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 $8.75.
Let's check the solution by substituting the values of c and d into the original equations:
Equation 1: 3c + 7d = 80
3(6.25) + 7(8.75) = 80
18.75 + 61.25 = 80
80 = 80 (True)
Equation 2: 6c + 3d = 63.75
6(6.25) + 3(8.75) = 63.75
37.5 + 26.25 = 63.75
63.75 = 63.75 (True)
The solution satisfies both equations, which confirms that our values for c and d are correct.
Edwin bought 3 cd's and 7 dvds for 80$. 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.
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