x + y = 28

Question

y = 2x - 5

y = Marci; x = Jennifer

Jennifer and Marci are selling cookies for a fundraiser. They sold 28 boxes of cookies together. Marci sold 5 less than twice the number that Jennifer sold. How many boxes of cookies did each girl sell? Use system of equations to solve the problem.
Responses
A Marci sold 11 boxes and Jennifer sold 17 boxes.Marci sold 11 boxes and Jennifer sold 17 boxes.
B Marci sold 13 boxes and Jennifer sold 15 boxes.Marci sold 13 boxes and Jennifer sold 15 boxes.
C Marci sold 15 boxes and Jennifer sold 13 boxes.Marci sold 15 boxes and Jennifer sold 13 boxes.
D Marci sold 17 boxes and Jennifer sold 11 boxes.Marci sold 17 boxes and Jennifer sold 11 boxes.
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Answer 1

To solve the problem, we have the following system of equations based on the information given:

1. \( x + y = 28 \) (The total number of boxes sold by Jennifer and Marci)
2. \( y = 2x - 5 \) (Marci sold 5 less than twice the number sold by Jennifer)

Now, we will substitute the second equation into the first.

Substituting \( y = 2x - 5 \) into \( x + y = 28 \):

\[ x + (2x - 5) = 28 \]

Combine like terms:

\[ 3x - 5 = 28 \]

Add 5 to both sides:

\[ 3x = 33 \]

Now, divide by 3:

\[ x = 11 \]

Now that we have \( x \) (the number of boxes sold by Jennifer), we can find \( y \) (the number sold by Marci) using the first equation:

\[ y = 28 - x = 28 - 11 = 17 \]

So, Jennifer sold 11 boxes and Marci sold 17 boxes.

Thus, the final answer is:

D: Marci sold 17 boxes and Jennifer sold 11 boxes.