Question

Electricity Rates During summer months in 2018 , Omaha Public Power District charged residential customers a monthly service charge of \(\$ 25,\) plus a usage charge of 10.064 per kilowatt-hour \((\mathrm{kWh})\). If one customer's monthly summer bills ranged from a low of \(\$ 140.69\) to a high of \(\$ 231.23\), over what range did usage vary (in \(\mathrm{kWh}\) )?

Short answer

The usage varied between approximately 1149.63 kWh and 2049.29 kWh.

Step-by-step solution

- Understand the Charges

There are two types of charges: a fixed monthly service charge of \(25, and a usage charge of \)0.10064 per kilowatt-hour (kWh).

- Set Up Equations for the Bills

Let the usage in kWh be denoted by u. The total bill B can be calculated as: \[ B = 25 + 0.10064u \]We have two equations corresponding to the minimum and maximum bills:\[ 140.69 = 25 + 0.10064u_{min} \]\[ 231.23 = 25 + 0.10064u_{max} \]

- Solve for Minimum Usage

Isolate \(u_{min}\) in the first equation:\[ 140.69 = 25 + 0.10064u_{min} \]Subtract 25 from both sides:\[ 115.69 = 0.10064u_{min} \]Divide both sides by 0.10064 to get:\[ u_{min} \approx \frac{115.69}{0.10064} \approx 1149.63 \text{ kWh} \]

- Solve for Maximum Usage

Isolate \(u_{max}\) in the second equation:\[ 231.23 = 25 + 0.10064u_{max} \]Subtract 25 from both sides:\[ 206.23 = 0.10064u_{max} \]Divide both sides by 0.10064 to get:\[ u_{max} \approx \frac{206.23}{0.10064} \approx 2049.29 \text{ kWh} \]

- State the Range of Usage

The usage varied between approximately 1149.63 kWh and 2049.29 kWh.

Key concepts

Linear Equations

Linear equations form the backbone of many problem-solving scenarios in mathematics and real life. A linear equation is an equation that makes a straight line when it is graphed. It is usually in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In the electricity usage problem, the linear equation \( B = 25 + 0.10064u \) represents the total bill (B) as a function of the electricity usage (u) in kilowatt-hours. The constant term (25) is the fixed monthly service charge, and the coefficient of \( u \) (0.10064) is the rate per kilowatt-hour. By understanding and setting up the correct linear equation, you can easily find the relationship between variables such as billing cost and electricity usage.

Problem-solving Steps

Solving mathematical problems step-by-step is crucial for understanding and accuracy. Let’s break down the electricity usage problem.

• Step 1 - Understand the Charges: Identify the fixed charge and the variable charge per kilowatt-hour.
• Step 2 - Set Up Equations: Use the general formula for the total bill to create equations for both the minimum and maximum bills observed.
• Step 3 - Solve for Minimum Usage: Isolate the variable representing usage in the equation for the minimum bill and solve it.
• Step 4 - Solve for Maximum Usage: Similarly, isolate and solve the variable for the maximum bill equation.
• Step 5 - State the Range: Finally, compile the results to state the range of the electricity usage.

Following these steps ensures that each part of the problem is addressed methodically, reducing the chance of mistakes.
A structured approach also helps in understanding the problem better and building strong problem-solving skills.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate and solve for a specific variable. In our exercise, we perform algebraic manipulation to determine the minimum and maximum usage values in kilowatt-hours. For example, to solve for \( u_{min} \) from the equation \( 140.69 = 25 + 0.10064u_{min} \):

• Step 1: Subtract 25 from both sides to get \( 115.69 = 0.10064u_{min} \).
• Step 2: Divide both sides by 0.10064 to isolate \( u_{min} \), resulting in \( u_{min} \approx \frac{115.69}{0.10064} \approx 1149.63 \).

This step-by-step manipulation process simplifies the equation and isolates the variable. Doing this correctly is essential for solving linear equations and other algebraic problems accurately. It provides clarity and makes complex problems seem manageable.

Real-world Applications

Understanding linear equations in the context of real-world applications can make the concepts more relatable and easier to grasp. The electricity usage problem is a perfect example. Such calculations are used regularly to determine monthly utility costs based on usage.

• Electricity Bills: Utility companies use similar linear equations to calculate how much a customer should be billed each month.
• Budgeting: Knowing how to compute these costs helps households budget their monthly expenses more accurately.
• Energy Conservation: By understanding the relationship between electricity usage and cost, consumers can make informed decisions to use energy more efficiently.

These applications illustrate why learning linear equations and problem-solving techniques is valuable. They extend beyond the classroom and have practical uses that affect daily life and financial planning directly.