Problem 97 - Algebra 1: Concepts and Skills

Question

The admission price for a high school basketball game is \(\$2\) for students and \(\$3\) for adults. At one game, 324 tickets were sold and \(\$764\) was collected. How many students and adults attended the game?

Step-by-Step Solution

Answer

208 students and 116 adults attended the game.

Step 1: Establishing the equations

Let's take the number of students as 's' and the number of adults as 'a'. There are two pieces of information that can be converted into equations: 1. The total number of tickets sold: \(s + a = 324\) 2. The total sum of money collected: \(2s + 3a = 764\)

Step 2: Solve the system of equations

Start by multiplying the first equation by 2: \(2s + 2a = 648\). Compare it with the second equation: \(2s + 3a = 764\). Now let's subtract the first equation from the second, resulting in: \(a = 116\). Substitute \(a = 116\) into the first equation, that gives us: \(s + 116 = 324\). Which reduces to: \(s = 208\). Hence, the solution of the system of equations is \(s = 208\) and \(a = 116\).

Step 3: Interpret the solution

The obtained solution to the system of equations corresponds to the number of students and adults who attended the game. So, 208 students and 116 adults attended the game.

Key Concepts

Linear Equations

In mathematics, a linear equation is an algebraic equation where each term is either a constant or the product of a constant and (the first power of) a single variable. Linear equations can be simple to understand because they represent a straight line when graphed on a coordinate plane.

Let's consider the high school basketball game ticket sales problem. We have two linear equations from the problem:

• The first equation represents the total number of tickets sold: \(s + a = 324\), indicating a direct relationship between the number of student tickets (s) and adult tickets (a).
• The second equation comes from the total money collected: \(2s + 3a = 764\). This describes how much money is collected from the student and adult tickets.

These equations are linear because each variable is to the first power, and their graphs are straight lines. The goal is to find the value of 's' (students) and 'a' (adults) that satisfy both equations simultaneously.

Algebraic Problem Solving

Algebraic problem solving involves finding unknown values that make a set of equations true. It's a step-by-step process that requires manipulation of equations to arrive at a solution.

Here's how it works with our ticket sales problem:

• We first establish the equations based on the given information.
• We then manipulate these equations to find the values of s and a.
• Finally, we interpret the results in the context of the problem to make sure they make sense (e.g., you can't sell a negative number of tickets).

It's important to understand each step and the logic behind it to successfully solve similar algebraic problems. Critical thinking and attention to the problem's details are key.

Substitution Method

The substitution method is a way of solving systems of equations by finding the value of one variable and then substituting that value into another equation to solve for another variable.

In our example, after determining that \(a = 116\) by using the elimination method, we substitute 116 for 'a' in the first equation:

Solving for 's' After Substitution

\(s + 116 = 324\), which simplifies to \(s = 324 - 116\), or \(s = 208\). Now we have the number of students who purchased tickets. By substituting and then simplifying, we've isolated one variable and solved for it effectively.

Elimination Method

The elimination method involves aligning two equations and eliminating one variable to solve for the other. This method often involves adding or subtracting whole equations from each other.

For the ticket sales problem, we aligned the two equations and subtracted the first from the second:

Elimination in Action

• First aligned equation: \(2s + 2a = 648\)
• Second aligned equation: \(2s + 3a = 764\)

By subtracting the first from the second, we eliminated 's', giving us the equation \(a = 116\). This is a crucial step in solving for both variables without dealing with fractions or more complex algebraic manipulations.