Problem 271 - Elementary Algebra

Question

Laurie was completing the treasurer's report for her son's Boy Scout troop at the end of the school year. She didn't remember how many boys had paid the \(\$ 15\) full-year registration fee and how many had paid the \(\$ 10\) partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If \(\$ 250\) was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?

Step-by-Step Solution

Answer

14 boys paid the full-year fee and 4 boys paid the partial-year fee.

Step 1: Define Variables

Let the number of boys who paid the full-year fee be denoted as \( x \) and the number who paid the partial-year fee be denoted as \( y \).

Step 2: Write Equations based on the Problem

From the problem, we have two pieces of information:1. The number of boys who paid for a full-year is ten more than those who paid for a partial-year: \( x = y + 10 \).2. The total amount collected from the fees is \( \$250 \). This gives the equation: \( 15x + 10y = 250 \).

Step 3: Substitute the First Equation into the Second Equation

From Step 2, we know that \( x = y + 10 \). Substitute \( x \) in the second equation: \[ 15(y + 10) + 10y = 250 \].

Step 4: Simplify the Equation

Distribute the terms inside the parentheses: \[ 15y + 150 + 10y = 250 \]. Combine like terms to get: \[ 25y + 150 = 250 \].

Step 5: Solve for y

To find \( y \), first subtract 150 from both sides: \[ 25y = 100 \]. Then, divide both sides by 25: \[ y = 4 \]. So, 4 boys paid the partial-year fee.

Step 6: Solve for x

Using the equation \( x = y + 10 \), substitute \( y = 4 \) to find \( x \): \[ x = 4 + 10 = 14 \]. So, 14 boys paid the full-year fee.

Step 7: Verify the Solution

Verify the solution by checking the total amount collected: \( 14 \times 15 + 4 \times 10 = 210 + 40 = 250 \), which is correct.

Key Concepts

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised only to the first power. In the Laurie example, we form two linear equations: one that relates the number of boys who paid for the full-year fee to those who paid for the partial-year fee, and another that accounts for the total revenue collected. This is typical in word problems where relationships and totals are expressed. Generally, linear equations look like:

• Ax + By = C

In the given problem:

• The first linear equation is: x = y + 10
• The second linear equation is: 15x + 10y = 250

Linear equations are foundational in algebra because they describe a straight line when graphed on a coordinate plane.

Problem-Solving

Problem-solving in algebra often involves translating a word problem into mathematical equations. It's crucial to thoroughly understand the problem statement and extract relevant numerical relationships. In our example, we follow these steps:

• Identify and define variables: Let x be the number of boys paying full-year fees and y be those paying partial-year fees.
• Write down the equations based on given relationships and totals.
• Substitute one equation into another to reduce the number of variables.
• Solve for one variable, then use that solution to find the other variable.

After forming and solving the equations, always verify your answer by substituting values back into the original relationships. This ensures the solution satisfies all conditions of the problem.

System of Equations

A system of equations involves solving two or more equations simultaneously. In our problem, we have a system:

• Equation 1: x = y + 10
• Equation 2: 15x + 10y = 250

These equations involve two variables, x and y. To solve, we use a method such as substitution or elimination. In the example:
First, express x in terms of y from Equation 1 and substitute it into Equation 2. This reduces the system to a single equation with one variable. Solutions to systems of equations represent the point(s) where the equations intersect when graphed. Systems of equations are extremely useful in multiple disciplines, including economics, engineering, and social sciences.

Algebraic Substitution

Algebraic substitution is a powerful method for solving systems of equations. The process involves replacing one variable with an equivalent expression derived from another equation. Here's how it's used in the example:

• From Equation 1, we get: x = y + 10
• Substitute this into Equation 2: 15(y + 10) + 10y = 250

This substitution simplifies the problem by forming a single equation with one variable. After substitution, solve for the variable (in this case, y), then use this value in one of the original equations to solve for the other variable (x). This technique is efficient and often used when one equation is already solved for one variable. Proper substitution can simplify complex systems and make the solution more accessible.