Meghan's Radio Ad Sales Solving A System Of Equations
Introduction
In this article, we delve into a mathematical problem involving Meghan, a radio station advertisement salesperson. Meghan sells both 30-second and 60-second ad slots, each with its own pricing structure. The challenge lies in determining the number of each type of ad Meghan sold, given the total number of ads and the total revenue generated. This problem provides a practical application of linear equations and systems of equations, showcasing how mathematical models can be used to solve real-world scenarios. We will explore the problem setup, the equations involved, and the step-by-step solution process, providing a comprehensive understanding of the underlying mathematical concepts.
Problem Statement
Meghan's radio ad sales involve selling two types of advertisements a 30-second ad, which costs $20 per play, and a 60-second ad, which costs $35 per play. In a particular sales period, Meghan sold a total of 12 ads, generating a revenue of $315. The task is to determine the number of 30-second ads and 60-second ads Meghan sold. To solve this, we can set up a system of linear equations, where $x$ represents the number of 30-second ads and $y$ represents the number of 60-second ads. This problem exemplifies a classic application of algebra in a business context, requiring the formulation and solution of simultaneous equations to arrive at the answer. Understanding this problem not only enhances mathematical skills but also provides insights into how mathematical modeling can be applied in sales and marketing.
Setting up the Equations
To effectively solve Meghan's radio ad sales problem, it is crucial to translate the given information into mathematical equations. Let's break down the process step by step. We've already defined our variables: let $x$ represent the number of 30-second ads sold, and let $y$ represent the number of 60-second ads sold. The first piece of information we have is that Meghan sold a total of 12 ads. This gives us our first equation:
This equation represents the total number of ads sold, combining both 30-second and 60-second slots. The second piece of information is the total revenue Meghan generated, which was $315. We know that each 30-second ad costs $20, so the total revenue from 30-second ads is $20x$. Similarly, each 60-second ad costs $35, so the total revenue from 60-second ads is $35y$. The sum of these revenues must equal the total revenue, giving us our second equation:
This equation represents the total revenue generated from both types of ads. Now, we have a system of two linear equations with two variables:
This system of equations forms the foundation for solving the problem. The next step involves choosing a method to solve this system, such as substitution or elimination, to find the values of $x$ and $y$. Mastering the art of setting up these equations is fundamental in solving a wide range of mathematical problems, especially those encountered in real-world applications.
Solving the System of Equations
With the system of equations established for Meghan's radio ad sales, the next crucial step is to solve for the variables $x$ and $y$, representing the number of 30-second and 60-second ads, respectively. There are several methods to solve such a system, but we will focus on the substitution method and the elimination method.
Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. From the first equation, $x + y = 12$, we can easily solve for $x$:
Now, we substitute this expression for $x$ into the second equation, $20x + 35y = 315$:
Expanding and simplifying this equation, we get:
Now that we have the value of $y$, we can substitute it back into the equation $x = 12 - y$ to find $x$:
Thus, using the substitution method, we find that Meghan sold 7 thirty-second ads and 5 sixty-second ads.
Elimination Method
The elimination method involves manipulating the equations so that the coefficients of one variable are opposites, allowing us to eliminate that variable when the equations are added together. To eliminate $x$, we can multiply the first equation, $x + y = 12$, by -20:
Now we add this modified equation to the second equation, $20x + 35y = 315$:
As with the substitution method, we find that $y = 5$. We can substitute this value back into either of the original equations to solve for $x$. Using the first equation, $x + y = 12$:
Again, we find that Meghan sold 7 thirty-second ads and 5 sixty-second ads. Both the substitution and elimination methods lead to the same solution, affirming the accuracy of our calculations. This ability to solve systems of equations is a cornerstone of algebra and is applicable in various fields beyond mathematics, such as economics, engineering, and computer science. The consistent result obtained through both methods reinforces the reliability and versatility of these algebraic techniques.
Interpretation of the Solution
Having solved the system of equations, we now arrive at the crucial step of interpreting the solution within the context of Meghan's radio ad sales problem. The solution we found is $x = 7$ and $y = 5$. Recall that we defined $x$ as the number of 30-second ads Meghan sold and $y$ as the number of 60-second ads she sold. Therefore, the solution tells us that Meghan sold 7 thirty-second advertisements and 5 sixty-second advertisements.
To ensure our solution is correct and makes sense in the real-world scenario, we can verify it against the original problem conditions. We know that Meghan sold a total of 12 ads. Our solution indicates that 7 thirty-second ads plus 5 sixty-second ads equals 12 ads, which satisfies the first condition. We also know that Meghan generated a total revenue of $315. Let's calculate the revenue from our solution:
Revenue from 30-second ads: $7 \text{ ads} \times $20/\text{ad} = $
Revenue from 60-second ads: $5 \text{ ads} \times $35/\text{ad} = $
Total revenue: $$140 + $175 = $
This matches the second condition provided in the problem, confirming that our solution is accurate. In the context of Meghan's job selling radio advertisements, this result is valuable information. It provides a breakdown of the types of ads she sold, which can be useful for tracking sales performance, understanding customer preferences, and making informed decisions about future sales strategies. For instance, if Meghan notices that 30-second ads are more popular, she might focus on promoting those more aggressively. Conversely, if 60-second ads generate more revenue, she might incentivize their sales. This problem demonstrates how mathematical solutions can translate into practical insights in a business setting. The ability to interpret mathematical results and apply them to real-world situations is a key skill in various professions, from sales and marketing to finance and management. The accuracy of our solution, verified against the problem's conditions, underscores the importance of careful problem-solving and the practical utility of algebraic techniques.
Conclusion
In conclusion, the problem involving Meghan's radio ad sales provides a compelling example of how mathematical concepts, specifically systems of linear equations, can be applied to solve real-world business scenarios. By translating the given information into a system of equations, we were able to determine the number of 30-second and 60-second ads Meghan sold. The process involved setting up the equations, choosing an appropriate solution method (in this case, both substitution and elimination), and interpreting the results in the context of the problem.
We began by defining the variables, with $x$ representing the number of 30-second ads and $y$ representing the number of 60-second ads. We then formulated two equations based on the given information: the total number of ads sold and the total revenue generated. This step highlighted the importance of understanding the problem statement and identifying the key relationships between the variables. Next, we explored two common methods for solving systems of equations: substitution and elimination. Both methods yielded the same solution, demonstrating their reliability and the consistency of mathematical principles. The substitution method involved solving one equation for one variable and substituting that expression into the other equation, while the elimination method involved manipulating the equations to eliminate one variable through addition or subtraction. The solution we obtained, $x = 7$ and $y = 5$, indicated that Meghan sold 7 thirty-second ads and 5 sixty-second ads. Crucially, we verified this solution against the original problem conditions to ensure its accuracy. By calculating the revenue generated from each type of ad and summing them up, we confirmed that the total revenue matched the given value of $315. This step emphasized the importance of checking solutions to avoid errors and ensure the results are meaningful in the real-world context. Finally, we discussed the practical implications of the solution. Understanding the breakdown of ad sales can help Meghan and her team make informed decisions about sales strategies, marketing efforts, and customer preferences. This illustrates the broader relevance of mathematical problem-solving in various fields, from business and finance to engineering and science.
Overall, this exercise in solving Meghan's radio ad sales problem not only reinforces algebraic skills but also highlights the power of mathematical modeling in understanding and addressing real-world challenges. It underscores the value of a systematic approach to problem-solving, from defining variables and setting up equations to solving them and interpreting the results. This comprehensive process equips individuals with the tools and mindset needed to tackle a wide range of quantitative problems in various domains. The ability to translate real-world situations into mathematical models and derive meaningful insights from the solutions is a valuable asset in today's data-driven world.
