Problem 45 - Introductory Algebra for College Students

Question

An automobile repair shop charged a customer 448 dollars listing 63 dollars for parts and the remainder for labor. If the cost of labor is 35 dollars per hour, how many hours of labor did it take to repair the car?

Step-by-Step Solution

Answer

So, it took 11 hours to repair the car.

Step 1: Calculate the Total Labor Cost

First, determine the cost spent on labor. This can be found by subtracting the cost of parts from the total cost. So, the total labor cost would be 448 dollars (total cost) - 63 dollars (cost of parts). Therefore, the total labor cost = 385 dollars.

Step 2: Calculate the Number of Hours

To find out how many hours this labor cost corresponds to, we need to divide the total labor cost by the cost of labor per hour. Mathematically, that's 385 dollars (total labor cost) / 35 dollars (cost per hour). This results in 11 hours.

Key Concepts

Linear Equations

Linear equations form the foundation of algebra and are ubiquitous in both textbook examples and real-world applications. They are characterized by variables that are raised to the power of one, resulting in a straight-line graph when plotted. In our exercise, the situation can be modeled by the linear equation \( L = 448 - 63 \text{, where } L \text{ represents the total labor cost.} \) This equation simplifies the relationship between the total cost, the cost of parts, and the labor cost. The cost of labor per hour, 35 dollars per hour, serves as the constant slope in this context.

The beauty of a linear equation lies in its simplicity for problem-solving. By performing basic arithmetic operations—subtraction in step one and division in step two—we've moved from a word problem to a numerical solution, clearly illustrating the concept of a linear equation in a practical scenario.

Problem Solving in Algebra

The process of problem solving in algebra involves translating a written scenario into a mathematical context, and then applying algebraic techniques to arrive at a solution. It's a critical skill in both academics and in the real world. To tackle the given problem efficiently, we first identified the relevant quantities: total cost, cost of parts, and labor cost. The next step was to express these relationships using algebraic expressions. For instance, to find the total labor cost, we crafted the equation \( 448 - 63 = L, \) and to find the number of hours of labor, \( L / 35 = H \text{, where } H \text{ is the number hours of labor} \).

Through problem solving, we learn how to break down complex real-world problems into manageable algebraic expressions, enhancing our analytical skills and providing a structured pathway to solutions.

Real-world Algebra Applications

Algebra is not just an abstract concept but also a handy tool that helps us solve real-world problems. In our example, the real-world application is evident in determining the labor hours in an automobile repair scenario. By applying algebraic methods to calculate labor costs, we translate a typical question encountered by customers and business owners into a solvable algebraic problem. This example demonstrates the utility of algebra in everyday life—from budgeting household expenses to making business decisions.

Incorporating real-world examples into algebraic problem solving helps students see the relevance of mathematics in their daily lives. It also enhances their understanding and retention of mathematical concepts, as they can directly relate the abstract equations to tangible outcomes or decisions that have practical implications.