Problem 24 - Algebra A Combined Function

Question

Joan Gundersen rented the same car model twice from Hertz, which rents this car model for a daily fee plus an additional charge per mile driven. Joan recalls that the car rented for 5 days and driven for 300 miles cost her \(\$ 178\), while the same model car rented for 4 days and driven for 500 miles cost \(\$ 197\). Find the daily fee, and find the mileage charge.

Step-by-Step Solution

Answer

Daily fee is $23; mileage charge is $0.21 per mile.

Step 1: Define Variables

Let's define two variables: \( d \) for the daily rental fee and \( m \) for the mileage charge per mile.

Step 2: Formulate Equations

We formulate the equations based on the information given: 1. For 5 days and 300 miles: \( 5d + 300m = 178 \) 2. For 4 days and 500 miles: \( 4d + 500m = 197 \)

Step 3: Solve the System of Equations

We have a system of two linear equations: \( 5d + 300m = 178 \)\( 4d + 500m = 197 \)Let's use the substitution or elimination method to solve this system.

Step 4: Using the Elimination Method

We multiply the first equation by 4 and the second by 5 to eliminate \( d \):\[ 20d + 1200m = 712 \]\[ 20d + 2500m = 985 \]Subtract the first equation from the second:\[ 1300m = 273 \]\[ m = \frac{273}{1300} = 0.21 \]

Step 5: Substitute Back to Find \( d \)

Substitute \( m = 0.21 \) back into one of the original equations:Use equation 1: \( 5d + 300(0.21) = 178 \)\[ 5d + 63 = 178 \]\[ 5d = 115 \]\[ d = \frac{115}{5} = 23 \]

Step 6: Verify the Solution

Substitute \( d = 23 \) and \( m = 0.21 \) into the second equation to verify:\( 4d + 500m = 197 \)\[ 4(23) + 500(0.21) = 92 + 105 = 197 \] The solution is verified as correct.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent a straight line and have two main components: the constants and the variables. In the context of the scenario provided, linear equations help us model the relationship between the number of days Joan rents the car, the miles she drives, and the total cost of the rental.

These equations take the form of:

• Constants: These are numbers without variables, like the total rental cost (\(178 and \)197 in the equations).
• Variables: Represent unknown values that need to be solved (e.g., daily fee and mileage charge).

A linear equation generally appears as \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables. In our example:

• \( d \) is the daily rate,
• \( m \) stands for the mileage rate per mile,
• The sum of these products is equal to the total cost.

Understanding linear equations is crucial because they allow us to solve for unknowns in various practical situations, like rental fees.

Substitution Method

The substitution method is a widely used approach to solve systems of linear equations by expressing one variable in terms of another and then substituting this expression into another equation. This breaks down the complex problem into simpler steps, ensuring clarity and correctness.

Here's a simple breakdown of how substitution works:

• First, solve one of the linear equations for one of the variables. For instance, express \( d \) in terms of \( m \) or vice versa.
• Second, substitute this expression into the other equation. Replace the variable in the second equation with the expression found in the first step.
• The result will be an equation with one variable, which can be solved directly.

This method is especially handy when one of the equations is easy to manipulate to express one variable.

Elimination Method

The elimination method is another approach to solve a system of linear equations, aimed at removing one variable to simplify the process. The goal is to alter the equations so that adding or subtracting them eliminates one variable, making it straightforward to solve the remaining equation.

Here's how elimination typically works:

• First, manipulate the equations by multiplication so that the coefficients of one of the variables are the same.
• Subtract or add these equations to eliminate one of the variables.
• What remains is a single linear equation with one variable, which can be easily solved.

This method often suits situations where substitution can appear cumbersome. In Joan’s car rental example, multiplying each equation by certain numbers allowed us to eliminate the daily fee \( d \) by ensuring both equations had equal coefficients for \( d \). Once one variable was solved, it was substituted back to determine the other.

Variable Definition

Variables are symbols used in equations to represent unknown values that we aim to find. In mathematics, properly defining variables is a critical step. It lays the foundation for forming equations and solving problems efficiently.

In the car rental problem, we defined:

• \( d \): The daily rental fee for the car.
• \( m \): The charge per mile driven.

Choosing appropriate variables helps streamline the equation-formation process. Clear definitions prevent confusion and make substitution and elimination methods much more manageable. By defining variables, we're able to convert a real-world problem into a mathematical model that can be solved systematically, unraveling unknowns that inform our decision-making and application.