Explanation Given: Total amount of money available with the company to buy cars is $3,160,000. Cost of each compact car is $18,000, cost of each midsize car is $28,000 and cost of each large car is $48,000. Total cars that has to be bought is 120 and total rental money that is earned is $7,600 where the rent for compact, midsize and large size car, per day are $50, $70 and $90, respectively. 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 compact, midsize and large size cars be x, y and z , respectively. As the total number of cars that has to be bought is 120, the first equation becomes x + y + z = 120 . Cost of each compact car is $18,000, cost of each midsize car is $28,000 and cost of each large car is $48,000 and the total money available is $3,160,000. Thus, the second equation is obtained as 18 , 000 x + 28 , 000 y + 48 , 000 z = 3 , 160 , 000 . The third equation will be 50 x + 70 y + 90 z = 7 , 600 , as the total money earned is $7,600. Hence, the system of equations becomes, { x + y + z = 120 18 , 000 x + 28 , 000 y + 48 , 000 z = 3 , 160 , 000 50 x + 70 y + 90 z = 7 , 600 Solve the system of equations as follows. By above formula, compute the determinant of the 3 × 3 matrix. D = | 1 1 1 18 , 000 28 , 000 48 , 000 50 70 90 | = ( ( 1 ) [ 28 , 000 ⋅ 90 − 48 , 000 ⋅ ( 70 ) ] − ( 1 ) [ 18 , 000 ⋅ 90 − 48 , 000 ⋅ 50 ] + ( 1 ) [ 18 , 000 ⋅ 70 − 28 , 000 ⋅ 50 ] ) = ( 2 , 520 , 000 − 3 , 360 , 000 ) − ( 1 , 620 , 000 − 2 , 400 , 000 ) + ( 1 , 260 , 000 − 1 , 400 , 000 ) = − 840 , 000 + 780 , 000 − 140 , 000 That is, D = − 200 , 000

College Algebra in Context, Books a la Carte Edition plus MyLab Math with Pearson eText -- Access Card Package (5th Edition)

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ISBN: 9780134397016

Author: Ronald J. Harshbarger, Lisa S. Yocco

Publisher: PEARSON

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Chapter 7.5, Problem 33E

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Chapter 7.5, Problem 32E

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Chapter 7 Solutions

College Algebra in Context, Books a la Carte Edition plus MyLab Math with Pearson eText -- Access Card Package (5th Edition)

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 each type of cars that should be bought by the car rental company.

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