Solve For The Cost Of Each Item At The Local Craft Store For Brenda, Sheena, And Cody Using The Given Equations.${ \begin{align*} 3s + 4c + 5m &= 24.40 \ 6s + 5c + 2m &= 30.40 \ 3s + 2c + M &= 13.40 \end{align*} }$Type The Correct Answer
Introduction
In the world of mathematics, solving a system of linear equations is a fundamental concept that has numerous real-world applications. In this article, we will explore a scenario where three friends, Brenda, Sheena, and Cody, visit a local craft store to purchase various items. The store owner provides them with a set of equations representing the total cost of each item. Our goal is to solve for the cost of each item using the given equations.
The Problem
The local craft store has three equations representing the total cost of each item:
{ \begin{align*} 3s + 4c + 5m &= 24.40 \\ 6s + 5c + 2m &= 30.40 \\ 3s + 2c + m &= 13.40 \end{align*} \}
where represents the cost of a scarf, represents the cost of a candle, and represents the cost of a mug.
Step 1: Write the Augmented Matrix
To solve this system of linear equations, we will first write the augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms.
| 3 4 5 | 24.40 |
| 6 5 2 | 30.40 |
| 3 2 1 | 13.40 |
Step 2: Perform Row Operations
To solve the system of linear equations, we will perform row operations on the augmented matrix. The goal is to transform the matrix into row-echelon form, where all the entries below the leading entry in each row are zeros.
| 1 0 0 | 8.00 |
| 0 1 0 | 4.00 |
| 0 0 1 | 1.00 |
Step 3: Solve for the Variables
Now that we have the row-echelon form of the matrix, we can solve for the variables. We will start by solving for the variable .
m = 1.00
Next, we will solve for the variable .
c = 4.00
Finally, we will solve for the variable .
s = 8.00
Conclusion
In this article, we solved a system of linear equations representing the total cost of each item at the local craft store. We used the augmented matrix and row operations to transform the matrix into row-echelon form, and then solved for the variables. The final answer is:
- The cost of a scarf is $8.00.
- The cost of a candle is $4.00.
- The cost of a mug is $1.00.
Discussion
This problem is a great example of how solving a system of linear equations can be applied to real-world scenarios. In this case, the local craft store owner can use the equations to determine the cost of each item and make informed decisions about pricing and inventory management. Additionally, this problem demonstrates the importance of using technology, such as calculators or computer software, to solve systems of linear equations.
Real-World Applications
Solving systems of linear equations has numerous real-world applications, including:
- Business and Finance: Solving systems of linear equations can be used to determine the cost of goods, calculate profit margins, and make informed decisions about investments.
- Science and Engineering: Solving systems of linear equations can be used to model complex systems, such as electrical circuits, mechanical systems, and population dynamics.
- Computer Science: Solving systems of linear equations can be used to develop algorithms for solving linear systems, which is a fundamental problem in computer science.
Conclusion
In conclusion, solving a system of linear equations is a fundamental concept in mathematics that has numerous real-world applications. In this article, we solved a system of linear equations representing the total cost of each item at the local craft store. We used the augmented matrix and row operations to transform the matrix into row-echelon form, and then solved for the variables. The final answer is:
- The cost of a scarf is $8.00.
- The cost of a candle is $4.00.
- The cost of a mug is $1.00.
Solving a System of Linear Equations: A Real-World Application at the Local Craft Store - Q&A =====================================================================================
Introduction
In our previous article, we solved a system of linear equations representing the total cost of each item at the local craft store. We used the augmented matrix and row operations to transform the matrix into row-echelon form, and then solved for the variables. In this article, we will provide a Q&A section to help clarify any questions or concerns you may have.
Q&A
Q: What is a system of linear equations?
A: A system of linear equations is a set of two or more linear equations that are solved simultaneously to find the values of the variables.
Q: How do I know if a system of linear equations has a solution?
A: To determine if a system of linear equations has a solution, we need to check if the system is consistent or inconsistent. If the system is consistent, it means that there is a solution. If the system is inconsistent, it means that there is no solution.
Q: What is the augmented matrix?
A: The augmented matrix is a matrix that includes the coefficients of the variables and the constant terms.
Q: How do I perform row operations on the augmented matrix?
A: To perform row operations on the augmented matrix, we need to follow these steps:
- Multiply a row by a non-zero constant.
- Add a multiple of one row to another row.
- Swap two rows.
Q: What is row-echelon form?
A: Row-echelon form is a matrix that has the following properties:
- All the entries below the leading entry in each row are zeros.
- The leading entry in each row is to the right of the leading entry in the previous row.
- The leading entry in each row is a 1.
Q: How do I solve for the variables?
A: To solve for the variables, we need to follow these steps:
- Solve for the variable with the leading entry in the first row.
- Substitute the value of the variable into the equation and solve for the next variable.
- Repeat the process until all the variables are solved.
Q: What are some real-world applications of solving systems of linear equations?
A: Solving systems of linear equations has numerous real-world applications, including:
- Business and Finance: Solving systems of linear equations can be used to determine the cost of goods, calculate profit margins, and make informed decisions about investments.
- Science and Engineering: Solving systems of linear equations can be used to model complex systems, such as electrical circuits, mechanical systems, and population dynamics.
- Computer Science: Solving systems of linear equations can be used to develop algorithms for solving linear systems, which is a fundamental problem in computer science.
Conclusion
In conclusion, solving a system of linear equations is a fundamental concept in mathematics that has numerous real-world applications. In this article, we provided a Q&A section to help clarify any questions or concerns you may have. We hope that this article has been helpful in understanding the concept of solving systems of linear equations.
Frequently Asked Questions
Q: What is the difference between a system of linear equations and a system of nonlinear equations?
A: A system of linear equations is a set of linear equations that are solved simultaneously to find the values of the variables. A system of nonlinear equations is a set of nonlinear equations that are solved simultaneously to find the values of the variables.
Q: How do I determine if a system of linear equations is consistent or inconsistent?
A: To determine if a system of linear equations is consistent or inconsistent, we need to check if the system has a solution. If the system has a solution, it is consistent. If the system does not have a solution, it is inconsistent.
Q: What is the importance of solving systems of linear equations?
A: Solving systems of linear equations is important because it has numerous real-world applications, including business and finance, science and engineering, and computer science.
Q: How do I use technology to solve systems of linear equations?
A: There are several ways to use technology to solve systems of linear equations, including using calculators, computer software, and online tools.