man has 14 coins in his pocket,...Ch. 10.1 - Admission Fees The admission fee at an amusement...Ch. 10.1 - Gas Station A gas station sells regular gas for...Ch. 10.1 - Fruit Stand A fruit stand sells two varieties of...Ch. 10.1 - Airplane Speed A man flies a small airplane from...Ch. 10.1 - Boat Speed A boat on a river travels downstream...Ch. 10.1 - Nutrition A researcher performs an experiment to...Ch. 10.1 - Coffee Blends A customer in a coffee shop...Ch. 10.1 - Mixture Problem A chemist has two large containers...Ch. 10.1 - Mixture Problem A biologist has two brine...Ch. 10.1 - Investments A woman invests a total of 20,000 in...Ch. 10.1 - Investments A man invests his savings in two...Ch. 10.1 - Prob. 71ECh. 10.1 - Prob. 72ECh. 10.1 - Prob. 73ECh. 10.1 - Area of a Triangle Find the area of the triangle...Ch. 10.1 - Prob. 75ECh. 10.2 - These exercises refer to the following system:...Ch. 10.2 - Prob. 2ECh. 10.2 - Prob. 3ECh. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - Prob. 7ECh. 10.2 - Prob. 8ECh. 10.2 - Prob. 9ECh. 10.2 - 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Prob. 47ECh. 10.3 - If a system of linear equations has infinitely...Ch. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.3 - Prob. 20ECh. 10.3 - Prob. 21ECh. 10.3 - Prob. 22ECh. 10.3 - Prob. 23ECh. 10.3 - Prob. 24ECh. 10.3 - Prob. 25ECh. 10.3 - Prob. 26ECh. 10.3 - Prob. 27ECh. 10.3 - Prob. 28ECh. 10.3 - Prob. 29ECh. 10.3 - Prob. 30ECh. 10.3 - Prob. 31ECh. 10.3 - Prob. 32ECh. 10.3 - Prob. 33ECh. 10.3 - Prob. 34ECh. 10.3 - Prob. 35ECh. 10.3 - Prob. 36ECh. 10.3 - Prob. 37ECh. 10.3 - Prob. 38ECh. 10.3 - Prob. 39ECh. 10.3 - Prob. 40ECh. 10.3 - Prob. 41ECh. 10.3 - Prob. 42ECh. 10.3 - Prob. 43ECh. 10.3 - Prob. 44ECh. 10.3 - Prob. 45ECh. 10.3 - Prob. 46ECh. 10.3 - Prob. 47ECh. 10.3 - Prob. 48ECh. 10.3 - 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Prob. 31ECh. 10.7 - Prob. 32ECh. 10.7 - Prob. 33ECh. 10.7 - Prob. 34ECh. 10.7 - Prob. 35ECh. 10.7 - Prob. 36ECh. 10.7 - Prob. 37ECh. 10.7 - Prob. 38ECh. 10.7 - Prob. 39ECh. 10.7 - Prob. 40ECh. 10.7 - Prob. 41ECh. 10.7 - Prob. 42ECh. 10.7 - Prob. 43ECh. 10.7 - Prob. 44ECh. 10.7 - Prob. 45ECh. 10.7 - Prob. 46ECh. 10.7 - Prob. 47ECh. 10.7 - Prob. 48ECh. 10.8 - The system of equations {2yx2=0yx=4 is graphed...Ch. 10.8 - Prob. 2ECh. 10.8 - Prob. 3ECh. 10.8 - Prob. 4ECh. 10.8 - Prob. 5ECh. 10.8 - Prob. 6ECh. 10.8 - Prob. 7ECh. 10.8 - Prob. 8ECh. 10.8 - Prob. 9ECh. 10.8 - Prob. 10ECh. 10.8 - Prob. 11ECh. 10.8 - Prob. 12ECh. 10.8 - Prob. 13ECh. 10.8 - Prob. 14ECh. 10.8 - Prob. 15ECh. 10.8 - Prob. 16ECh. 10.8 - Prob. 17ECh. 10.8 - Prob. 18ECh. 10.8 - Prob. 19ECh. 10.8 - Prob. 20ECh. 10.8 - Prob. 21ECh. 10.8 - Prob. 22ECh. 10.8 - Prob. 23ECh. 10.8 - Prob. 24ECh. 10.8 - Prob. 25ECh. 10.8 - Prob. 26ECh. 10.8 - Prob. 27ECh. 10.8 - Prob. 28ECh. 10.8 - Prob. 29ECh. 10.8 - Prob. 30ECh. 10.8 - Prob. 31ECh. 10.8 - 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