Problem 56 - Algebra A Combined Function

Question

Mark Keaton's workout consists of jogging for 3 miles, and then riding his bike for 5 miles at a speed 4 miles per hour faster than he jogs. If his total workout time is 1 hour, find his jogging speed and his biking speed.

Step-by-Step Solution

Answer

Jogging speed is 6 mph, and biking speed is 10 mph.

Step 1: Understand the Problem

We are given two activities: jogging for 3 miles and biking for 5 miles. The biking speed is 4 mph faster than the jogging speed. We need to find the jogging speed and biking speed, knowing that the total workout time is 1 hour.

Step 2: Set Up Variables

Let \( x \) be the jogging speed in miles per hour. Therefore, the biking speed is \( x + 4 \) miles per hour.

Step 3: Use Time Formula for Each Activity

Recall the time formula: \( \, \text{time} = \frac{\text{distance}}{\text{speed}} \, \). For jogging, \( \text{time} = \frac{3}{x} \, \), and for biking, \( \text{time} = \frac{5}{x+4} \, \).

Step 4: Set Up the Total Time Equation

The total workout time is 1 hour. Therefore, \( \frac{3}{x} + \frac{5}{x+4} = 1 \).

Step 5: Solve the Equation

Multiply the entire equation by \( x(x+4) \) to eliminate the fractions: \[ 3(x+4) + 5x = x(x+4). \] Simplify and solve the equation: \[ 3x + 12 + 5x = x^2 + 4x, \] leading to \[ x^2 - 4x - 12 = 0. \]

Step 6: Factor the Quadratic Equation

Factor the quadratic equation \( x^2 - 4x - 12 = 0 \) as \( (x - 6)(x + 2) = 0 \).

Step 7: Find Solutions for the Quadratic

This gives solutions \( x = 6 \) and \( x = -2 \). However, speed cannot be negative, so \( x = 6 \).

Step 8: Determine Biking Speed

The jogging speed is \( x = 6 \) mph. The biking speed is \( x + 4 = 10 \) mph.

Key Concepts

Jogging Speed

Jogging speed plays a crucial role in solving algebra word problems. Here, jogging speed refers to the speed at which Mark Keaton jogs during his workout. The problem involves setting up variables to define this speed for calculation purposes.

When approaching a word problem involving speed, distance, and time, we often need to:

• Define the jogging speed as an initial unknown variable, typically represented as \( x \).
• Use the basic time formula, \( ext{time} = \frac{distance}{speed} \), to express the time taken to jog a given distance.

Since Mark jogs for 3 miles, the expression for jogging time is \( \frac{3}{x} \), where \( x \) is the jogging speed in miles per hour.

This foundational setup is essential for linking the conditions of the problem, especially when combined with other activities such as biking. Properly assigning and manipulating these variables leads to finding numeric solutions that meet all given criteria.

Biking Speed

Biking speed is another vital component of this problem. It is directly related to the jogging speed. In the context of the exercise, Mark bikes at a speed that is 4 mph faster than his jogging speed.

To incorporate biking into the algebraic expression, consider the following:

• If jogging speed is \( x \), then biking speed becomes \( x + 4 \) mph, due to its 4 mph increase over jogging speed.
• Apply the time formula for biking, \( \frac{5}{x+4} \), representing the time needed to cover 5 miles.

This step is crucial for establishing the relationship between jogging and biking speeds. Combining these two expressions allows you to form an equation grounded in time, bridging the connection between these activities under the constraint of Mark's total workout duration of 1 hour.

Quadratic Equations

Quadratic equations frequently appear in problems involving multiple relationships between distances, speeds, and times. In this exercise, solving the problem naturally involves forming a quadratic equation.

The process includes:

• Setting up a total time equation: \( \frac{3}{x} + \frac{5}{x+4} = 1 \).
• Clearing denominators by multiplying through by a common expression, in this case, \( x(x+4) \).
• Simplifying and rearranging terms to form a standard quadratic equation: \( x^2 - 4x - 12 = 0 \).
• Factoring into \( (x - 6)(x + 2) = 0 \) to find potential solutions for \( x \).

The solution \( x = 6 \) represents the jogging speed because, logically, speed cannot be negative, thus eliminating \( x = -2 \).

This concept underscores how quadratic equations provide powerful tools in algebra to solve complex real-world problems, revealing values that align with given contexts. Proper understanding and application of these methods are key to accurately solving for unknown variables.