Problem 74 - College Algebra with Corequisite Support

Question

For the following exercises, use a system of linear equations with two variables and two equations to solve. CDs cost $$\$ 5.96$$ more than DVDS at All Bets Are Off Electronics. How much would \(6 \mathrm{CDs}\) and 2 DVDs cost if 5 CDs and 2 DVDS cost $$\$ 127.73 ?$$

Step-by-Step Solution

Answer

6 CDs and 2 DVDs cost $147.68.

Step 1: Define Variables

Let's define the variables for the problem. Let \( x \) represent the cost of a CD and \( y \) represent the cost of a DVD. We are given that CDs cost $5.96 more than DVDs. Therefore, we can write this relationship as \( x = y + 5.96 \).

Step 2: Create Equation from Cost Information

From the information given, 5 CDs and 2 DVDs cost $127.73. This can be represented as an equation: \( 5x + 2y = 127.73 \).

Step 3: Substitute Variable Expression

Using the relationship \( x = y + 5.96 \), substitute \( x \) in the equation \( 5x + 2y = 127.73 \): \( 5(y + 5.96) + 2y = 127.73 \).

Step 4: Expand and Simplify the Equation

Expand the substituted equation: \( 5y + 29.8 + 2y = 127.73 \). Simplify it to get \( 7y + 29.8 = 127.73 \).

Step 5: Solve for Cost of DVD

Subtract 29.8 from both sides to solve for \( y \): \( 7y = 97.93 \). Then divide both sides by 7: \( y = 13.99 \). Therefore, each DVD costs $13.99.

Step 6: Calculate Cost of CD

Use the expression \( x = y + 5.96 \) to calculate the cost of a CD. Since \( y = 13.99 \), substitute this value into the equation: \( x = 13.99 + 5.96 = 19.95 \). Thus, each CD costs $19.95.

Step 7: Calculate Total Cost for 6 CDs and 2 DVDs

Finally, calculate the cost for 6 CDs and 2 DVDs: \( 6x + 2y \). Substitute \( x = 19.95 \) and \( y = 13.99 \): \( 6(19.95) + 2(13.99) = 119.7 + 27.98 = 147.68 \).

Key Concepts

Variable Definition

Defining variables is a key step in solving problems involving systems of equations. In this exercise, we used variables to represent the costs of CDs and DVDs, which are essential pieces of the puzzle we are trying to solve. By using variables, we can translate words into mathematical expressions and equations.

• Letting \( x \) denote the cost of one CD.
• Letting \( y \) denote the cost of one DVD.

These assignments help us set up the relationships and eventually solve for these unknown costs. It's like giving a name to something unknown so we can talk about it and manipulate it in our equations. By defining these variables, we can formulate the relationships and constraints of the problem into mathematical terms.

Cost Equations

Once the variables are defined, the next step is to create equations that represent the relationships and constraints in the problem. In this exercise, we have one key relationship: CDs cost \(5.96 more than DVDs. This is expressed as a cost equation:

\( x = y + 5.96 \)

Additionally, another equation is formed from the given cost information about purchasing 5 CDs and 2 DVDs for \)127.73:

\( 5x + 2y = 127.73 \)

These equations are tools that describe the situation mathematically. They allow us to use algebraic methods to find the unknown costs. Cost equations are a reflection of the problem's conditions, giving a structure to the numbers and allowing us to explore the solution.

Substitution Method

The substitution method is a strategy used for solving systems of linear equations. It's particularly useful when one equation is already solved for one variable, or can easily be solved. The idea is to solve one of the equations in terms of one variable and substitute this expression into the other equation.

In this exercise, we used the expression from the cost relationship \( x = y + 5.96 \) to substitute for \( x \) in the second equation:\[ 5(y + 5.96) + 2y = 127.73 \]

This substitution step reduces the problem to a single equation with one variable, making it much simpler to solve. The flexibility of this method makes it a powerful tool in algebra for breaking down complex problems into manageable steps.

Solving Linear Equations

Once we have a single equation with one variable—thanks to substitution—we can focus on solving it. In this exercise, after substituting for \( x \), the equation is:

\[ 5y + 29.8 + 2y = 127.73 \]

Simplifying this gives \( 7y + 29.8 = 127.73 \). Solving involves basic operations like subtraction and division:

• Subtract 29.8 from both sides: \( 7y = 97.93 \)
• Divide each side by 7 to find \( y: \) \( y = 13.99 \)

Once we solve for \( y \), we can use it to find \( x \) through the relationship we had from the beginning. Solving linear equations is about isolating the variable—getting it "alone"—through a series of straightforward arithmetic steps. By following these orderly steps, we can mathematically determine the costs given in the original problem.