Question

select the correct answer.
the total cost of a vacation rental depends on when it is being rented: during the week, on the weekend, or over a holiday. the total cost for the three different groups to stay at the rental is shown in the table.

	group #1	group #2	group #3
weekend nights	2	1	1
holiday	0	1	1
total cost	$614	$555	$733

create and solve a system of linear equations to represent this situation. then select the true statement.

a. the solution to this system is viable.

b. the solution to this system is nonviable because it results in a fractional amount of money.

c. the solution to this system is nonviable because it results in a fractional number of nights.

d. the solution to this system is nonviable because it results in a negative amount of money.

Explanation:

Step1: Define Variables

Let \( x \) = cost per weeknight, \( y \) = cost per weekend night, \( z \) = cost per holiday night.

Step2: Set Up Equations

For Group #1: \( 4x + 2y + 0z = 614 \) → \( 4x + 2y = 614 \) (Equation 1)
For Group #2: \( 3x + 1y + 1z = 555 \) (Equation 2)
For Group #3: \( 5x + 1y + 1z = 733 \) (Equation 3)

Step3: Eliminate \( z \) (Equation 2 & 3)

Subtract Equation 2 from Equation 3:
\( (5x + y + z) - (3x + y + z) = 733 - 555 \)
\( 2x = 178 \) → \( x = 89 \)

Step4: Solve for \( y \) (Equation 1)

Substitute \( x = 89 \) into Equation 1:
\( 4(89) + 2y = 614 \)
\( 356 + 2y = 614 \)
\( 2y = 258 \) → \( y = 129 \)

Step5: Solve for \( z \) (Equation 2)

Substitute \( x = 89 \), \( y = 129 \) into Equation 2:
\( 3(89) + 129 + z = 555 \)
\( 267 + 129 + z = 555 \)
\( 396 + z = 555 \) → \( z = 159 \)

Step6: Analyze Viability

Costs \( x = 89 \), \( y = 129 \), \( z = 159 \) are positive, so the solution is viable.

Answer:

A. The solution to this system is viable.