Problem 300 - Elementary Algebra

Question

In the following exercises, solve. It takes Darline 20 minutes to drive to work in light traffic. To come home, when there is heavy traffic, it takes her 36 minutes. Her speed in light traffic is 24 miles per hour faster than her speed in heavy traffic. Find her speed in light traffic and in heavy traffic.

Step-by-Step Solution

Answer

Speed in heavy traffic: 30 mph. Speed in light traffic: 54 mph.

Step 1: Define Variables

Let the speed in heavy traffic be denoted as \( x \) mph. Her speed in light traffic will then be \( x + 24 \) mph, since she drives 24 mph faster in light traffic.

Step 2: Use the Relationship Between Time, Speed, and Distance

We are given that it takes 20 minutes (which is \( \frac{1}{3} \) hour) in light traffic and 36 minutes (which is \( \frac{3}{5} \) hour) in heavy traffic. We use the formula \( \text{Distance} = \text{Speed} \times \text{Time} \). Therefore, the distance to work is \( d = (x + 24) \times \frac{1}{3} \). The distance from work is \( d = x \times \frac{3}{5} \). Since the distances are the same, we set the equations equal.

Step 3: Set Up the Equation

Set the two expressions for distance equal to each other: \[ (x + 24) \times \frac{1}{3} = x \times \frac{3}{5} \]

Step 4: Clear the Fractions by Multiplying

Multiply both sides by the least common multiple of the denominators (15) to clear the fractions:\[ 15 \left( (x + 24) \times \frac{1}{3} \right) = 15 \left( x \times \frac{3}{5} \right) \] \[ 5(x + 24) = 9x \]

Step 5: Solve the Equation

Distribute and solve for \( x \): \[ 5x + 120 = 9x \] Subtract \( 5x \) from both sides: \[ 120 = 4x \] Divide by 4: \[ x = 30 \]

Step 6: Determine Speeds

Her speed in heavy traffic is 30 mph. Her speed in light traffic is \( 30 + 24 = 54 \) mph.

Key Concepts

solving equations

Solving algebraic equations is a fundamental skill in mathematics. It involves finding the value(s) of the variable(s) in an equation that make the equation true. In our problem, we need to find Darline's speed during her commute in different traffic conditions. We started by defining the speed in heavy traffic as a variable, say, \( x \), and using known relations to set up an equation. By equating the expressions for distance in light and heavy traffic and solving the equation, we determined Darline's speed in both scenarios.
Algebraic solutions often involve several steps:

• Define the variable
• Set up the equation
• Manipulate the equation to isolate the variable
• Solve for the variable

This same approach is applied when the equations are more complex or involve more than one variable.

variables in algebra

In algebra, variables are symbols (like \( x \)) that represent unknown values. They are essential for expressing general mathematical relationships. In our example, we used \( x \) to represent Darline's speed in heavy traffic. This step allows us to create a formula to find the speed under different conditions.

Here are a few tips when working with variables:

• Choose a symbol (like \( x \)) for your variable.
• Use the relationships described in the problem to write equations involving the variable.
• Solve the equation to find the value of the variable.

By representing unknown quantities as variables, we can systematically solve problems through logical steps, making even complex problems manageable.

distance-speed-time relationship

The relationship between distance, speed, and time is a key concept in many algebra problems and can be expressed with the formula: \[ \text{Distance} = \text{Speed} \times \text{Time} \] This formula allows us to calculate one of these aspects if the other two are known. In our example, Darline's journey to work involves different traffic conditions, which affects her speed. By applying this relationship, we created equations for her travel distances in light and heavy traffic.

To recap:

• Identify the time traveled and the speed.
• Multiply these two values to find the distance.
• If the distances are equal, set the expressions for distance equal to solve for speed.

This method systematically breaks down problems involving movement and time, making them easier to solve.

contextual problem-solving

Contextual problems involve real-world scenarios, making abstract algebra concepts more tangible and relatable. In Darline's case, the problem deals with her commute in different traffic conditions. Understanding how to model these scenarios using algebra is a powerful skill. First, identify the quantities involved and their relationships, then translate this information into algebraic equations.

Here’s how to approach contextual problems:

• Understand the situation being described.
• Identify and define the variables representing unknown quantities.
• Write equations based on the problem context.
• Solve the equations to find the solution.

Contextual problem-solving helps bridge the gap between theoretical math and practical application, demonstrating how algebra is used in everyday life.