Problem 32 - Intermediate Algebra

Question

Moving Houses. A house mover towed a historic Victorian home 45 miles to locate it on a new site. On his return, without the heavy house in tow, his average speed was 30 mph faster and the trip was 2 hours shorter. How fast did he drive in each direction?

Step-by-Step Solution

Answer

The speeds were 15 mph for towing and 45 mph on the return.

Step 1: Identify Variables

Let's denote the speed of the truck while towing the house as \( x \) mph. This is the speed we wish to find. The speed on the return trip is \( x + 30 \) mph, since the question mentions the return trip was at a speed 30 mph faster.

Step 2: Write Down the Equations for Each Trip

For the trip towing the house, the time is equal to \( \frac{45}{x} \). For the return trip without the house, the time is \( \frac{45}{x+30} \).

Step 3: Set Up Equation Based on Time Difference

The return trip takes 2 hours less than the trip with the house. So, the equation is: \[ \frac{45}{x} - \frac{45}{x+30} = 2 \]

Step 4: Clear Fractions and Solve for x

Multiply through by \( x(x+30) \) to clear the fractions: \[ 45(x + 30) - 45x = 2x(x + 30) \] Expanding and simplifying gives: \[ 45x + 1350 - 45x = 2x^2 + 60x \] \[ 1350 = 2x^2 + 60x \]

Step 5: Rearrange and Factor Quadratic Equation

Rearrange to form a standard quadratic equation: \[ 2x^2 + 60x - 1350 = 0 \] Divide through by 2 to simplify: \[ x^2 + 30x - 675 = 0 \] Next, factor the quadratic equation: \[ (x + 45)(x - 15) = 0 \]

Step 6: Solve the Factored Equation

Setting each factor to zero gives the potential solutions: \( x + 45 = 0 \) or \( x - 15 = 0 \). Thus, \( x = -45 \) or \( x = 15 \). Since a negative speed doesn't make sense in this context, \( x = 15 \) mph.

Step 7: Calculate Return Speed

Now calculate the return speed, which is \( x + 30 = 15 + 30 \), so 45 mph.

Key Concepts

Understanding the Distance-Speed-Time Relationship

When solving problems like moving a house and considering different speeds, it's essential to understand the relationship between distance, speed, and time. This relationship is structured in the formula:

• Distance = Speed × Time
• Time = \( \frac{\text{Distance}}{\text{Speed}} \)
• Speed = \( \frac{\text{Distance}}{\text{Time}} \)

In the context of the problem, the house mover towed the house for a distance of 45 miles. His speed while towing was \( x \) mph, meaning:

• Time towing = \( \frac{45}{x} \)

The return journey was also 45 miles, but his speed was 30 mph faster, resulting in:

• Time returning = \( \frac{45}{x+30} \)

By comparing the times spent on each journey, we can form an equation expressing how much shorter the return trip took.

Techniques for Factoring Quadratic Equations

Factoring quadratic equations is a key skill in algebra used to find unknown values. In this scenario, once the initial equation is set up and simplified, it results in a standard quadratic form:

1. Identify the quadratic equation, like \( 2x^2 + 60x - 1350 = 0 \).
2. Simplify by dividing through by a common factor, if possible, like dividing by 2 here, leading to \( x^2 + 30x - 675 = 0 \).

To factor the equation, look for two numbers that multiply to -675 and add to 30. Once identified, they can be used to write the equation as the product of two binomials:

• (x + 45)(x - 15) = 0

This shows possible solutions for x when each binomial is set to zero.

Mastering Algebraic Problem Solving

Algebraic problem solving often involves translating real-world scenarios into mathematical equations. Start by identifying what you know and what you need to find.

• Use variables to represent unknowns, such as \( x \) for the speed in this problem.
• Formulate equations based on given relationships, like time differences based on speed changes.

Once equations are set, algebraic techniques like factoring or substituting values help solve for unknown variables. In this exercise, solving for \( x \) involved handling a quadratic equation through factoring. This process turns a complex real-world scenario into a solvable mathematical problem, demonstrating how algebra transforms words into equations and solutions.