Problem 44 - Algebra 2

Question

The Friendly Fix-It Company charges a base fee of \(\$ 45\) for any in-home repair. In addition, the technician charges \(\$ 30\) per hour. Write an equation for the cost \(c\) of an in-home repair of \(h\) hours.

Step-by-Step Solution

Answer

The cost equation is \( c = 45 + 30h \).

Step 1: Identify Costs

First, identify the components of the cost. The Friendly Fix-It Company charges a base fee of \( \\(45 \), regardless of the hours worked. In addition to the base fee, they charge \( \\)30 \) per hour for the technician's time.

Step 2: Understand the Variables

We need to write an equation for the cost of a repair. Let \( c \) represent the total cost in dollars, and \( h \) represent the number of hours the technician works.

Step 3: Combine the Costs

The total cost \( c \) is the sum of the base fee and the cost for the hours worked. The equation representing this will combine the constant base fee and the hourly charge.

Step 4: Construct the Equation

The equation for the cost is constructed by adding the base fee to the product of the hourly rate and the number of hours: \( c = 45 + 30h \).

Key Concepts

Algebraic Expressions

Algebraic expressions are a fundamental concept in algebra involving variables, constants, and arithmetic operations. In the context of the Friendly Fix-It Company problem, we create an algebraic expression to represent the cost of in-home repairs.
The expression combines fixed costs with variable costs depending on the number of hours worked. Here, the base fee is a constant value, \( \\(45 \), which is always part of the expression irrespective of the hours worked.
Meanwhile, \( h \), the variable, stands for the number of hours worked, and it is associated with the hourly rate \( \\)30 \).
So, combining these gives the expression \( 30h \), which changes with different instances of \( h \), reflecting the variable part of the cost. The complete algebraic expression, \( c = 45 + 30h \), represents the total cost of the repair.

Cost Analysis

Cost analysis is the process of breaking down and examining the costs related to a particular service or product. In this case, we analyze the cost of an in-home repair by the Friendly Fix-It Company.
We begin by identifying two primary components: a fixed base fee of \( \\(45 \), and a variable cost of \( \\)30 \) per hour of service.

• **Fixed Base Fee:** This fee is constant, ensuring that the company covers initial expenses needed for any repair service.
• **Variable Hourly Rate:** The fee for each hour provides sufficient flexibility to account for work duration.

By summing the base fee with the hourly fee calculated as \( 30h \), we perform a complete cost analysis to determine the total expense, making it easier for customers to anticipate costs based on service time.

Problem Solving in Algebra

Problem solving in algebra involves translating real-world scenarios into mathematical models using equations. This requires understanding the problem, identifying variables, and formulating expressions.
In our example, we started by comprehending the problem—a repair cost calculation that depends on a fixed fee and a variable number of hours worked.
By identifying the variables: \( c \) (the total cost) and \( h \) (the number of hours), we constructed an equation that accurately represents the situation: \( c = 45 + 30h \).
This process of creating and manipulating equations is crucial in algebra, enabling us to solve practical problems, whether it's a straightforward repair cost calculation or more complex analyses.
This approach helps in making informed decisions and understanding underlying patterns in everyday situations.