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Difficulty: MediumLinear Equations in One and Two Variables

A car rental agency charges a fixed daily rate plus a constant fee per mile driven. A customer who rented a car for 3 days and drove 150 miles paid a total of 180.Anothercustomerwhorentedthesamemodelofcarfor5daysanddrove200milespaidatotalof180. Another customer who rented the same model of car for 5 days and drove 200 miles paid a total of 270. What is the fixed daily rate, in dollars, charged by the rental agency?

Answer: 30 dollars

Answer

The fixed daily rate charged by the rental agency is 30 dollars.
Formulating the equations 3d+150m=1803d + 150m = 180 and 5d+200m=2705d + 200m = 270 and solving for the daily rate dd yields d=30d = 30.

Step-by-Step Solution

1
Define variables and formulate the system of linear equations.
Let dd be the fixed daily rate (in dollars) and mm be the cost per mile driven (in dollars). The scenario translates to:
Equation 1: 3d+150m=1803d + 150m = 180
Equation 2: 5d+200m=2705d + 200m = 270
Total charge is the linear combination of daily fixed costs and per-mile variable costs.
2
Simplify Equation 1 to express dd in terms of mm.
Dividing Equation 1 by 3 yields d+50m=60d + 50m = 60, so d=60−50md = 60 - 50m.
Simplifying equations reduces computation error when using substitution.
3
Substitute the expression for dd into Equation 2 to solve for mm.
5(60−50m)+200m=270⇒300−250m+200m=270⇒−50m=−30⇒m=0.605(60 - 50m) + 200m = 270 \Rightarrow 300 - 250m + 200m = 270 \Rightarrow -50m = -30 \Rightarrow m = 0.60
Substitution eliminates variable dd, leaving a linear equation in one variable.
4
Calculate the fixed daily rate dd.
d=60−50(0.60)=60−30=30d = 60 - 50(0.60) = 60 - 30 = 30
Substituting m=0.60m = 0.60 gives the exact daily fixed rate.

Key Concept

Setting up and solving a system of two linear equations in two variables using elimination or substitution.
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Topics

Linear Equations in One and Two Variables

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