Problem 32 - College Algebra

Question

A car rental agency charges \(\$ 180\) per week plus \(\$ 0.25\) per mile to rent a car. How many miles can you travel in one week for \(\$ 395 ?\)

Step-by-Step Solution

Answer

You can travel 860 miles in one week for \$395.

Step 1: Identify and Setup the Equation

First, identify the fixed and variable costs in the problem. The fixed cost is \$180, this is the base cost of renting a car for a week regardless of how many miles are driven. The variable cost is \$0.25 per mile, this is an additional cost added based on how many miles are driven in a week. The total cost for the week is \$395. The equation to represent this is \$180 + \$0.25x = \$395 where x represents the number of miles driven.

Step 2: Solve for x

Next, solve this equation for x, to find how many miles can be driven. First, subtract $180 from both sides of the equation to isolate \$0.25x: \$0.25x = \$395 - \$180 = \$215. Then divide every side by \$0.25 to find x: x = \$215 / \$0.25 = 860.

Step 3: Interpret the result

The solution x = 860 means that 860 miles can be driven in one week for the total cost of \$395.

Key Concepts

Algebra Word Problems

When encountering algebra word problems, it's crucial to understand how to dissect the written information and translate it into a solvable mathematical expression.

Firstly, carefully read the problem to identify the key pieces of information. Look for fixed values, which are numbers that do not change, and variable values, which can vary based on certain conditions. In our exercise, the car rental problem presents a typical scenario where a real-life situation needs to be expressed using an algebraic equation.

To approach such problems, follow these tips:

• Identify the constants in the situation, like the \(180 weekly rental fee.
• Spot the variable costs or rates, which in this instance is \)0.25 per mile driven.
• Formulate an equation using a placeholder, or variable, to represent the unknown quantity you're solving for.
• Combine the constants and variable expressions based on the situation described.

By converting the word problem into an algebraic equation, we create a bridge between the textual description and the numeric solution.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra, involving finding the value of the variable that makes the equation true.

Every linear equation, at its core, represents a straight line when plotted on a coordinate plane. A linear equation in one variable takes the general form of \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. To solve these equations:

• Isolate the variable on one side of the equation using inverse operations.
• Perform the same operation on both sides to maintain the equality.
• Once the variable is isolated, solve for the variable by performing the necessary arithmetic operation.

Remember to check your solution by plugging it back into the original equation to ensure it makes the equation true. In the car rental agency problem, isolating the variable and solving it defines the number of miles that fit the rental cost constraints.

Variable Cost Calculations

Variable costs play a crucial role in budgeting and finance, fluctuating directly with the level of production or activity. Such costs are different from fixed costs, which remain constant regardless of activity volume.

In our exercise, the cost per mile represented a variable cost. Calculating variable costs involves multiplying the variable rate by the quantity of the variable activity. For real-world application, consider the following steps:

• Determine the variable cost per unit of activity, for example, the $0.25 cost per mile.
• Multiply this cost by the number of activity units to find the total variable cost.
• Add any fixed costs to calculate the total expense.

Understanding how to calculate variable costs helps in making informed financial decisions, such as determining whether you can afford to travel a certain distance within a budget.