Problem 63 - Algebra for College Students

Question

Set up an equation and solve each problem. A group of students agreed that each would contribute the same amount to buy their favorite teacher an \(\$ 80\) birthday gift. At the last minute, 2 of the students decided not to chip in. This increased the amount that the remaining students had to pay by \(\$ 2\) per student. How many students actually contributed to the gift?

Step-by-Step Solution

Answer

8 students actually contributed to the gift.

Step 1: Define Variables

Let's define the variables we will use in this problem. Let \( x \) represent the original number of students who agreed to contribute to the gift. Let \( y \) be the amount of money each originally agreed to contribute.

Step 2: Set Up Original Contribution Equation

The total amount needed for the gift is \( \$80 \). Therefore, we have the equation \( x \times y = 80 \), where \( x \) is the original number of students and \( y \) is the amount each contributes.

Step 3: New Contribution Per Student Equation

After 2 students dropped out, the remaining students had to pay an additional \( \$2 \) each. Thus, the new number of students is \( x - 2 \), and each has to pay \( y + 2 \). The equation becomes \((x - 2) \times (y + 2) = 80\).

Step 4: Solve the System of Equations

With the two equations: 1. \( xy = 80 \) 2. \( (x - 2)(y + 2) = 80 \), substitute \( y \) from the first equation into the second: \( y = \frac{80}{x} \) Substitute into the second equation: \( (x - 2)\left(\frac{80}{x} + 2\right) = 80 \). Simplify:\( 80 - \frac{160}{x} + 2x - 4 = 80 \). Solve:\( 2x - \frac{160}{x} = 4 \). Multiply through by \( x \):\( 2x^2 - 160 = 4x \). Rearrange:\( 2x^2 - 4x - 160 = 0 \). Divide by 2:\( x^2 - 2x - 80 = 0 \). Solve using the quadratic formula:\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1, b=-2, c=-80 \).\( x = \frac{2 \pm \sqrt{4 + 320}}{2} \),\( x = \frac{2 \pm \sqrt{324}}{2} \),\( x = \frac{2 \pm 18}{2} \). Solutions are \( x = 10 \) or \( x = -8 \). Since we cannot have negative students, \( x = 10 \).

Step 5: Calculate Remaining Contributing Students

If originally 10 students planned to contribute, after 2 decided not to, the number of students who actually contributed is \( x - 2 = 10 - 2 = 8 \).

Key Concepts

System of Equations

When solving algebra word problems, a **system of equations** can serve as a powerful tool. A system of equations is a set of two or more equations with the same variables. In real-life scenarios, like splitting the cost of a gift, systems are extremely useful.
In this exercise, we had two main equations to focus on:

• The original contribution equation: \( x \times y = 80 \) where \( x \) and \( y \) represent the number of students and their individual contributions, respectively.
• The adjusted contribution equation: \( (x - 2)(y + 2) = 80 \), accounting for two students dropping out and others having to pay more.

By solving this system, we found consistent solutions between the equations, allowing us to determine the correct values for our variables. This method ensures accuracy and helps identify all possible solutions.

Quadratic Formula

The **quadratic formula** is a vital tool when solving equations of the form \( ax^2 + bx + c = 0 \). It provides solutions based on the coefficients \( a \), \( b \), and \( c \).
In this problem, after forming the equation \( x^2 - 2x - 80 = 0 \), we used the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values \( a=1 \), \( b=-2 \), and \( c=-80 \), we calculated:
\[ x = \frac{2 \pm \sqrt{324}}{2} \]
This results in the potential solutions \( x = 10 \) or \( x = -8 \). It’s important to fisregard negative values when counting people or items since you can't have a negative count. This step allowed us to determine the original number of students.

Problem-Solving Strategies

Effective **problem-solving strategies** in algebra word problems involve several key steps:

• **Understanding the problem:** Begin by fully comprehending what the question is asking. Identify all the key components, like the total cost and changes in contributors.
• **Defining variables:** Choose variables that represent the unknowns in the problem. In this instance, \( x \) for the original number of students and \( y \) for their contribution.
• **Creating equations:** Formulate equations that model the situation. Each equation should represent a part of the problem scenario.
• **Solving the equations:** Utilize algebraic methods, such as substitution or the quadratic formula, to solve the equations.
• **Verifying the solution:** Always check that the solution answers the question and makes sense in the real-world context.

This step-by-step approach fosters clear thinking and helps ensure that all aspects of the problem are addressed.

Defining Variables

**Defining variables** is a crucial first step in solving algebra word problems. By turning unknowns into variables, you translate the word problem into mathematical terms.
In our exercise:

• We let \( x \) symbolize the "original number of students". This provided a simple way to reference the total number of people originally planning to contribute.
• We defined \( y \) as the "amount each agreed to contribute". Assigning this role to \( y \) clarified how costs were divided.

By clearly establishing each variable’s purpose, you prevent confusion further into problem-solving. This allows you to focus on forming and solving the equations, rather than on the meanings of the numbers you use.