Problem 17 - Algebra A Combined Function

Question

Ann Marie Jones has been pricing Amtrak train fares for a group trip to New York. Three adults and four children must pay \(\$ 159 .\) Two adults and three children must pay \(\$ 112\). Find the price of an adult's ticket, and find the price of a child's ticket.

Step-by-Step Solution

Answer

Adult's ticket is $50.60, child's ticket is $3.60.

Step 1: Define Variables

Let's define two variables to represent the unknowns in this problem. Let \( a \) be the price of an adult's ticket, and \( c \) be the price of a child's ticket.

Step 2: Set Up Equations

From the problem, we have two scenarios: one with 3 adults and 4 children, and one with 2 adults and 3 children. We can set up two equations based on the total fare calculations:1. For 3 adults and 4 children: \( 3a + 4c = 159 \)2. For 2 adults and 3 children: \( 2a + 3c = 112 \)

Step 3: Solve Equation System

We will solve the system of equations using substitution or elimination. Here, we'll use elimination:Multiply the second equation by 2 to align the children term (3c) with the first:\[ 4a + 6c = 224 \]Now, multiply the first equation by 3:\[ 9a + 12c = 477 \]Subtract the new version of the second equation from this equation to eliminate \( c \):\[ (9a + 12c) - (4a + 6c) = 477 - 224 \]\[ 5a + 6c - 6c = 253 \]\[ 5a = 253 \]\[ a = 50.6 \]

Step 4: Solve for Price of Child's Ticket

Use the value of \( a \) to find \( c \) using the original second equation:\( 2(50.6) + 3c = 112 \)\( 101.2 + 3c = 112 \)\( 3c = 112 - 101.2 \)\( 3c = 10.8 \)\( c = 3.6 \)

Key Concepts

System of Equations

In algebra, a system of equations is a set of two or more equations with the same set of unknowns. When you are trying to solve the system, you're looking for the values of these unknowns that satisfy all equations simultaneously. In our problem, we have two equations and two unknowns (variables):

• Equation 1: Represents the total fare for 3 adults and 4 children: \( 3a + 4c = 159 \).
• Equation 2: Represents the total fare for 2 adults and 3 children: \( 2a + 3c = 112 \).

When solving systems of equations, it's important to check if the equations relate the variables in a meaningful way. Once set up correctly, these equations can be used to find the solution either by substitution, elimination, or by using matrices.
In our exercise, we'll be using the elimination method, which can be very effective when equations are properly set up and organized. This approach helps in systematically reducing the number of variables.

Elimination Method

The elimination method is a common and powerful technique used to solve a system of equations. It involves eliminating one of the variables from the equations, by adding or subtracting the equations from each other. This allows us to focus on solving for one variable at a time.
Let's see how the elimination method is applied in our exercise:

• Multiply the second equation, \( 2a + 3c = 112 \), by 2, to get \( 4a + 6c = 224 \).
• Multiply the first equation, \( 3a + 4c = 159 \), by 3 to obtain \( 9a + 12c = 477 \).
• Subtract the modified second equation from the modified first equation: \((9a + 12c) - (4a + 6c) = 477 - 224 \).
• This results in \( 5a = 253 \), leading us to find \( a = 50.6 \).

Using elimination, we've successfully found that the price for an adult ticket \( a \) is \( 50.6 \). This method is particularly useful when dealing with linear systems with at least two equations.

Variables Definition

The proper definition and understanding of variables are critical when solving algebra word problems. Variables act as placeholders for unknown values that we need to determine through the problem-solving process.

• Define each variable clearly to establish what each represents. In this scenario, let \( a \) denote the price of an adult's ticket.
• Let \( c \) represent the price of a child's ticket. Naming your variables appropriately helps in tracking what each letter stands for and prevents confusion.

Starting with clearly defined variables is like having a roadmap for your problem. It tells you exactly what you're solving for, and it provides a base for setting up your equations correctly. In this exercise, without defining \( a \) and \( c \), it would be difficult to proceed methodically through the problem.
Once defined, these variables can then be substituted into your equations, allowing you to systematically approach the solution.