Solving Cookie Sales With System Of Equations

Introduction

In this article, we will delve into a practical problem involving the application of a system of equations. This problem revolves around Jillian, who is raising funds for her basketball team by selling boxes of cookies. She offers two sizes: a 10 oz box priced at $3.50 and a 16 oz box priced at $5.00. Over the course of a week, Jillian manages to collect $97.50 from selling a total of 24 boxes. Our task is to determine the system of equations that accurately models this scenario and to subsequently find out how many boxes of each size Jillian sold. This exercise not only showcases the utility of mathematical modeling in real-world situations but also reinforces the importance of understanding and manipulating algebraic equations. This comprehensive exploration will provide a step-by-step approach to solving this problem, ensuring a clear understanding of the underlying mathematical principles and their application.

Setting Up the Equations

To solve this problem effectively, we need to translate the given information into a mathematical framework. The core concept here is to represent the unknowns—the number of 10 oz boxes and the number of 16 oz boxes—with variables. Let's denote the number of 10 oz boxes as x and the number of 16 oz boxes as y. Our primary objective is to establish two distinct equations that capture the two crucial pieces of information provided in the problem statement: the total number of boxes sold and the total revenue collected.

Equation 1: Total Number of Boxes

The first piece of information we have is that Jillian sold a total of 24 boxes. This means that the sum of the number of 10 oz boxes (x) and the number of 16 oz boxes (y) must equal 24. We can express this relationship in a simple linear equation:

x + y = 24

This equation serves as the foundation for our system, providing a direct link between the quantities of the two types of boxes sold. It is a straightforward representation of the total sales volume and sets the stage for further analysis.

Equation 2: Total Revenue

The second key piece of information is the total revenue Jillian collected, which amounts to $97.50. This revenue is generated from the sales of both the 10 oz boxes, priced at $3.50 each, and the 16 oz boxes, priced at $5.00 each. To formulate the second equation, we need to consider the revenue generated from each type of box individually and then combine them to match the total revenue.

The revenue from the 10 oz boxes can be calculated by multiplying the price per box ($3.50) by the number of boxes sold (x), resulting in 3.50x. Similarly, the revenue from the 16 oz boxes is the price per box ($5.00) multiplied by the number of boxes sold (y), giving us 5.00y. The sum of these two revenues must equal the total revenue of $97.50. Therefore, the second equation can be written as:

3. 50x + 5.00y = 97.50

This equation represents the financial aspect of the problem, linking the quantities of boxes sold to the total income generated. It adds a layer of complexity to the system, incorporating the pricing of each box type.

The System of Equations

Together, these two equations form a system of linear equations:

1. x + y = 24
2. 50x + 5.00y = 97.50

This system of equations provides a complete mathematical model of the problem. It encapsulates all the essential information and allows us to solve for the unknowns, x and y, which represent the number of 10 oz and 16 oz boxes sold, respectively. The next step in solving this problem involves choosing an appropriate method to solve this system, such as substitution or elimination, which we will explore in the following sections. This rigorous setup is crucial for accurately determining the solution and understanding the dynamics of Jillian's cookie sales.

Solving the System of Equations

Now that we have successfully established the system of equations that models Jillian's cookie sales, the next critical step is to solve this system to determine the values of x and y. This will tell us exactly how many boxes of each size Jillian sold. There are several methods available to solve a system of linear equations, including substitution, elimination, and graphical methods. For this particular problem, we will focus on the substitution and elimination methods, as they are particularly well-suited for this type of algebraic problem.

Method 1: Substitution

The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one variable, which can then be easily solved. Let's apply this method to our system:

1. x + y = 24
2. 50x + 5.00y = 97.50

Step 1: Solve the First Equation for x

We can easily solve the first equation for x by subtracting y from both sides:

x = 24 - y

This gives us an expression for x in terms of y. This is a crucial step, as it allows us to replace x in the second equation with this expression.

Step 2: Substitute into the Second Equation

Next, we substitute the expression for x (24 - y) into the second equation:

3. 50(24 - y) + 5.00y = 97.50

This substitution is the heart of the method, effectively reducing the problem to a single equation with one unknown. Now, we can solve this equation for y.

Step 3: Solve for y

Expanding and simplifying the equation, we get:

84 - 3.50y + 5.00y = 97.50

Combining like terms, we have:

1. 50y = 13.50

Dividing both sides by 1.50, we find:

y = 9

This tells us that Jillian sold 9 boxes of the 16 oz size.

Step 4: Solve for x

Now that we have the value of y, we can substitute it back into the equation x = 24 - y to find the value of x:

x = 24 - 9 x = 15

So, Jillian sold 15 boxes of the 10 oz size.

Method 2: Elimination

The elimination method involves manipulating the equations so that the coefficients of one variable are the same (or additive inverses) in both equations. Then, by adding or subtracting the equations, that variable is eliminated, leaving a single equation with one variable. Let's apply this method to our system:

1. x + y = 24
2. 50x + 5.00y = 97.50

Step 1: Multiply the First Equation by a Constant

To eliminate x, we can multiply the first equation by -3.50:

-3.50(x + y) = -3.50(24) -3.50x - 3.50y = -84

Now, we have the modified system:

1. -3.50x - 3.50y = -84
2. 50x + 5.00y = 97.50

Step 2: Add the Equations

Adding the two equations, we eliminate x:

(5.00x - 3.50x) + (5.00y - 3.50y) = 97.50 - 84

1. 50y = 13.50

Step 3: Solve for y

Dividing both sides by 1.50, we find:

y = 9

Again, we find that Jillian sold 9 boxes of the 16 oz size.

Step 4: Solve for x

Substitute y = 9 back into the first original equation:

x + 9 = 24 x = 15

Thus, Jillian sold 15 boxes of the 10 oz size.

Conclusion

Through both the substitution and elimination methods, we have arrived at the same solution: Jillian sold 15 boxes of the 10 oz cookies and 9 boxes of the 16 oz cookies. This demonstrates the power and versatility of systems of equations in solving real-world problems. By carefully setting up the equations and applying appropriate algebraic techniques, we can effectively model and solve a variety of situations. This example not only helps in understanding the specific scenario of Jillian's cookie sales but also reinforces the broader applicability of mathematical modeling in problem-solving.

This exercise underscores the importance of understanding algebraic methods and their practical applications. Whether it's in business, finance, or everyday situations, the ability to translate real-world scenarios into mathematical models and solve them is an invaluable skill. The detailed walkthrough provided here should serve as a helpful guide for anyone looking to enhance their problem-solving capabilities using systems of equations. The successful resolution of this problem highlights the significance of both the setup and the solution phases in mathematical modeling, ensuring a comprehensive understanding of the problem and its solution.