Explanation Given: Let x be the number of bills in denomination of $1. Let y be the number of bills in denomination of $5. Let z be the number of bills in denomination of $10. Let w be the number of bills in denomination of $20. The server has a total of $95. The equation is, x + 5 y + 10 z + 20 w = 95 → ( 1 ) Total number of paper bills is 26. The equation is, x + y + z + w = 26 → ( 2 ) The number of $5 is 4 times the number of $10 bills. The equation is, y = 4 z y − 4 z = 0 → ( 3 ) The number of $1 bills is one less than twice the number of $5 bills. The equation is, x = 2 y − 1 x − 2 y = − 1 → ( 4 ) From the equations, ( 1 ) , ( 2 ) , ( 3 ) , a n d ( 4 ) we get the system of linear equations. x + 5 y + 10 z + 20 w = 95 x + y + z + w = 26 y − 4 z = 0 x − 2 y = − 1 The augmented matrix is, [ 1 1 0 1 5 1 1 − 2 10 1 − 4 0 20 1 0 0 95 26 0 − 1 ] R 2 → − R 1 + R 2 [ 1 1 0 1 5 1 1 − 2 10 1 − 4 0 20 1 0 0 95 26 0 − 1 ] = [ 1 1 − 1 0 1 5 1 − 5 1 − 2 10 1 − 10 − 4 0 20 1 − 20 0 0 95 26 − 95 0 − 1 ] = [ 1 0 0 1 5 − 4 1 − 2 10 − 9 − 4 0 20 − 19 0 0 95 − 69 0 − 1 ] R 4 → R 4 − R 1 [ 1 0 0 1 5 − 4 1 − 2 10 − 9 − 4 0 20 − 19 0 0 95 − 69 0 − 1 ] = [ 1 0 0 1 − 1 5 − 4 1 − 2 − 5 10 − 9 − 4 0 − 10 20 − 19 0 0 − 20 95 − 69 0 − 1 − 95 ] = [ 1 0 0 0 5 − 4 1 − 7 10 − 9 − 4 − 10 20 − 19 0 − 20 95 − 69 0 − 96 ] R 2 ↔ R 3 [ 1 0 0 0 5 − 4 1 − 7 10 − 9 − 4 − 10 20 − 19 0 − 20 95 − 69 0 − 96 ] = [ 1 0 0 0 5 1 − 4 − 7 10 − 4 − 9 − 10 20 0 − 19 − 20 95 0 − 69 − 96 ] R 1 → R 1 − 5 R 2 [ 1 0 0 0 5 1 − 4 − 7 10 − 4 − 9 − 10 20 0 − 19 − 20 95 0 − 69 − 96 ] = [ 1 − 5 ( 0 ) 0 0 0 5 − 5 ( 1 ) 1 − 4 − 7 10 − 5 ( − 4 ) − 4 − 9 − 10 20 − 5 ( 0 ) 0 − 19 − 20 95 − 5 ( 0 ) 0 − 69 − 96 ] = [ 1 − 0 0 0 0 5 − 5 1 − 4 − 7 10 + 20 − 4 − 9 − 10 20 − 0 0 − 19 − 20 95 − 0 0 − 69 − 96 ] = [ 1 0 0 0 0 1 − 4 − 7 30 − 4 − 9 − 10 20 0 − 19 − 20 95 0 − 69 − 96 ] R 3 → R 3 + 4 R 2 [ 1 0 0 0 0 1 − 4 − 7 30 − 4 − 9 − 10 20 0 − 19 − 20 95 0 − 69 − 96 ] = [ 1 0 0 + 4 ( 0 ) 0 0 1 − 4 + 4 ( 1 ) − 7 30 − 4 − 9 + 4 ( − 4 ) − 10 20 0 − 19 + 4 ( 0 ) − 20 95 0 − 69 + 4 ( 0 ) − 96 ] = [ 1 0 0 + 0 0 0 1 − 4 + 4 − 7 30 − 4 − 9 − 16 − 10 20 0 − 19 + 0 − 20 95 0 − 69 + 0 − 96 ] = [ 1 0 0 0 0 1 0 − 7 30 − 4 − 25 − 10 20 0 − 19 − 20 95 0 − 69 − 96 ] R 4 → R 4 + 7 R 2 [ 1 0 0 0 0 1 0 − 7 30 − 4 − 25 − 10 20 0 − 19 − 20 95 0 − 69 − 96 ] = [ 1 0 0 0 + 7 ( 0 ) 0 1 0 − 7 + 7 ( 1 ) 30 − 4 − 25 − 10 + 7 ( − 4 ) 20 0 − 19 − 20 + 7 ( 0 ) 95 0 − 69 − 96 + 7 ( 0 ) ] = [ 1 0 0 0 0 1 0 0 30 − 4 − 25 − 10 − 28 20 0 − 19 − 20 + 0 95 0 − 69 − 96 + 0 ] = [ 1 0 0 0 0 1 0 0 30 − 4 − 25 − 38 20 0 − 19 − 20 95 0 − 69 − 96 ] R 3 → − R 3 25

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN: 9781305658004

Author: Ron Larson

Publisher: Cengage Learning

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Chapter 1.2, Problem 48E

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To find:

The number of denomination.

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Chapter 1.2, Problem 47E

Chapter 1.2, Problem 49E

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The number of denomination.

Solution Summary: The author explains that the server has a total of 95. The equation is, x+5y+10z+20w=95to (1).

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To find: The number of denomination.

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Chapter 1 Solutions

To find:

The number of denomination.