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.

Answer 1

To solve the system of equations, we can substitute the value of y from the second equation into the first equation:

x + (2x - 5) = 28
3x - 5 = 28
3x = 33
x = 11

Substituting the x value back into the second equation to find y:

y = 2(11) - 5
y = 17

So Jennifer sold 11 boxes and Marci sold 17 boxes.

The answer is option D: Marci sold 17 boxes and Jennifer sold 11 boxes.

Answer 2

A discount store is selling 5 small tables with 8 chairs for $115. Three tables with 5 chairs cost $70.
Question 1
Which system of linear equations could be used to find the cost of each table (x) and the cost of each chair (y)?
Responses
A 3x + 8y = $70; 8x + 3y = $1153x + 8y = $70; 8x + 3y = $115
B 5x + 8y = $115; 3x + 5y = $705x + 8y = $115; 3x + 5y = $70
C 5x + 8y = $115; 2x − 5y = $705x + 8y = $115; 2x − 5y = $70
D 8x + 5y = $115; 5x + 3y = $708x + 5y = $115; 5x + 3y = $70
Question 2
Determine the cost of each table (x) and the cost of each chair (y).
Responses
A x = $15; y = $5x = $15; y = $5
B x = $10; y = $5x = $10; y = $5
C x = $5; y = $10x = $5; y = $10
D x = $12; y = $3