Problem 87 - Introductory and Intermediate Algebra for College Students 4th

Question

In Massachusetts, speeding fines are determined by the formula $$F=10(x-65)+50,$$ where \(F\) is the cost, in dollars, of the fine if a person is caught driving \(x\) miles per hour. Use this formula to solve. If a fine comes to \(\$ 250,\) how fast was that person driving?

Step-by-Step Solution

Answer

The driver was driving at 85 miles per hour.

Step 1: Substituting the known values

The first step is to substitute the value of the fine \(F = 250\) dollars into the given formula \(F = 10(x-65)+50\). Doing so gives us: \(250 = 10(x-65)+50\).

Step 2: Solving for 'x'

The second step involves simplification and rearrangement of the equation so as to isolate 'x'. This is an important step as it allows us to find the corresponding value of 'x' i.e., speed in this context. Let's start by deducting 50 from both sides of the equation in order to remove the '+50' on the right side. This leaves us with \(200 = 10(x-65)\). The next step is to divide both sides of the equation by 10 so that 'x' is left alone on the one side of the equation. This simplification gives us: \[20 = x - 65\]. Lastly, we add 65 to both sides of the equation to isolate 'x'. This results in: \(x = 85\).

Key Concepts

Solving Linear Equations

Understanding how to solve linear equations is a foundational skill in algebra that allows us to find unknown values. The equation presented in the exercise, \(F=10(x-65)+50\), is a linear equation because it can be simplified to a form where \(x\) is raised to the power of one.

In practice, solving linear equations involves a series of steps aimed at isolating the variable in question. For instance, in our example where a person receives a fine of \$250, we substitute the fine into our equation, giving us \(250 = 10(x-65)+50\). The goal is to solve for \(x\), which represents the speed of the driver. We use inverse operations to undo the addition, subtraction, multiplication, and division present in the equation, effectively 'peeling back layers' until \(x\) is by itself. Here's a breakdown of these steps:

• First subtract 50 from both sides to get \(200 = 10(x - 65)\).
• Divide both sides by 10 to simplify further, which gives \(20 = x - 65\).
• Finally, add 65 to both sides to find the driver's speed, resulting in \(x = 85\) miles per hour.

By following these orderly steps, we ensure that the equation remains balanced, and we're able to find the accurate value of the unknown variable.

Mathematical Modeling

Mathematical modeling is the art of translating real-world scenarios into mathematical terms to make them easier to analyze and solve. It plays a significant role in numerous fields, from physics to finance. In our exercise, the mathematical model is the formula \(F=10(x-65)+50\) that represents the fine for speeding in Massachusetts.

This model reflects a direct relationship between the speed \(x\) and the fine \(F\). The structure of the equation—particularly the \(10(x-65)\) part—shows that for every mile per hour over 65, the fine increases by $10. This linear model provides clarity and predictability for both law enforcers and drivers regarding the consequences of speeding. When a fine is issued, the model allows us to work backwards—to deconstruct the situation and find out the speed at which the person was driving, as we executed in the 'Solving Linear Equations' section.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operation symbols that represent a specific quantity. In the context of our example, the expression \(10(x-65)+50\) is an algebraic expression that calculates the fine \(F\), where \(x\) is a variable representing the driver's speed.

An important facet of working with algebraic expressions is the ability to manipulate and simplify them. Here, we apply distributive, associative, and commutative properties to combine like terms and simplify the equations. It's these algebraic principles that empower us to transform the model \(F=10(x-65)+50\) into a simpler form—ultimately allowing us to solve for \(x\). Recognizing and understanding these expressions is crucial as they are the building blocks for more complex mathematical concepts.