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High School Math Algebra High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

To find the error made in solving a system of linear equations in stating that the system has no solution and to find the solution of the system. The solution of the given system of equations is 4 , − 2 . This error made is that friend has to take more input values to find the corresponding output values in order to find the solution of given system of equations. Given information: Your friend used a graphing calculator to solve a system of linear equations, shown below. After using the TABLE feature, your friend says that the system has no solutions. Explain what your friend did wrong. What is the solution of the system. 2 x + y = 6 y = 6 − 2 x 3 x + 2 y = 8 y = 8 − 3 x 2 Calculation: Solver stated that there is no solution to the given system of equations. He has to take different values of x to find the corresponding values of y as shown below: The system of equations is y = 6 − 2 x and y = 8 − 3 x 2 : Find y for x = 4 in the first given equation: y = 6 − 2 x ⇒ y = 6 − 2 4 ⇒ y = 6 − 8 ⇒ y = − 2 Find y for x = 4 in the second given equation: y = 8 − 3 x 2 ⇒ y = 8 − 3 4 2 ⇒ y = 8 − 12 2 ⇒ y = − 2 Therefore, the solution to the system of equations is 4 , − 2 .

To find the error made in solving a system of linear equations in stating that the system has no solution and to find the solution of the system. The solution of the given system of equations is 4 , − 2 . This error made is that friend has to take more input values to find the corresponding output values in order to find the solution of given system of equations. Given information: Your friend used a graphing calculator to solve a system of linear equations, shown below. After using the TABLE feature, your friend says that the system has no solutions. Explain what your friend did wrong. What is the solution of the system. 2 x + y = 6 y = 6 − 2 x 3 x + 2 y = 8 y = 8 − 3 x 2 Calculation: Solver stated that there is no solution to the given system of equations. He has to take different values of x to find the corresponding values of y as shown below: The system of equations is y = 6 − 2 x and y = 8 − 3 x 2 : Find y for x = 4 in the first given equation: y = 6 − 2 x ⇒ y = 6 − 2 4 ⇒ y = 6 − 8 ⇒ y = − 2 Find y for x = 4 in the second given equation: y = 8 − 3 x 2 ⇒ y = 8 − 3 4 2 ⇒ y = 8 − 12 2 ⇒ y = − 2 Therefore, the solution to the system of equations is 4 , − 2 .

Solution Summary: The author explains how the graphing calculator solves a system of linear equations, and the solution of the system.

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

15th Edition

ISBN: 9780133281163

Author: Prentice Hall

Publisher: Prentice Hall

See similar textbooks

Expert Solution & Answer

Chapter 3.1, Problem 41PPSE

Solution

To find: To find the error made in solving a system of linear equations in stating that the system has no solution and to find the solution of the system.

The solution of the given system of equations is 4, −2 . This error made is that friend has to take more input values to find the corresponding output values in order to find the solution of given system of equations.

Given information: Your friend used a graphing calculator to solve a system of linear equations, shown below. After using the TABLE feature, your friend says that the system has no solutions. Explain what your friend did wrong. What is the solution of the system.

2x+y=6 y=6−2x

3x+2y=8 y=8−3x2

Calculation:

Solver stated that there is no solution to the given system of equations. He has to take different values of x to find the corresponding values of y as shown below:

The system of equations is y=6−2x and y=8−3x2 :

Find y for x=4 in the first given equation:

y=6−2x⇒y=6−24⇒y=6−8⇒y=−2

Find y for x=4 in the second given equation:

y=8−3x2⇒y=8−342⇒y=8−122⇒y=−2

Therefore, the solution to the system of equations is 4, −2 .

Chapter 3.1, Problem 40PPSE

Chapter 3.1, Problem 42PPSE

Chapter 3 Solutions

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Ch. 3.1 - Prob. 1LC Ch. 3.1 - Prob. 2LC Ch. 3.1 - Prob. 3LC Ch. 3.1 - Prob. 4LC Ch. 3.1 - Prob. 5LC Ch. 3.1 - Prob. 6LC Ch. 3.1 - Prob. 7PPSE Ch. 3.1 - Prob. 8PPSE Ch. 3.1 - Prob. 9PPSE Ch. 3.1 - Prob. 10PPSE

Ch. 3.1 - Prob. 11PPSE Ch. 3.1 - Prob. 12PPSE Ch. 3.1 - Prob. 13PPSE Ch. 3.1 - Prob. 14PPSE Ch. 3.1 - Prob. 15PPSE Ch. 3.1 - Prob. 16PPSE Ch. 3.1 - Prob. 17PPSE Ch. 3.1 - Prob. 18PPSE Ch. 3.1 - Prob. 19PPSE Ch. 3.1 - Prob. 20PPSE Ch. 3.1 - Prob. 21PPSE Ch. 3.1 - Prob. 22PPSE Ch. 3.1 - Prob. 23PPSE Ch. 3.1 - Prob. 24PPSE Ch. 3.1 - Prob. 25PPSE Ch. 3.1 - Prob. 26PPSE Ch. 3.1 - Prob. 27PPSE Ch. 3.1 - Prob. 28PPSE Ch. 3.1 - Prob. 29PPSE Ch. 3.1 - Prob. 30PPSE Ch. 3.1 - Prob. 31PPSE Ch. 3.1 - Prob. 32PPSE Ch. 3.1 - Prob. 33PPSE Ch. 3.1 - Prob. 34PPSE Ch. 3.1 - Prob. 35PPSE Ch. 3.1 - Prob. 36PPSE Ch. 3.1 - Prob. 37PPSE Ch. 3.1 - Prob. 38PPSE Ch. 3.1 - Prob. 39PPSE Ch. 3.1 - Prob. 40PPSE Ch. 3.1 - Prob. 41PPSE Ch. 3.1 - Prob. 42PPSE Ch. 3.1 - Prob. 43PPSE Ch. 3.1 - Prob. 44PPSE Ch. 3.1 - Prob. 45PPSE Ch. 3.1 - Prob. 46PPSE Ch. 3.1 - Prob. 47PPSE Ch. 3.1 - Prob. 48PPSE Ch. 3.1 - Prob. 49PPSE Ch. 3.1 - Prob. 50PPSE Ch. 3.1 - Prob. 51PPSE Ch. 3.1 - Prob. 52PPSE Ch. 3.1 - Prob. 53PPSE Ch. 3.1 - Prob. 54PPSE Ch. 3.1 - 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