Question

You buy a car with 35,000 miles on it and each year you drive 7000 miles. (a) Write a formula for the mileage at the end of the \(m^{\text {th }}\) year. (b) If the car becomes unusable after 140,000 miles, how many years does it last?

Short answer

Question: Write a formula for the mileage at the end of the mth year and determine how many years it takes for a car with an initial mileage of 35,000 miles to reach 140,000 miles if it is driven 7,000 miles per year. Answer: The formula for mileage at the end of the mth year is M(m) = 35,000 + (7,000 × m). The car would last for 15 years before it becomes unusable.

Step-by-step solution

Setup the equation

To find the mileage at the end of the \(m^{\text {th }}\) year, we'll add the initial mileage of the car to the total miles driven up till that year.

Write the mileage formula

Let \(M(m)\) represent the mileage at the end of the \(m^{\text {th }}\) year. The mileage formula would be: \(M(m) = 35000 + (7000 \times m)\) Now, let's move to part (b). (b) If the car becomes unusable after 140,000 miles, how many years does it last?

Express the problem as an equation

In this part, we need to find the number of years it takes for the car to reach 140,000 miles. So, we will set the mileage formula equal to 140,000 miles. \(140000 = 35000 + (7000 \times m)\)

Solve the equation

Now, let's solve the equation for 'm': \(140000 - 35000 = 7000 \times m\) \(105000 = 7000 \times m\) \(m = \frac{105000}{7000}\) \(m = 15\) The car would last for 15 years before it becomes unusable.

Key concepts

Linear Functions

Understanding linear functions is essential when working with problems like determining mileage over time. A linear function can be thought of as a straight line on a graph. It increases or decreases at a constant rate. This kind of function is often expressed in the form of a formula, such as \(y = mx + c\), where:

• \(y\) is the dependent variable (in this case, the mileage).
• \(x\) represents the independent variable (here, it’s the years).
• \(m\) is the slope of the line, revealing how much \(y\) changes for each unit increase in \(x\) (the miles driven per year).
• \(c\) is the y-intercept, which indicates the value of \(y\) when \(x\) is zero (the car’s initial mileage).

In our example, the linear function for determining mileage every year is given by:\[M(m) = 35000 + 7000m\]This function clearly defines how the car’s mileage increases by 7000 for every additional year of driving, starting from 35,000 miles.

Problem Solving

Problem-solving in algebra involves breaking down a problem into smaller, more manageable steps. In this exercise, we tackle two related questions systematically. First, we wrote the equation for the car's mileage at the end of any given year using a linear function. This gives us a mathematical model to predict future mileage. Simplifying the process ensures clarity and reduces the chance of errors.The second problem takes our function and sets it to a given output—when the car reaches 140,000 miles. We translate this real-world desire into an algebraic equation:\[140000 = 35000 + 7000m\]Solving for \(m\) involves isolating the variable by performing straightforward arithmetic operations. We subtract the initial mileage from the total desired mileage to find the total miles driven:\[105000 = 7000m\]Then, we divide by the miles driven per year to determine the number of years:\[m = \frac{105000}{7000} = 15\]Successfully solving this gives us a practical solution.

Real-World Applications

Algebraic equations, especially linear functions, find myriad applications in daily life. They offer solutions for scenarios where change happens at a constant rate. Consider how knowing the mileage of your car helps in planning for replacements or maintenance. This example shows how crucial it is to predict when a car might become unusable based on mileage. While the car mileage example illustrates a scenario many can relate to, linear equations apply broadly. For businesses, they might represent costs or inventory over time. For finance, they could model savings growth. Anytime someone needs to make decisions based on a progressive change—whether it be accumulating miles, increasing costs, or measuring physical growth—linear functions become invaluable. Through problems like these, students learn the importance of math in real-world decision-making. Understanding such applications heightens the value of algebra as a tool for predicting and managing future states.