Using linear systems to solve problems (1)

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SUMMARY

The discussion centers on solving a linear system to determine the distance at which the cost of renting a car from two agencies becomes equal. Rent-a-Heap charges \$50 per day plus \$0.12/km, while Kurt's Rent-a-Car charges \$40 per day plus \$0.20/km. The equation to find the distance (x) where costs are equal is derived from the linear equation format y = mx + b, leading to the formulation Ax + By = C. The solution involves setting the total costs equal and solving for x.

PREREQUISITES
  • Understanding of linear equations and systems
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  • Basic algebra skills, including solving for variables
  • Knowledge of the slope-intercept form of a line (y = mx + b)
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  • Learn how to set up and solve linear equations in two variables
  • Explore real-world applications of linear systems in cost analysis
  • Study the graphical representation of linear equations
  • Investigate methods for solving systems of equations, such as substitution and elimination
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Students, mathematicians, and professionals involved in cost analysis, budgeting, or any field requiring the application of linear systems to solve practical problems.

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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
 

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