Answer The systems of linear equations which are equivalent are 4 x − 5 y = 3 2 x + 15 y = − 1 4 x − 5 y = 3 − 4 x − 30 y = 2 12 x − 15 y = 9 2 x + 15 y = − 1 Explanation Given information: The systems of linear equations 4 x − 5 y = 3 2 x + 15 y = − 1 ……………………… (1) 4 x − 5 y = 3 − 4 x − 30 y = 2 ……………………… (2) 4 x − 5 y = 3 4 x + 30 y = − 1 ……………………… (3) 12 x − 15 y = 9 2 x + 15 y = − 1 ……………………… (4) Calculation: The 2 n d equation of (2) can be written as, − 4 x − 30 y = 2 ⇒ − 2 ( 2 x + 15 y ) = 2 ⇒ 2 x + 15 y = 2 − 2 ⇒ 2 x + 15 y = − 1 So, the system of linear equation (2) is equivalent to, 4 x − 5 y = 3 2 x + 15 y = − 1 Thus, (2) is equivalent to (1). The 2 n d equation of (3) can be written as, 4 x + 30 y = − 1 ⇒ 2 ( 2 x + 15 y ) = − 1 ⇒ 2 x + 15 y = − 1 2 So, the system of linear equation (3) is equivalent to, 4 x − 5 y = 3 2 x + 15 y = − 1 2 Thus, (3) is not equivalent to (1). The 1 s t equation of (4) can be written as, 12 x − 15 y = 9 ⇒ 3 ( 4 x − 5 y ) = 9 ⇒ 4 x − 5 y = 9 3 ⇒ 4 x − 5 y = 3 So, the system of linear equation (4) is equivalent to, 4 x − 5 y = 3 2 x + 15 y = − 1 Thus, (4) is equivalent to (1). Conclusion: The systems of linear equations which are equivalent are 4 x − 5 y = 3 2 x + 15 y = − 1 4 x − 5 y = 3 − 4 x − 30 y = 2 12 x − 15 y = 9 2 x + 15 y = − 1

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

1st Edition

ISBN: 9781680331141

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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Question

Chapter 5, Problem 9CA

To determine

To find:The systems of linear equations which are equivalent.

Expert Solution & Answer

Answer to Problem 9CA

The systems of linear equations which are equivalent are

4x−5y=32x+15y=−1

4x−5y=3−4x−30y=2

12x−15y=92x+15y=−1

Explanation of Solution

Given information:The systems of linear equations

4x−5y=32x+15y=−1 ……………………… (1)

4x−5y=3−4x−30y=2 ……………………… (2)

4x−5y=34x+30y=−1 ……………………… (3)

12x−15y=92x+15y=−1 ……………………… (4)

Calculation:The 2ⁿ^d equation of (2) can be written as,

−4x−30y=2⇒−2(2x+15y)=2⇒2x+15y=2−2⇒2x+15y=−1

So, the system of linear equation (2) is equivalent to,

4x−5y=32x+15y=−1

Thus, (2) is equivalent to (1).

The 2ⁿ^d equation of (3) can be written as,

4x+30y=−1⇒2(2x+15y)=−1⇒2x+15y=−12

So, the system of linear equation (3) is equivalent to,

4x−5y=32x+15y=−12

Thus, (3) is not equivalent to (1).

The 1^s^t equation of (4) can be written as,

12x−15y=9⇒3(4x−5y)=9⇒4x−5y=93⇒4x−5y=3

So, the system of linear equation (4) is equivalent to,

4x−5y=32x+15y=−1

Thus, (4) is equivalent to (1).

Conclusion:The systems of linear equations which are equivalent are

4x−5y=32x+15y=−1

4x−5y=3−4x−30y=2

12x−15y=92x+15y=−1

Chapter 5, Problem 8CA

Chapter 5, Problem 10CA

Chapter 5 Solutions

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E

Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.2 - Prob. 1E Ch. 5.2 - Prob. 2E Ch. 5.2 - Prob. 3E Ch. 5.2 - Prob. 4E Ch. 5.2 - Prob. 5E Ch. 5.2 - Prob. 6E Ch. 5.2 - Prob. 7E Ch. 5.2 - Prob. 8E Ch. 5.2 - Prob. 9E Ch. 5.2 - Prob. 10E Ch. 5.2 - Prob. 11E Ch. 5.2 - Prob. 12E Ch. 5.2 - Prob. 13E Ch. 5.2 - Prob. 14E Ch. 5.2 - Prob. 15E Ch. 5.2 - Prob. 16E Ch. 5.2 - Prob. 17E Ch. 5.2 - Prob. 18E Ch. 5.2 - Prob. 19E Ch. 5.2 - Prob. 20E Ch. 5.2 - Prob. 21E Ch. 5.2 - Prob. 22E Ch. 5.2 - Prob. 23E Ch. 5.2 - Prob. 24E Ch. 5.2 - 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