Real-World Systems of Equations Quick Check

Question

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Question
The senior class at Hartford High School is planning their prom. They’ve narrowed the location down to two choices. The Regal Hotel charges, C
, $500 for the ballroom rental fee and $75 per student, s
, for the meal. The Greenhill Country Club charges, C
, $1,000 for the ballroom rental fee and $65 per student, s
, for the meal. Create a pair of linear equations that represent this real-world problem.(1 point)
Responses

Regal Hotel: C=500s+75
; Greenhill Country Club: C=1,000s+65
Regal Hotel: upper C equals 500 s plus 75 ; Greenhill Country Club: upper C equals 1,000 s plus 65

Regal Hotel: C=75s+1,000
; Greenhill Country Club: C=65s+500
Regal Hotel: upper C equals 75 s plus 1,000 ; Greenhill Country Club: upper C equals 65 s plus 500

Regal Hotel: C=1,000s+75
; Greenhill Country Club: C=500s+65
Regal Hotel: upper C equals 1,000 s plus 75 ; Greenhill Country Club: upper C equals 500 s plus 65

Regal Hotel: C=75s+500
; Greenhill Country Club: C=65s+1,000

Answer 1

To set up the equations, we need to define the total cost \( C \) as a function of the number of students \( s \).

For the Regal Hotel, the total cost consists of a fixed rental fee of $500 plus $75 per student, so the equation is: \[ C = 500 + 75s \]

For the Greenhill Country Club, the total cost consists of a fixed rental fee of $1,000 plus $65 per student, so the equation is: \[ C = 1000 + 65s \]

Therefore, the correct pair of linear equations representing this real-world problem is:

• Regal Hotel: \( C = 75s + 500 \)
• Greenhill Country Club: \( C = 65s + 1000 \)

None of the provided options match this formulation directly in terms of the variable placement, but the closest correct relationship based on the fixed costs and cost per student is:

Regal Hotel: C = 75s + 500; Greenhill Country Club: C = 65s + 1000 (Note that I flipped the terms to reflect the constant and variable correctly).