Problem 35 - Prealgebra

Question

Long-Distance Charges The cost of a long-distance phone call is $$\$ 0.41$$ for the first minute and $$\$ 0.32$$ for each additional minute. If the total charge for a long-distance call is $$\$ 5.21$$ how many minutes was the call?

Step-by-Step Solution

Answer

The call lasted 16 minutes.

Step 1: Define Variables

Let \( x \) be the total number of minutes of the call. The cost for the first minute is a fixed \( \\(0.41 \). Each additional minute after the first costs \( \\)0.32 \).

Step 2: Create the Equation

The total cost of the call is given as \( \$5.21 \). Therefore, we can set up the equation based on the costs: \( 0.41 + 0.32(x - 1) = 5.21 \). The term \( x - 1 \) accounts for the additional minutes after the first minute.

Step 3: Simplify and Solve the Equation

Start by solving the equation: \[ 0.41 + 0.32(x - 1) = 5.21 \].First, distribute the \( 0.32 \): \[ 0.41 + 0.32x - 0.32 = 5.21 \].Combine like terms: \[ 0.32x + 0.09 = 5.21 \].Subtract \( 0.09 \) from both sides: \[ 0.32x = 5.12 \].Divide both sides by \( 0.32 \) to solve for \( x \): \[ x = \frac{5.12}{0.32} = 16 \].

Step 4: Verify the Solution

Calculate the cost for a 16-minute call. The cost is \( 0.41 \) for the first minute plus \( 0.32 \times 15 \) for the remaining 15 minutes.Calculate: \[ 0.32 \times 15 = 4.80 \].Total: \[ 0.41 + 4.80 = 5.21 \].The calculated total matches the given total, verifying the solution.

Key Concepts

Solving Linear Equations

When solving linear equations, the goal is to find the value of the variable that satisfies the equation. In this exercise, we are trying to figure out how many minutes a phone call lasted based on the cost.
To solve linear equations effectively:

• Start by defining variables. Here, we let \( x \) represent the total number of minutes.
• Create an equation based on the problem description. We knew the first minute cost \( \\(0.41 \) and each additional minute cost \( \\)0.32 \).
• Solve the equation by isolating the variable. This involves distributing any coefficients, combining like terms, and moving constants to one side of the equation.

In this example, we form the equation \( 0.41 + 0.32(x - 1) = 5.21 \). Simplifying involves distributing the 0.32 across \((x - 1)\) and solving for \( x \). The final step is to verify if your solution is correct by plugging it back into the original context.

Cost Calculations

Cost calculations for services like phone calls often involve different rates for the first minute compared to subsequent minutes. To manage such costs, it's vital to categorize each part of the call:

• The first minute has a fixed charge. For instance, the initial minute is always \( \\(0.41 \).
• Additional minutes carry a different rate, here, \( \\)0.32 \) per minute after the first.

Adding these costs involves multiplying the cost per minute with the number of minutes, then summing them together.
Precisely for this problem, the calculation starts with the base cost of \( \$0.41 \) and adds \( 0.32 \times 15 \) for the remaining 15 minutes. It's a straightforward multiplication that determines how much each additional minute contributes to the overall bill.

Problem-Solving Steps

Problem-solving requires a clear, systematic approach to finding solutions. Here is the step-by-step approach taken in the exercise:

• **Identify Variables and Constants**: Start by knowing what you have to find (variable) and what's given (constants).
• **Set Up an Equation**: Translate the problem statement into mathematical language.
• **Solve the Equation**: Simplify terms, isolate the variable, and calculate its value.
• **Verify the Solution**: Substitute back to ensure the found solution satisfies the original problem constraints.

For the phone call cost example, the linear equation was setup to reflect the different cost rates and found the total minutes by distributing and simplifying coefficients and constants. Verification finally ensures that the equation setup and simplification is accurate and that the step-by-step calculation corresponds to the problem statement.