Explanation Given: Total number of seats available in the theater is 3,600 with cost of ticket $40, $70 and $100. Total money that has to be earned by the owner from the full house is $192,000. The owner wants the seats that cost $40 per unit should be twice the sum of the seats that cost $70 and $100. Definition used: Cramer’s rule: The solution to { a 1 x + b 1 y + c 1 z = d 1 a 2 x + b 2 y + c 2 z = d 2 a 3 x + b 3 y + c 3 z = d 3 is x = D x D , y = D y D and z = D z D , where D = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | , D = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | , D = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | , D = | a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 | D ≠ 0 . If D = 0 and at least one of the determinants in the numerators, D x , D y and D z is not zero, then the system is inconsistent and the system has no solution. If D = 0 and all the determinants in the numerators, D x , D y and D z are zero, then the system of equations is dependent and the system has infinitely many solutions. Formula used: The determinant of the matrix [ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ] is det [ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 ] = | a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 | = a 1 b 2 c 3 + b 1 c 2 a 3 + c 1 a 2 b 3 − a 3 b 2 c 1 − b 3 c 2 a 1 − c 3 a 2 b 1 . Calculation: Let the number of seats in section C, B and A be x, y and z , respectively. As the total number of seats in the theater is 3,600, the first equation becomes x + y + z = 3 , 600 . The second equation is obtained as 40 x + 70 y + 100 z = 192 , 000 , as the total money that has to be earned by the owner from the full house is $192,000. The owner wants the seats that cost $40 per unit should be twice the sum of the seats that cost $70 and $100. Thus, the third equation will be x = 2 ( y + z ) that is x − 2 y − 2 z = 0 . Hence, the system of equations becomes, { x + y + z = 3 , 600 40 x + 70 y + 100 z = 192 , 000 x − 2 y − 2 z = 0 Solve the system of equations as follows. By above formula, compute the determinant of the 3 × 3 matrix. D = | 1 1 1 40 70 100 1 − 2 − 2 | = ( 1 ) [ 70 ⋅ ( − 2 ) − 100 ⋅ ( − 2 ) ] − ( 1 ) [ 40 ⋅ ( − 2 ) − 100 ( 1 ) ] + ( 1 ) [ 40 ( − 2 ) − 70 ( 1 ) ] = 1 ( − 140 + 200 ) − 1 ( − 80 − 100 ) + 1 ( − 80 − 70 ) = 60 + 180 − 150 That is, D = 90

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- for College Algebra in Context

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ISBN: 9780135909621

Author: HARSHBARGER, Ronald, YOCCO, Lisa

Publisher: PEARSON

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Chapter 7.5, Problem 32E

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Chapter 7 Solutions

MyLab Math with Pearson eText -- 24-Month Standalone Access Card -- for College Algebra in Context

Ch. 7.1 - In Exercises 14, solve the systems of equations....Ch. 7.1 - Prob. 2E Ch. 7.1 - Prob. 3E Ch. 7.1 - Prob. 4E Ch. 7.1 - Prob. 5E Ch. 7.1 - Prob. 6E Ch. 7.1 - Prob. 7E Ch. 7.1 - Prob. 8E Ch. 7.1 - Prob. 9E Ch. 7.1 - Prob. 10E

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The number of seats in each section.

Solution Summary: The author explains the C, B, and A type seats are 2,400, 800 and 400 respectively. The owner wants the seats that cost 40 per unit to be twice the sum of 70 and 100.

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To find: The number of seats in each section.

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Chapter 7 Solutions