MHB Using linear systems to solve problems (1)

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To determine when the cost of renting a car from Rent-a-Heap and Kurt's Rent-a-Car is the same, a linear equation can be set up based on their pricing structures. Rent-a-Heap charges $50 per day plus $0.12 per kilometer, while Kurt's Rent-a-Car charges $40 per day plus $0.20 per kilometer. The equations can be expressed as C = 50 + 0.12D for Rent-a-Heap and C = 40 + 0.20D for Kurt's. Setting these equations equal allows for solving the distance D at which both rental costs are identical. This approach utilizes linear systems to find the solution effectively.
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Sook-Lee wants to rent a car for a day so she can visit her sister at university. She has called two car rental agencies. Rent-a-Heap charges \$50 per day, plus \$0.12/km. Kurt's Rent-a-Car charges \$40 per day, plus \$0.20/km. At what distance will the cost of renting of a car be the same from companies?
 
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We are warned that it is a linear system. There should be a lot of y - mx + b going on. Maybe Ax + By = C.

Rules #1 - Name Stuff!

D = Daily Charge
M = Mileage Charge

Now what?
 
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* y = mx + b
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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