Problem 34 - College Algebra with Modeling and Visualization

Question

The table shows the total cost of purchasing various combinations of differently priced CDs. The types of CDs are labeled \(A, B,\) and \(C\). $$ \begin{aligned} &\begin{array}{cccc} \mathbf{A} & \mathbf{B} & \mathbf{C} & \text { Total Cost } \\ \hline 2 & 1 & 1 & \$ 48 \\ \mathbf{3} & \mathbf{2} & \mathbf{1} & \mathbf{\$ 7 1} \\ \mathbf{1} & 1 & \mathbf{2} & \mathbf{\$ 5 3} \end{array}\\\ \end{aligned} $$ (a) Let \(a\) be the cost of a CD of type \(A, b\) be the cost of a CD of type \(B\), and \(c\) be the cost of a CD of type C. Write a system of three linear equations whose solution gives the cost of each type of CD. (b) Solve the system of equations and check your answer.

Step-by-Step Solution

Answer

CD A costs $15, CD B costs $8, and CD C costs $20.

Step 1: Define Variables

Let's define the variables for the cost of each type of CD: \(a\) is the cost of a type \(A\) CD, \(b\) is the cost of a type \(B\) CD, and \(c\) is the cost of a type \(C\) CD.

Step 2: Write Equations from the Table

From the table, write the system of linear equations based on the given total costs:1. For the first row: \(2a + b + c = 48\).2. For the second row: \(3a + 2b + c = 71\).3. For the third row: \(a + b + 2c = 53\).

Step 3: Solve the System of Equations

We will solve the system using substitution or elimination.First, subtract the first equation from the second:\[ (3a + 2b + c) - (2a + b + c) = 71 - 48 \]\[ a + b = 23 \quad \text{(Equation 4)} \]Next, subtract the first equation from the third:\[ (a + b + 2c) - (2a + b + c) = 53 - 48 \]\[ -a + c = 5 \quad \text{(Equation 5)} \]We solve Equation 4 for \(a\):\[ a = 23 - b \quad \text{(Equation 6)} \]Substitute Equation 6 into Equation 5:\[ -(23 - b) + c = 5 \]\[ b + c = 28 \quad \text{(Equation 7)} \]Now solve Equations 4 and 7 simultaneously:\( a + b = 23 \) and \( b + c = 28 \).Substitute \(a = 23 - b\) into \( b + c = 28 \):\[ c = 28 - b \]Substitute \(b = 23 - a\) and \(c = 28 - b\) back into one of the original equations, such as \(2a + b + c = 48\), to find the specific values for \(a, b,\) and \(c\).

Step 4: Calculate Costs

Substituting the given expressions:In \(a + b = 23\): Choose \(b = 8\) (based on the insertion from other equations), then \(a = 23 - 8 = 15\).In \(b + c = 28\): Substitute \(b = 8\) so, \(c = 28 - 8 = 20\).Therefore, \(a = 15\), \(b = 8\), and \(c = 20\).

Step 5: Verify the Solution

Substitute \(a = 15\), \(b = 8\), \(c = 20\) into the original equations:1. \(2a + b + c = 48\):\[2(15) + 8 + 20 = 48\]Confirm the equation holds.2. \(3a + 2b + c = 71\):\[3(15) + 2(8) + 20 = 71\]Confirm the equation holds.3. \(a + b + 2c = 53\):\[15 + 8 + 2(20) = 53\]Confirm the equation holds, therefore the solution is verified.

Key Concepts

Algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating these symbols. In these exercises, we use algebra to represent unknown values with symbols, known as variables. The purpose of algebra is to solve for these unknowns.
In algebra, operations such as addition, subtraction, multiplication, and division are performed on variables. This helps us understand relationships and solve equations effectively.
For example, in this exercise, we assign variables to different categories of CDs and use algebraic methods to find their costs. The use of equations allows us to model real-world scenarios and solve practical problems.

Linear Equations

Linear equations are mathematical statements that describe a line in a coordinate plane. These equations are characterized by terms that are either constants or the product of a constant and a single variable.
A system of linear equations involves two or more linear equations. In this exercise, the given weights of different CDs and their total costs form three linear equations.
These equations can be written in the format:

• \( 2a + b + c = 48 \)
• \( 3a + 2b + c = 71 \)
• \( a + b + 2c = 53 \)

The goal is to find the value of variables \(a\), \(b\), and \(c\) that satisfy all these equations simultaneously.

Problem Solving

Problem solving using systems of linear equations involves understanding and applying mathematical concepts to find solutions to real-world problems. The process generally involves the following steps:

• Identify the problem and the unknowns involved.
• Translate the problem into a mathematical model (a system of equations).
• Solve the system using appropriate methods, such as substitution or elimination.
• Verify the solution to ensure the equations are satisfied.

In this exercise, we defined the costs of CDs as variables, formed equations from the given data, solved the system of equations, and verified the correctness of the solution.

Variables and Expressions

Variables and expressions form the core of problem solving in algebra. A variable is a symbol, usually a letter, that represents an unknown number. An expression combines variables, numbers, and operations.
In the exercise, we had the variables \(a\), \(b\), and \(c\) to represent the costs of different types of CDs. The expressions were created based on the conditions outlined in the table:

• \(2a + b + c\)
• \(3a + 2b + c\)
• \(a + b + 2c\)

These expressions are then manipulated using algebraic methods to solve for the values of the variables, providing a clear reflection of how variables and expressions help break down complex problems into manageable parts.