Problem 26 - College Algebra with Corequisite Support

Question

For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $$\$ 75 / w$$ k plus $$\$.10 / \mathrm{mi}$$ driven. Plan \(\mathrm{B}\) charges $$\$ 100 / \mathrm{wk}$$ plus $$\$ .05 / \mathrm{mi}$$ driven. Find the number of miles that would generate the same cost for both plans.

Step-by-Step Solution

Answer

The number of miles that makes both plans cost the same is 500 miles.

Step 1: Identify Cost Equations

For Plan A, the cost is given by \( C_A = 75 + 0.10x \), where \( x \) is the number of miles driven. For Plan B, the cost is \( C_B = 100 + 0.05x \). We need to find \( x \) such that \( C_A = C_B \).

Step 2: Set Up Equation for Equality

Since we want the costs of both plans to be equal, set the equations equal to each other: \( 75 + 0.10x = 100 + 0.05x \).

Step 3: Eliminate Constant Terms

Subtract 75 from both sides of the equation to eliminate the constant term on the left side: \( 0.10x = 25 + 0.05x \).

Step 4: Eliminate Variable Coefficient

Subtract \( 0.05x \) from both sides to isolate \( x \) on the left side: \( 0.05x = 25 \).

Step 5: Solve for x

Divide both sides by 0.05 to solve for \( x \): \( x = \frac{25}{0.05} \).

Step 6: Compute the Final Answer

Perform the division: \( x = 500 \). Thus, 500 miles make the cost of both plans equal.

Key Concepts

Linear Equations

Linear equations are a fundamental concept in algebra and are used to describe a relationship between two variables in a straight line when plotted on a graph. In this context, each plan offered by the truck rental agency is represented by a linear equation that models cost. These equations incorporate fixed weekly fees and variable costs based on the number of miles traveled. For example: - **Plan A:** The linear equation is formulated as \( C_A = 75 + 0.10x \), where 75 represents the weekly fee in dollars, and \( 0.10x \) is the cost dependent on miles driven. - **Plan B:** Similarly, \( C_B = 100 + 0.05x \), where 100 is the weekly fee and \( 0.05x \) is the mileage cost.Each equation has two components: a constant and a coefficient multiplied by a variable. The constant represents the fixed cost, while the mileage rate is reflected in the coefficient multiplying the variable \( x \), which denotes the number of miles. Linear equations make it easier to understand how costs depend on miles driven, providing a simple yet powerful way to compare and analyze different pricing models.

Cost Analysis

Cost analysis involves examining the expenses associated with different choices and determining the conditions under which those expenses are equal or optimal. In this problem, we are tasked with finding when the costs of two different rental plans are equal. This comparison uses the concept of setting equations equal to one another, effectively performing a cost crossover analysis. To find the point at which the costs of Plan A and Plan B align, we set the respective cost equations equal to each other: - \( C_A = C_B \) translates to \( 75 + 0.10x = 100 + 0.05x \). By solving this equation, we determine the number of miles \( x \) where the cost of renting a truck with each plan becomes the same. This type of analysis is crucial in making informed decisions, especially in scenarios involving various financial commitments or business models.

Algebraic Solution

The algebraic solution to finding equal costs involves a step-by-step method to isolate and solve for the variable representing the number of miles. This approach requires manipulating the linear equations to determine the exact point where both plans cost the same. Here's how it's achieved:- **Set Up Equation:** Begin by setting the two cost equations equal: \( 75 + 0.10x = 100 + 0.05x \).- **Simplify:** Subtract 75 from both sides to get \( 0.10x = 25 + 0.05x \). Then, subtract \( 0.05x \) from both sides to isolate the variable term, resulting in \( 0.05x = 25 \).- **Solve for \( x \):** Finally, divide both sides by 0.05 to solve for \( x \), leading to \( x = \frac{25}{0.05} \).This results in \( x = 500 \), meaning that 500 miles driven would result in the same cost for both plans. Algebraic solutions like this allow for clear and concise problem-solving, giving us exact values that are crucial in decision-making scenarios.