U4A Discussion: Two Variable Equations and... Below are your two lines. Total Cost 50 40- -30- -20- -10 #21. What is the solution to this system of equations? 0 < 34 of 36 Next > Edit my response Your combined table for both companies is below: Total Cost for Kites- Total Cost for Windy Kites Purchased R-Fun Kites 0 $10 $16 1 $16 $20 4 7 10 0 5 10 Number of kites

1 answer

To find the solution to your system of equations, let's set up the equations based on the given information.

From your description, you have a "Total Cost" for "R-Fun Kites" and "Windy Kites." It appears that the costs depend on the number of kites purchased.

Let's create the equations based on your data:

  1. R-Fun Kites Cost:

    • Total Cost (R-Fun) = $10 * (Number of R-Fun Kites) + Fixed Cost
  2. Windy Kites Cost:

    • Total Cost (Windy) = $16 * (Number of Windy Kites) + Fixed Cost

Given Table:

  • Distribution of Kites and their costs:
    • R-Fun Kites:
      • 0 kites: $10
      • 1 kite: $16
      • 4 kites: $40
      • 7 kites: $70
      • 10 kites: $100
    • Windy Kites:
      • 0 kites: $16
      • 1 kite: $20
      • 4 kites: $40
      • 7 kites: $56
      • 10 kites: $76

Finding the Intersection:

To find the solution to the system of equations, we will see where the cost to purchase the same number of kites from both companies becomes equal.

Let's analyze the costs:

  1. For purchasing 0 kites:

    • R-Fun: $10
    • Windy: $16
  2. For purchasing 1 kite:

    • R-Fun: $16
    • Windy: $20
  3. For purchasing 4 kites:

    • R-Fun: $40
    • Windy: $40
  4. For purchasing 7 kites:

    • R-Fun: $70
    • Windy: $56
  5. For purchasing 10 kites:

    • R-Fun: $100
    • Windy: $76

The Solution:

From the table above, we can observe that at 4 kites, the total cost for both R-Fun Kites and Windy Kites is $40. Therefore, the solution to the system of equations (the point where their costs are equal) is at:

Number of Kites: 4; Total Cost: $40

Let me know if you need any additional help!