Edwin bought 3 cd's and 7 dvds for 80$. Jessica bought 6 cds and 3 dvds for 63.75$. Write a system of linear equations that could be used to deterime the cost of one cd and the cost of one dvd. Solve the linear equation & check your work.
Let's denote the cost of one CD as "c" and the cost of one DVD as "d".
Based on the given information, we can create the following system of linear equations:
Equation 1: 3c + 7d = 80
This equation represents Edwin's purchase of 3 CDs and 7 DVDs amounting to $80.
Equation 2: 6c + 3d = 63.75
This equation represents Jessica's purchase of 6 CDs and 3 DVDs amounting to $63.75.
Now, let's solve this system of equations:
Multiplying Equation 1 by 3 and Equation 2 by 7 to eliminate c:
9c + 21d = 240
42c + 21d = 446.25
Subtracting Equation 1 from Equation 2:
42c + 21d - (9c + 21d) = 446.25 - 240
42c + 21d - 9c - 21d = 206.25
33c = 206.25
Dividing both sides of the equation by 33:
c = 206.25 / 33
c ≈ $6.25
Substituting c back into Equation 1 to find d:
3(6.25) + 7d = 80
18.75 + 7d = 80
7d = 80 - 18.75
7d = 61.25
Dividing both sides of the equation by 7:
d = 61.25 / 7
d ≈ $8.75
Therefore, the cost of one CD (c) is approximately $6.25, and the cost of one DVD (d) is approximately $8.75.
Let's check the solution by substituting the values of c and d into the original equations:
Equation 1: 3c + 7d = 80
3(6.25) + 7(8.75) = 80
18.75 + 61.25 = 80
80 = 80 (True)
Equation 2: 6c + 3d = 63.75
6(6.25) + 3(8.75) = 63.75
37.5 + 26.25 = 63.75
63.75 = 63.75 (True)
The solution satisfies both equations, which confirms that our values for c and d are correct.