Question

Sewer Bills The village of Oak Lawn charges homeowners \(\$ 23.55\) per quarter- year for sewer usage, plus \(\$ 0.40\) per 1000 gallons of water metered. In 2018 , one homeowner's quarterly bill ranged from a high of \(\$ 36.75\) to a low of \(\$ 30.35 .\) Over what range did metered water usage vary?

Short answer

The metered water usage varied from 17 to 33 thousand gallons.

Step-by-step solution

Understand the bill calculation formula

The total quarterly sewer bill is calculated using the formula: Total Bill = Quarterly Charge + (Cost per 1000 gallons of water * Number of 1000 gallons). Here, the quarterly charge is \(23.55, and the cost per 1000 gallons is \)0.40.

Set up the equations for the given bills

Given the highest bill (36.75) and the lowest bill (30.35), set up two equations: 36.75 = 23.55 + 0.40x (high usage), 30.35 = 23.55 + 0.40y (low usage), where x and y represent the number of 1000 gallons of water.

Solve for high water usage (x)

To find x, rearrange the high usage equation: 36.75 = 23.55 + 0.40x. Subtract 23.55 from both sides: 36.75 - 23.55 = 0.40x. Simplify: 13.20 = 0.40x. Then, solve for x by dividing both sides by 0.40: x = 13.20 / 0.40 = 33.

Solve for low water usage (y)

To find y, rearrange the low usage equation: 30.35 = 23.55 + 0.40y. Subtract 23.55 from both sides: 30.35 - 23.55 = 0.40y. Simplify: 6.80 = 0.40y. Then, solve for y by dividing both sides by 0.40: y = 6.80 / 0.40 = 17.

Determine the range of water usage

The range of metered water usage varies from low usage of 17 thousand gallons to high usage of 33 thousand gallons.

Key concepts

Linear Equations

Linear equations are equations that make a straight line when graphed. They take the general form of \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable we solve for.

In our sewer bill problem, we have two linear equations:

• 36.75 = 23.55 + 0.40x (for high water usage)
• 30.35 = 23.55 + 0.40y (for low water usage)

These equations help us understand and calculate how much water was used based on the total bill. We use these equations to find the values of \( x \) and \( y \), which represent water usage in units of thousands of gallons.

Solving for Variables

To solve for variables in a linear equation, we need to isolate the variable on one side of the equation. Here's how we solved for \( x \) and \( y \) in the example:

For high water usage (x):

• We start with 36.75 = 23.55 + 0.40x
• Subtract 23.55 from both sides: 36.75 - 23.55 = 0.40x
• This simplifies to: 13.20 = 0.40x
• Finally, divide by 0.40: \( x = \frac{13.20}{0.40} = 33 \)

For low water usage (y):

• We start with 30.35 = 23.55 + 0.40y
• Subtract 23.55 from both sides: 30.35 - 23.55 = 0.40y
• This simplifies to: 6.80 = 0.40y
• Finally, divide by 0.40: \( y = \frac{6.80}{0.40} = 17 \)

Through these steps, we isolate the variables, making it easier to find the values of \( x \) and \( y \).

Range Calculation

Calculating the range means finding the difference between the highest and lowest values in a set. In our case, we want the range of metered water usage.

We found that the highest water usage \( x \) is 33 thousand gallons and the lowest water usage \( y \) is 17 thousand gallons.

To determine the range, we simply subtract the lowest value from the highest value: \( 33 - 17 = 16 \).
This tells us that the range of metered water usage varies by 16 thousand gallons. Now we know that the homeowner's water consumption ranged from 17 thousand to 33 thousand gallons.

Algebraic Formulas

Algebraic formulas are equations used to represent relationships between different quantities. They often involve variables that we solve for.

In our problem, we used an algebraic formula to calculate the total sewer bill:

\[ \text{Total Bill} = \text{Quarterly Charge} + (\text{Cost per 1000 gallons} \times \text{Number of 1000 gallons}) \]

For the quarterly charge of \( 23.55 \) and the cost per 1000 gallons being \( 0.40 \), our formulas were:

• High usage bill: 36.75 = 23.55 + 0.40x
• Low usage bill: 30.35 = 23.55 + 0.40y

By substituting the known values into these formulas, we were able to set up our equations and solve for the unknown water usage amounts. Using algebraic formulas helps break down complex real-world problems into structured steps.