Problem 58 - College Algebra

Question

For the following exercises, use this scenario: The cost of renting a car is \(\$ 45 / \mathrm{wk}\) plus \(\$ 0.25 / \mathrm{mi}\) traveled during that week. An equation to represent the cost would be \(y=45+0.25 x\), where \(x\) is the number of miles traveled. If your cost were \(\$ 63.75\), how many miles were you charged for traveling?

Step-by-Step Solution

Answer

You were charged for traveling 75 miles.

Step 1: Understand the Equation

The given equation is \( y = 45 + 0.25x \), where \( y \) represents the total cost, \( 45 \) is the weekly rental cost, and \( 0.25x \) is the cost per mile for \( x \) miles traveled.

Step 2: Substitute the Known Value

We need to find out how many miles \( x \) were traveled when the cost \( y \) is \( 63.75 \). Replace \( y \) in the equation with \( 63.75 \):\[ 63.75 = 45 + 0.25x \]

Step 3: Isolate the Variable Term

Subtract \( 45 \) from both sides of the equation to isolate the \( 0.25x \) term:\[ 63.75 - 45 = 0.25x \] This simplifies to:\[ 18.75 = 0.25x \]

Step 4: Solve for \( x \)

Divide both sides of the equation by \( 0.25 \) to solve for \( x \):\[ x = \frac{18.75}{0.25} \]Calculating the division gives:\[ x = 75 \]

Step 5: Verify the Solution

Substitute \( x = 75 \) back into the original equation to ensure the solution is correct:\[ y = 45 + 0.25(75) \]Calculate \( 0.25 \times 75 = 18.75 \) and then add to \( 45 \):\[ y = 45 + 18.75 = 63.75 \]The calculation confirms the total cost of \$63.75.

Key Concepts

Equation Solving

Equation solving is a fundamental mathematical process. It involves finding the value of a variable that makes an equation true. In our exercise, we're working with the equation: \[y = 45 + 0.25x\]This equation links the cost, \(y\), to the number of miles, \(x\), traveled. Solving the equation helps us figure out how many miles were driven based on a given cost.
To solve an equation:

• Identify what you know (in this case, the total cost \( y = 63.75 \)).
• Identify what you need to find (here, the number of miles \( x \)).
• The equation will guide you through balancing and finding the unknown.

Equation solving is all about logical steps to find the answer. You often begin by substituting known values into the equation, just like you replace \( y \) with \( 63.75 \) in our example to begin solving for \( x \).

Variable Isolation

Variable isolation is the key step in solving equations effectively. It simply means we want to get our variable of interest alone on one side of the equation. In our case, \( x \) represents miles traveled, and we aim to find its value by isolating it.
Imagine the equation as a scale—you want to keep it balanced while moving everything that is not the variable to the other side. The original equation in the exercise was:\[63.75 = 45 + 0.25x\]The key steps here are:

• Subtract \( 45 \) from both sides to get all terms involving \( x \) on one side: \( 63.75 - 45 = 18.75 \)
• You now have \( 18.75 = 0.25x \). We see that \( 0.25x \) is the term with our variable.
• Finally, divide by \( 0.25 \) to solve for \( x \): \( x = \frac{18.75}{0.25} \).

By performing these operations, \( x \) is isolated and its value is discovered, supported by the rules of algebra.

Cost Calculation

With linear equations, cost calculation becomes straightforward following some basic rules. The formula given, \( y = 45 + 0.25x \), helps calculate the total cost of renting a car at a given number of miles. Here's how the components break down:

• Basic rental cost: \(\\( 45\) per week; that's a fixed cost, meaning it doesn't change.
• Mileage cost: \(\\) 0.25\) per mile \(x\); it varies depending on how many miles you travel.

In the exercise, calculating the total cost means adding these two parts together. To reverse-engineer how many miles were driven given a total cost, like \( 63.75 \), we use the equation differently: First, the \(\$ 45\) is subtracted since it's constant and doesn't depend on miles. That leaves the variable part: \(0.25x = 18.75\). The remaining steps focus on isolating \( x \), as explained in variable isolation, above, solving for the miles.This approach lets you calculate costs or adjust plans for car travel flexibly, depending on the distance needed and your budget constraints.