Problem 72 - Elementary Algebra

Question

Use algebra to solve the following applications. Billy traveled 140 miles to visit his grandmother on the bus and then drove the 140 miles back in a rental car. The bus averages 14 miles per hour slower than the car. If the total time spent traveling was 4.5 hours, then what was the average speed of the bus?

Step-by-Step Solution

Answer

The average speed of the bus was 56 miles per hour.

Step 1: Define Variables

Let the average speed of the rental car be \( x \) miles per hour. Then, the average speed of the bus is \( x - 14 \) miles per hour due to it being 14 miles per hour slower than the car.

Step 2: Set Up the Equation Using Time

The time taken by the bus can be calculated as \( \frac{140}{x - 14} \) hours and the time taken by the car as \( \frac{140}{x} \) hours. Since the total time is 4.5 hours, we set up the equation: \( \frac{140}{x - 14} + \frac{140}{x} = 4.5 \).

Step 3: Solve for \( x \)

To solve the equation \( \frac{140}{x - 14} + \frac{140}{x} = 4.5 \), first find a common denominator, which is \( x(x - 14) \). This allows us to rewrite the equation as \( 140x + 140(x - 14) = 4.5x(x - 14) \). Simplify to get \( 280x - 1960 = 4.5x^2 - 63x \).

Step 4: Rearrange and Simplify the Equation

Rearrange the equation: \( 4.5x^2 - 63x - 280x + 1960 = 0 \), which simplifies to \( 4.5x^2 - 343x + 1960 = 0 \).

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for \( 4.5x^2 - 343x + 1960 = 0 \). Here, \( a = 4.5 \), \( b = -343 \), and \( c = 1960 \). Calculate the discriminant \( b^2 - 4ac = 343^2 - 4(4.5)(1960) \), which simplifies to \( 117649 - 35280 = 82369 \), and the square root is \( 287 \). Then solve for \( x \): \( x = \frac{343 \pm 287}{9} \).

Step 6: Choose the Valid Solution

We get two solutions from the quadratic formula: \( x = \frac{343 + 287}{9} \) and \( x = \frac{343 - 287}{9} \). Calculating both, we find \( x = 70 \) and \( x = 6.222 \). Since the speed of a car cannot be \( 6.222 \) miles per hour to return on time, choose \( x = 70 \).

Step 7: Calculate Bus Speed

The average speed of the bus is \( x - 14 = 70 - 14 = 56 \) miles per hour.

Key Concepts

Understanding Quadratic Equations

Quadratic equations play a crucial role in algebra. They are equations of the form \( ax^2 + bx + c = 0 \). Solving them often involves finding the values of \( x \) that make the equation true. Quadratic equations often arise when dealing with problems involving areas, trajectories, or any situation where there is a squared term.
In our given problem, a quadratic equation was essential to finding the speed of the rental car. We had to rearrange and simplify the equation to fit the standard quadratic form.
One reliable method for solving these equations is by using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula requires computing the discriminant \( b^2 - 4ac \) first. If the discriminant is positive, there are two real solutions; if zero, there is one real solution; and if negative, there are no real solutions. In our problem, the discriminant was positive, yielding two potential solutions for the car's speed.

Calculating Average Speed

Average speed is a fundamental concept in kinematics and algebra. It is calculated as the total distance traveled divided by the total time taken. In the context of our problem, we had to determine the average speed for both the bus and the car.

• The total distance for each vehicle was 140 miles.
• The car had an average speed \( x \) miles per hour.
• The bus, being slower, traveled at \( x - 14 \) miles per hour.

Using these speeds, we set up equations for the time each vehicle took. The sum of these times equaled the total time of 4.5 hours. Understanding these relationships was key to solving for the variables involved.

Systems of Equations in Problem Solving

Systems of equations are sets of equations with multiple variables, where each equation is reliant on the others. They are solved simultaneously to find a common solution. In our busy bus and car scenario, a system of equations provided a framework to connect times, speeds, and distances.

• We established one equation for the bus’s travel time and another for the car’s travel time.
• But instead of multiple equations, we deduced one from combining the travel times.

Though it initially seems like a single equation, this complexity is typical of larger systems of equations. Here, the key was forming one equation from these relational statements concerning times.

Applying Word Problems in Algebra

Word problems like the one presented are challenges in expressing real-world situations using algebraic equations. They test the ability to translate text to math, which is essential for effective problem-solving.
In our problem:

• We interpreted the travel scenario. Recognizing the difference in speed was critical.
• Variables were assigned to distinguish the speeds of the bus and the car.
• A narrative of distance and time turned into useful mathematical expressions.

The main task in such problems is to break down all given information into meaningful equations, solve for the unknowns, and ensure results make sense in context. Practice and familiarity with such problems improve efficiency and accuracy in solving them.