Problem 32 - Algebra A Combined Function

Question

The rate of growth of participation of women in sports has been increasing since Title IX was enacted in the 1970 s. In 2008 , the number of women participating in swimming was 1.1 million less than twice the number participating in running. If the total number of women participating in these two sports was 50.5 million, find the number of participants in each sport. (Source: Sporting Goods Association of America, 2009 report)

Step-by-Step Solution

Answer

17.2 million in running; 33.3 million in swimming.

Step 1: Define Variables

Let \( r \) represent the number of women participating in running, and \( s \) represent the number of women participating in swimming.

Step 2: Set Up Equations Based on Problem Statement

We are given two conditions:1. The number of women participating in swimming is 1.1 million less than twice those participating in running: \[ s = 2r - 1.1 \]2. The total number of participants in both sports is 50.5 million:\[ r + s = 50.5 \]

Step 3: Substitute to Solve

Substitute the expression for \( s \) from the first equation into the second equation:\[ r + (2r - 1.1) = 50.5 \]

Step 4: Simplify the Equation

Combine like terms:\[ 3r - 1.1 = 50.5 \]

Step 5: Isolate the Variable

Add 1.1 to both sides to isolate terms with \( r \):\[ 3r = 51.6 \]

Step 6: Solve for \( r \)

Divide both sides by 3 to find \( r \):\[ r = \frac{51.6}{3} = 17.2 \]Thus, 17.2 million women participated in running.

Step 7: Solve for \( s \)

Use the expression for \( s \):\[ s = 2r - 1.1 = 2(17.2) - 1.1 = 34.4 - 1.1 = 33.3 \]Thus, 33.3 million women participated in swimming.

Step 8: Verify the Solution

Check the total:\[ r + s = 17.2 + 33.3 = 50.5 \]Confirm other condition is met:\[ s = 2r - 1.1 = 34.4 - 1.1 = 33.3 \]Both conditions are satisfied, confirming the solution is correct.

Key Concepts

Linear Equations

Linear equations are fundamental in solving various real-world problems, including those found in Algebra word problems. A linear equation is an equation that represents a straight line when graphed. It involves variables with a maximum exponent of one. For example, in this problem, we have linear equations formed from the problem statement:

• The equation for swimming participants: \( s = 2r - 1.1 \)
• The equation for the total number of participants: \( r + s = 50.5 \)

Both equations are linear and involve the variables \( r \) and \( s \). Linear equations help in establishing relationships between different variables, allowing us to find unknown quantities through mathematical operations like addition, subtraction, multiplication, and division. Understanding how to interpret and set up linear equations from word problems is crucial for successfully solving them.

Variable Definition

In algebra, defining variables is an essential step in problem-solving. A variable is a symbol, usually a letter, that represents an unknown value in a mathematical expression or equation. By defining variables, we can transform word problems into algebraic equations, which are easier to manipulate mathematically. In this problem, we define:

• \( r \) to represent the number of women participating in running,
• \( s \) to represent the number of women participating in swimming.

With these variables, we convert the problem statement into mathematical expressions. Variable definition is key because it provides clarity and direction when setting up equations. Clear variable definitions ensure that each part of the word problem is accurately represented, which is necessary for a correct solution.

Substitution Method

The substitution method is a common technique used to solve systems of equations. It involves solving one of the equations for a variable and then substituting this expression into the other equation. In our exercise, we first solve the first equation \( s = 2r - 1.1 \) for \( s \), and then substitute it into the second equation \( r + s = 50.5 \). This process involves:

• Rewriting the second equation: \( r + (2r - 1.1) = 50.5 \)
• Simplifying it to \( 3r - 1.1 = 50.5 \)
• Solving for \( r \).

By using the substitution method, we simplify the number of variables and our system becomes a single equation with one variable, making it much easier to solve. It's an effective strategy for solving simultaneous equations when one of the equations is easily expressed in terms of a single variable.

Problem Solving Steps

Solving word problems in algebra involves a series of structured steps. Following them systematically helps ensure problems are understood and solved correctly. Here's a recap of the steps used in this exercise:

• Define Variables: Identify what each variable represents. For example, \( r \) for running participants, \( s \) for swimming participants.
• Set Up Equations: Use the relationships described in the problem to formulate equations. Here, we derive \( s = 2r - 1.1 \) and \( r + s = 50.5 \).
• Use the Substitution Method: Substitute the expression involving one variable into the other equation, simplifying the problem to one equation with one variable.
• Solve the Equation: Carry out mathematical operations to solve for the unknowns. In this example, solve \( 3r = 51.6 \) to find \( r \), then use it to find \( s \).
• Verify the Solution: Plug the values back into the original equations to ensure that all conditions of the problem are satisfied.

These steps provide a roadmap to approach and solve algebraic word problems effectively and confidently.