In The Total Cost Of 5 Notebooks And 4 Pencils Is 165; The Total Cost Of Another 6 Notebooks And 3 Pencils Is 180. How Much Does Each Pencil And Each Notebook Cost?

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Introduction

In this article, we will delve into a mathematical problem that involves solving a system of linear equations to determine the cost of each notebook and pencil. The problem states that the total cost of 5 notebooks and 4 pencils is 165, and the total cost of another 6 notebooks and 3 pencils is 180. Our goal is to find the individual costs of each notebook and pencil.

The Problem

Let's denote the cost of each notebook as N and the cost of each pencil as P. We can set up two equations based on the given information:

  • 5N + 4P = 165
  • 6N + 3P = 180

Breaking Down the Equations

We can start by analyzing the first equation: 5N + 4P = 165. This equation tells us that the total cost of 5 notebooks and 4 pencils is 165. We can see that the cost of the notebooks is 5 times the cost of one notebook (N), and the cost of the pencils is 4 times the cost of one pencil (P).

Similarly, the second equation: 6N + 3P = 180, tells us that the total cost of 6 notebooks and 3 pencils is 180. Again, we can see that the cost of the notebooks is 6 times the cost of one notebook (N), and the cost of the pencils is 3 times the cost of one pencil (P).

Solving the System of Equations

To solve this system of equations, we can use the method of substitution or elimination. In this case, we will use the elimination method.

First, let's multiply the first equation by 3 and the second equation by 4 to make the coefficients of P equal:

  • 15N + 12P = 495
  • 24N + 12P = 720

Now, we can subtract the first equation from the second equation to eliminate P:

  • (24N - 15N) + (12P - 12P) = 720 - 495
  • 9N = 225

Finding the Cost of One Notebook

Now that we have eliminated P, we can solve for N:

  • N = 225 / 9
  • N = 25

So, the cost of one notebook is 25.

Finding the Cost of One Pencil

Now that we have found the cost of one notebook, we can substitute it into one of the original equations to find the cost of one pencil. Let's use the first equation:

  • 5N + 4P = 165
  • 5(25) + 4P = 165
  • 125 + 4P = 165
  • 4P = 40
  • P = 10

So, the cost of one pencil is 10.

Conclusion

In this article, we have solved a system of linear equations to determine the cost of each notebook and pencil. We have found that the cost of one notebook is 25 and the cost of one pencil is 10. This problem is a great example of how mathematical concepts can be applied to real-world problems.

Real-World Applications

This problem has many real-world applications, such as:

  • Business: A company that sells notebooks and pencils can use this problem to determine the cost of each item and set prices accordingly.
  • Education: Students can use this problem to practice solving systems of linear equations and apply mathematical concepts to real-world problems.
  • Finance: A person who wants to buy notebooks and pencils can use this problem to determine the total cost and make informed purchasing decisions.

Final Thoughts

Q: What is the main concept behind solving the system of equations?

A: The main concept behind solving the system of equations is to eliminate one of the variables by multiplying the equations by necessary multiples such that the coefficients of one variable are the same in both equations.

Q: Why did we multiply the first equation by 3 and the second equation by 4?

A: We multiplied the first equation by 3 and the second equation by 4 to make the coefficients of P equal. This allowed us to eliminate P by subtracting the first equation from the second equation.

Q: What is the significance of eliminating P?

A: Eliminating P allowed us to solve for N, which is the cost of one notebook. Once we found the cost of one notebook, we could substitute it into one of the original equations to find the cost of one pencil.

Q: Can we use the substitution method to solve this system of equations?

A: Yes, we can use the substitution method to solve this system of equations. However, in this case, the elimination method was more straightforward and easier to apply.

Q: How can we apply this problem to real-world scenarios?

A: This problem can be applied to real-world scenarios such as business, education, and finance. For example, a company that sells notebooks and pencils can use this problem to determine the cost of each item and set prices accordingly.

Q: What are some common mistakes to avoid when solving systems of equations?

A: Some common mistakes to avoid when solving systems of equations include:

  • Not checking if the equations are linear
  • Not identifying the variables correctly
  • Not using the correct method (substitution or elimination)
  • Not checking for extraneous solutions

Q: Can we use this problem to solve for more than two variables?

A: Yes, we can use this problem to solve for more than two variables. However, the problem would need to be modified to include more equations and variables.

Q: How can we check if our solution is correct?

A: We can check if our solution is correct by plugging the values back into the original equations and verifying that they are true.

Q: What are some real-world applications of systems of equations?

A: Some real-world applications of systems of equations include:

  • Business: Solving systems of equations can help businesses determine the cost of goods, set prices, and make informed decisions.
  • Education: Solving systems of equations can help students practice problem-solving skills and apply mathematical concepts to real-world problems.
  • Finance: Solving systems of equations can help individuals and businesses make informed financial decisions and manage risk.

Q: Can we use technology to solve systems of equations?

A: Yes, we can use technology to solve systems of equations. Many graphing calculators and computer software programs can solve systems of equations quickly and accurately.

Q: What are some tips for solving systems of equations?

A: Some tips for solving systems of equations include:

  • Read the problem carefully and identify the variables
  • Choose the correct method (substitution or elimination)
  • Check for extraneous solutions
  • Verify the solution by plugging the values back into the original equations