Problem 29 - Introductory Algebra for College Students

Question

A car rental agency charges 200 dollars per week plus 0.15 dollars per mile to rent a car. How many miles can you travel in one week for 320 dollars ?

Step-by-Step Solution

Answer

You can travel 800 miles in one week for 320 dollars.

Step 1: Write down the equation

Start by writing an equation that captures the total cost of renting a car. The total cost is equal to the fixed cost and the variable cost added together. In this case, the fixed cost is 200 dollars, and the variable cost is 0.15 dollars times the number of miles driven, \(m\). Therefore, the equation is \(200 + 0.15m = 320\).

Step 2: Solve the equation for m

Next, solve this equation for \(m\). You can do this by subtracting 200 from both sides of the equation. This simplifies the equation to \(0.15m = 120\).

Step 3: Solve the equation

Finally, you can solve for \(m\) by dividing both sides of the equation by 0.15. This gives \(m = 120/0.15 = 800\).

Key Concepts

Equation Solving

Equation solving is a process where we find the value of a variable that makes an equation true. In our car rental problem, the equation we start with is \(200 + 0.15m = 320\). Here, we need to identify both the fixed part of the cost (200 dollars) and the variable part (0.15 dollars per mile).

• The first step is understanding how the total cost is structured: a fixed cost plus a variable cost that depends on the number of miles \(m\).
• To find \(m\), our goal is to isolate it on one side of the equation.

Begin by removing the fixed cost. Subtract 200 from both sides to simplify the equation to \(0.15m = 120\). Now, only the variable part remains, making it easier to solve for \(m\). Lastly, to isolate \(m\), divide both sides by 0.15, resulting in \(m = 800\). Therefore, you can drive 800 miles within the budget of 320 dollars.

Linear Equations

Linear equations are used to describe relationships between variables. These equations form a straight line when graphed, showing consistent rates of change. In our example, the equation \(200 + 0.15m = 320\) is linear, with \(m\) being the number of miles you can drive.

• This equation comprises a constant term (200) and a coefficient of the variable \(m\) (0.15), indicating a constant rate of cost increase per mile.
• When solving, you seek to balance both sides, ensuring the equality holds with the found value of \(m\).

Linear equations like this help simplify complex problems by expressing them as straightforward relationships. In real-life applications, they allow us to make predictions and handle constraints efficiently. Here, the linear relationship helps calculate how many miles you can travel for a given budget, making financial decisions easier.

Word Problems

Word problems translate real-world situations into mathematical equations. The key is extracting necessary information to set up an equation that can be solved. In this example, we're dealing with a car rental scenario involving costs and mileage.

• Understand the components: Identify the fixed charge and the per-mile charge from the problem statement.
• Convert these into an algebraic form: Create an equation reflecting these components.

Once the equation is established, as in \(200 + 0.15m = 320\), solving it involves standard techniques like those used in any algebraic equation. Word problems often appear daunting but breaking them into smaller pieces—like cost structure and equation formation—simplifies the process. Approaching these problems methodically allows you to handle even the most challenging scenarios reliably.