A Vendor Sells Hot Dogs And Bags Of Potato Chips.1. A Customer Buys 4 Hot Dogs And 5 Bags Of Potato Chips For $16.75. 2. Another Customer Buys 3 Hot Dogs And 3 Bags Of Potato Chips For $11.25.Find The Cost Of Each Item:- Let X X X Be

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Introduction

In this article, we will delve into a real-world problem involving the cost of hot dogs and potato chips. A vendor sells these items, and we are given two different transactions involving the purchase of hot dogs and potato chips. Our goal is to find the cost of each item, using the information provided in the transactions. We will use algebraic equations to solve this problem, making it a great example of how mathematics is used in everyday life.

The Problem

Let's take a closer look at the two transactions:

  1. A customer buys 4 hot dogs and 5 bags of potato chips for $16.75.
  2. Another customer buys 3 hot dogs and 3 bags of potato chips for $11.25.

We are asked to find the cost of each item, denoted by the variables x and y, where x represents the cost of a hot dog and y represents the cost of a bag of potato chips.

Setting Up the Equations

To solve this problem, we need to set up two equations based on the information provided in the transactions. Let's denote the cost of a hot dog as x and the cost of a bag of potato chips as y.

The first transaction can be represented by the equation:

4x + 5y = 16.75

This equation states that the total cost of 4 hot dogs and 5 bags of potato chips is $16.75.

The second transaction can be represented by the equation:

3x + 3y = 11.25

This equation states that the total cost of 3 hot dogs and 3 bags of potato chips is $11.25.

Solving the System of Equations

We now have a system of two equations with two variables. To solve this system, we can use the method of substitution or elimination. In this case, we will use the elimination method.

First, let's multiply the second equation by 5 to make the coefficients of y in both equations equal:

15x + 15y = 56.25

Now, we can subtract the first equation from this new equation to eliminate the variable y:

(15x + 15y) - (4x + 5y) = 56.25 - 16.75 11x + 10y = 39.50

Next, let's multiply the first equation by 3 to make the coefficients of x in both equations equal:

12x + 15y = 50.25

Now, we can subtract the second equation from this new equation to eliminate the variable x:

(12x + 15y) - (3x + 3y) = 50.25 - 11.25 9x + 12y = 39.00

Finding the Cost of Each Item

We now have two equations with two variables:

11x + 10y = 39.50 9x + 12y = 39.00

We can solve this system of equations by multiplying the first equation by 12 and the second equation by 10:

132x + 120y = 475.00 90x + 120y = 390.00

Now, we can subtract the second equation from this new equation to eliminate the variable y:

(132x + 120y) - (90x + 120y) = 475.00 - 390.00 42x = 85.00

Dividing both sides by 42, we get:

x = 2.02

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

4x + 5y = 16.75 4(2.02) + 5y = 16.75 8.08 + 5y = 16.75

Subtracting 8.08 from both sides, we get:

5y = 8.67

Dividing both sides by 5, we get:

y = 1.73

Conclusion

In this article, we used algebraic equations to solve a real-world problem involving the cost of hot dogs and potato chips. We set up two equations based on the information provided in the transactions and used the elimination method to solve the system of equations. We found that the cost of a hot dog is $2.02 and the cost of a bag of potato chips is $1.73.

Discussion

This problem is a great example of how mathematics is used in everyday life. By using algebraic equations, we can solve real-world problems and make informed decisions. The cost of hot dogs and potato chips may seem like a simple problem, but it requires a deep understanding of algebraic equations and the ability to solve systems of equations.

References

  • [1] "Algebraic Equations" by Math Open Reference
  • [2] "Systems of Equations" by Khan Academy

Keywords

  • Algebraic equations
  • Systems of equations
  • Real-world problems
  • Hot dogs
  • Potato chips
  • Cost
  • Mathematics
    Frequently Asked Questions: Solving the Cost of Hot Dogs and Potato Chips ====================================================================

Q: What is the main goal of the problem?

A: The main goal of the problem is to find the cost of each item, denoted by the variables x and y, where x represents the cost of a hot dog and y represents the cost of a bag of potato chips.

Q: How many transactions are given in the problem?

A: There are two transactions given in the problem. The first transaction involves a customer buying 4 hot dogs and 5 bags of potato chips for $16.75, and the second transaction involves another customer buying 3 hot dogs and 3 bags of potato chips for $11.25.

Q: What type of equations are used to solve the problem?

A: Algebraic equations are used to solve the problem. Specifically, a system of two equations with two variables is set up to represent the two transactions.

Q: What method is used to solve the system of equations?

A: The elimination method is used to solve the system of equations. This involves multiplying the equations by necessary multiples such that the coefficients of one of the variables are the same in both equations, and then subtracting one equation from the other to eliminate that variable.

Q: What is the cost of a hot dog?

A: The cost of a hot dog is $2.02.

Q: What is the cost of a bag of potato chips?

A: The cost of a bag of potato chips is $1.73.

Q: How can this problem be applied to real-life situations?

A: This problem can be applied to real-life situations where you need to find the cost of multiple items based on a set of transactions. For example, if you are a store owner and you want to know the cost of each item in your store, you can use this method to solve the problem.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

  • Not setting up the equations correctly
  • Not using the correct method to solve the system of equations
  • Not checking the solutions for consistency
  • Not considering the possibility of multiple solutions

Q: How can you verify the solutions to this problem?

A: You can verify the solutions to this problem by plugging the values of x and y back into the original equations and checking if they are true.

Q: What are some extensions of this problem?

A: Some extensions of this problem include:

  • Adding more transactions to the problem
  • Using different methods to solve the system of equations
  • Considering the possibility of multiple solutions
  • Applying the problem to real-life situations where there are multiple items and transactions.

Q: What are some common applications of this problem?

A: Some common applications of this problem include:

  • Store management
  • Inventory control
  • Cost accounting
  • Financial analysis.

Keywords

  • Algebraic equations
  • Systems of equations
  • Real-world problems
  • Hot dogs
  • Potato chips
  • Cost
  • Mathematics
  • Store management
  • Inventory control
  • Cost accounting
  • Financial analysis