Problem 58 - A Graphical Approach to College Algebra

Question

Speeding Fines Suppose that speeding fines are determined by \(y=10(x-65)+50, x>65,\) where \(y\) is the cost in dollars of the fine if a person is caught driving \(x\) miles per hour. (a) How much is the fine for driving 76 mph? (b) While balancing the checkbook, Johnny found a check that his wife Gwen had written to the Department of Motor Vehicles for a speeding fine. The check was written for \(\$ 100 .\) How fast was Gwen driving? (c) At what whole-number speed are tickets first given? (d) For what speeds is the fine greater than \(\$ 200 ?\)

Step-by-Step Solution

Answer

(a) $160, (b) 70 mph, (c) 66 mph, (d) speeds > 80 mph.

Step 1: Calculating the fine for 76 mph

Using the equation for speeding fines, substitute \( x = 76 \) into \( y = 10(x - 65) + 50 \).\[ y = 10(76 - 65) + 50 = 10(11) + 50 = 110 + 50 = 160\]Thus, the fine for driving 76 mph is \( \$160 \).

Step 2: Determining the speed for a $100 fine

Given \( y = 100 \), substitute into the equation and solve for \( x \):\[100 = 10(x - 65) + 50\]Simplify and solve:\[100 = 10x - 650 + 50 \100 = 10x - 600 \10x = 700 \x = 70\]Therefore, Gwen was driving at 70 mph.

Step 3: Determining the first speed that results in a ticket

Tickets are given for speeds where \( x > 65 \). Since \( x \) must be greater than 65, the smallest whole number speed is 66 mph.

Step 4: Determining speeds for a fine greater than $200

Set the inequality for the fine greater than \( \\(200 \):\[10(x - 65) + 50 > 200\]Simplify the inequality:\[10x - 650 + 50 > 200 \10x - 600 > 200 \10x > 800 \x > 80\]So, the fine is greater than \( \\)200 \) for speeds greater than 80 mph.

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. In our problem, the formula for speeding fines is an algebraic expression: \( y = 10(x - 65) + 50 \). This equation is used to determine the cost of a fine based on speed, \( x \), in miles per hour. Let's break down this expression step by step.

First, \( 10(x - 65) \) is a term that calculates a base fine determined by how much the speed \( x \) exceeds 65 mph. The \( 10 \) can be seen as a multiplier, indicating the cost per mile over the speed limit. Secondly, the constant \( 50 \) is added, which represents a fixed part of the fine everyone gets on top of their over-speeding penalty.

Expressions like this use different elements:

• **Variables:** These are symbols that represent unknown values. Here, \( x \) is the variable.
• **Coefficients:** These are numbers placed before the variable, which multiply the variable. The number 10 is a coefficient.
• **Constants:** Fixed numbers added or subtracted, like 50 in our expression.

Understanding algebraic expressions helps you to form equations accurately and solve them effectively, as seen when determining fines based on varying speeds.

Solving Inequalities

Solving inequalities involves finding the values of a variable that make the inequality true. In the problem, one key task was to determine the speeds that result in a fine greater than \(200. This is expressed as the inequality \( 10(x - 65) + 50 > 200 \). Let's analyze the steps to solve this inequality.

First, simplify the expression by breaking down the equation and removing constants:

• Start by distributing the 10: \( 10x - 650 + 50 \).
• Then simplify the right-hand side to reveal: \( 10x - 600 > 200 \).

Next, isolate \( x \) to find the solution:

• Add 600 to both sides to remove the constant: \( 10x > 800 \).
• Finally, divide by 10 to solve for \( x \): \( x > 80 \).

The solution to this inequality means that if a person's speed \( x \) is greater than 80 mph, the fine will exceed \)200. Tackling inequalities like this requires careful attention to operations, similar to solving equations, but with the consideration that the direction of inequalities can change if multiplied or divided by negative numbers.

Word Problems

Word problems require careful translation of real-world situations into mathematical expressions or equations. A common challenge is identifying relevant information and forming equations accordingly. The given exercise presents several real-world scenarios: calculating fines based on different speeds, or deducing a speed from a given fine amount.

For example, when asked how fast Gwen was driving after writing a \(100 check for a fine, the solution involved:

• Identifying the known element (the fine amount, \)100) and the unknown (speed, \( x \)).
• Substituting \( y = 100 \) into the equation and solving for \( x \).

This type of problem teaches how to decode information and transcribe it into a solvable equation, critical for solving both everyday questions and complex mathematical challenges.

In solving these problems, consider:

• Clearly defining what each variable represents.
• Organizing information logically before inserting it into equations.
• Double-checking calculations at each step to avoid errors.

Mastering word problems empowers students to apply mathematical methods to real-life dilemmas effectively.