Problem 97 - Intermediate Algebra

Question

Cost of Electricity. The cost of electricity in a city is given by the formula \(C=0.07 n+6.50,\) where \(C\) is the cost in dollars and \(n\) is the number of kilowatt hours used. Solve for \(n .\) Then find the number of kilowatt hours used each month by a homeowner whose checks to pay the monthly clectric bills are: \(\$ 49.97, \$ 76.50,\) and \(\$ 125\)

Step-by-Step Solution

Answer

Kilowatt hours are approximately 621, 1000, and 1693 for the bills $49.97, $76.50, and $125, respectively.

Step 1: Understand the Formula

The formula provided for the cost of electricity is \( C = 0.07n + 6.50 \), where \( C \) is the total cost and \( n \) is the number of kilowatt hours used. Our goal is to solve this formula for \( n \).

Step 2: Rearrange the Formula to Solve for n

To solve for \( n \), we need to isolate it on one side of the equation. Start by subtracting 6.50 from both sides of the equation: \( C - 6.50 = 0.07n \). Then divide both sides by 0.07 to get \( n \) alone: \( n = \frac{C - 6.50}{0.07} \).

Step 3: Calculate n for Each Bill

Using the formula \( n = \frac{C - 6.50}{0.07} \), plug in each bill amount for \( C \) to find \( n \).

Step 3.a: Calculate n for $49.97

\( n = \frac{49.97 - 6.50}{0.07} = \frac{43.47}{0.07} \approx 620.999 \). Therefore, \( n \approx 621 \) kilowatt hours.

Step 3.b: Calculate n for $76.50

\( n = \frac{76.50 - 6.50}{0.07} = \frac{70}{0.07} = 1000 \). Therefore, \( n = 1000 \) kilowatt hours.

Step 3.c: Calculate n for $125

\( n = \frac{125 - 6.50}{0.07} = \frac{118.5}{0.07} \approx 1692.857 \). Therefore, \( n \approx 1693 \) kilowatt hours.

Key Concepts

Cost Calculation

When you look at the cost of electricity, it's crucial to understand how the cost relates to usage. In this example, the cost is determined by a specific formula:

• The basic charge or fixed cost is \(6.50.
• In addition to this, every kilowatt-hour used costs \)0.07.

These numbers combined form the formula: \[ C = 0.07n + 6.50 \]where \( C \) is the cost in dollars, and \( n \) is the number of kilowatt hours used.
Each part of the formula contributes to the total electricity cost a homeowner pays. The fixed part, $6.50, is constant regardless of usage, while the 0.07\( n \) is a variable cost depending on the total energy consumption.

Solving Equations

An essential part of mathematics involves solving equations. Here, we needed to find a way to figure out how many kilowatt hours were used, given the cost. Solving for \( n \) requires rearranging the formula.
To start this process:

• We first subtract $6.50 from both sides of the equation, which gets rid of the fixed cost on the right side. So, it becomes:

\[C - 6.50 = 0.07n\]

• Next, to isolate \( n \), we divide both sides by 0.07:

\[n = \frac{C - 6.50}{0.07}\]
After these steps, we have a clear formula to find \( n \), which tells us how many kilowatt hours were consumed based on any given cost.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to make them easier to work with. This method helps us to isolate variables such as \( n \) in equations, as seen in the electricity cost formula.
Starting with the basic equation \( C = 0.07n + 6.50 \):

• Identify the terms to move. Here, we moved the 6.50 by subtracting it from both sides. This step is crucial as it simplifies the equation and focuses on the variable \( n \).
• Next, simplify even further by dividing the remaining equation by 0.07 to completely isolate \( n \).

These steps showcase how algebraic manipulation is a fundamental tool in math, making it possible to solve real-world problems by rearranging and solving equations.
This approach isn't limited only to cost calculation but is applicable across different areas where understanding relationships between variables is necessary.