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

To solve the system of equations x − y = − 4 3 x + 2 y = 7 by the method of elimination. The solution of the system of equations x − y = − 4 3 x + 2 y = 7 is found to be x = − 1 5 and y = 19 5 . Given the system of equations: x − y = − 4 3 x + 2 y = 7 Concept Used: Method of elimination in the case of linear equations in two variables: Given a system of linear equations in two variables. The method of elimination employs the elimination of one variable from the system of equations. Thus, the first step is to determine the variable that is to be eliminated. Then, the equations are multiplied with different non-zero constants and added together in a way so that the targeted variable is eliminated. In some cases, a simple addition would eliminate the variable. But, in general scalar multiplication is needed as said earlier. Calculation: Multiply the first equation x − y = − 4 with 2 on either side: 2 x − y = 2 − 4 2 x − 2 y = − 8 ... ( 1 ) . Multiply the second equation 3 x + 2 y = 7 with 1 on either side: 1 3 x + 2 y = 1 7 3 x + 2 y = 7 ... ( 2 ) . Perform the addition ( 1 ) + ( 2 ) : 2 x − 2 y = − 8 3 x + 2 y = 7 5 x + 0 y = − 1 + . Thus, it is found that: 5 x + 0 y = − 1 . Find the value of x from 5 x + 0 y = − 1 : 5 x + 0 y = − 1 5 x + 0 = − 1 5 x = − 1 x = − 1 5 . Thus, it is found that: x = − 1 5 . Substitute x = − 1 5 in the first equation x − y = − 4 and solve it for y : x − y = − 4 − 1 5 − y = − 4 − y = − 4 + 1 5 y = 4 − 1 5 y = 19 5 . Thus, it is found that: y = 19 5 . Thus, the solution of the system of equations x − y = − 4 3 x + 2 y = 7 is found to be x = − 1 5 and y = 19 5 . Conclusion: The solution of the system of equations x − y = − 4 3 x + 2 y = 7 is found to be x = − 1 5 and y = 19 5 .

To solve the system of equations x − y = − 4 3 x + 2 y = 7 by the method of elimination. The solution of the system of equations x − y = − 4 3 x + 2 y = 7 is found to be x = − 1 5 and y = 19 5 . Given the system of equations: x − y = − 4 3 x + 2 y = 7 Concept Used: Method of elimination in the case of linear equations in two variables: Given a system of linear equations in two variables. The method of elimination employs the elimination of one variable from the system of equations. Thus, the first step is to determine the variable that is to be eliminated. Then, the equations are multiplied with different non-zero constants and added together in a way so that the targeted variable is eliminated. In some cases, a simple addition would eliminate the variable. But, in general scalar multiplication is needed as said earlier. Calculation: Multiply the first equation x − y = − 4 with 2 on either side: 2 x − y = 2 − 4 2 x − 2 y = − 8 ... ( 1 ) . Multiply the second equation 3 x + 2 y = 7 with 1 on either side: 1 3 x + 2 y = 1 7 3 x + 2 y = 7 ... ( 2 ) . Perform the addition ( 1 ) + ( 2 ) : 2 x − 2 y = − 8 3 x + 2 y = 7 5 x + 0 y = − 1 + . Thus, it is found that: 5 x + 0 y = − 1 . Find the value of x from 5 x + 0 y = − 1 : 5 x + 0 y = − 1 5 x + 0 = − 1 5 x = − 1 x = − 1 5 . Thus, it is found that: x = − 1 5 . Substitute x = − 1 5 in the first equation x − y = − 4 and solve it for y : x − y = − 4 − 1 5 − y = − 4 − y = − 4 + 1 5 y = 4 − 1 5 y = 19 5 . Thus, it is found that: y = 19 5 . Thus, the solution of the system of equations x − y = − 4 3 x + 2 y = 7 is found to be x = − 1 5 and y = 19 5 . Conclusion: The solution of the system of equations x − y = − 4 3 x + 2 y = 7 is found to be x = − 1 5 and y = 19 5 .

Solution Summary: The author explains the method of elimination in the case of linear equations in two variables.

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 3.2, Problem 5LC

Solution

To solve the system of equations x−y=−43x+2y=7 by the method of elimination.

The solution of the system of equations x−y=−43x+2y=7 is found to be x=−15 and y=195 .

Given the system of equations:

x−y=−43x+2y=7

Concept Used:

Method of elimination in the case of linear equations in two variables:

Given a system of linear equations in two variables.

The method of elimination employs the elimination of one variable from the system of equations. Thus, the first step is to determine the variable that is to be eliminated. Then, the equations are multiplied with different non-zero constants and added together in a way so that the targeted variable is eliminated. In some cases, a simple addition would eliminate the variable. But, in general scalar multiplication is needed as said earlier.

Calculation:

Multiply the first equation x−y=−4 with 2 on either side:

2x−y=2−42x−2y=−8 ...(1) .

Multiply the second equation 3x+2y=7 with 1 on either side:

13x+2y=173x+2y=7 ...(2) .

Perform the addition (1) +(2) :

2x−2y=−83x+2y=75x+ 0y=−1+ .

Thus, it is found that:

5x+ 0y=−1 .

Find the value of x from 5x+ 0y=−1 :

5x+0y=−15x+0=−15x=−1x=−15 .

Thus, it is found that:

x=−15 .

Substitute x=−15 in the first equation x−y=−4 and solve it for y :

x−y=−4−15−y=−4−y=−4+15y=4−15y=195 .

Thus, it is found that:

y=195 .

Thus, the solution of the system of equations x−y=−43x+2y=7 is found to be x=−15 and y=195 .

Conclusion:

The solution of the system of equations x−y=−43x+2y=7 is found to be x=−15 and y=195 .

Chapter 3.2, Problem 4LC

Chapter 3.2, Problem 6LC

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

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