Question

1. mp modeling real life a drama club earns $1040 from a production by selling 64 adult tickets and 132 student tickets. an adult ticket costs twice as much as a student ticket.

write a system of linear equations that represents this situation.
b. what is the cost of each ticket?

1. open - ended write a system of linear equations that has the ordered pair (1, 6) as its solution.

choosing a solution method solve the system. explain your choice of method.

1. $y - x = 4$

$x + y = 6$

1. $0.5x + y = 4$

$0.5x - y = -1$

1. $y = 2x + 5$

$y = -3x$

1. critical thinking a system consists of two different proportional relationships.

what is the solution of the system? justify your answer.

Explanation:

Problem 17

Let \(x\) = cost of student ticket, \(y\) = cost of adult ticket.

Step1: Define variables & total earnings

Total revenue from tickets:
\(64y + 132x = 1040\)

Step2: Relate adult/student ticket cost

Adult ticket is twice student ticket:
\(y = 2x\)

Step3: Substitute \(y=2x\) into first equation

Substitute to solve for \(x\):
\(64(2x) + 132x = 1040\)
\(128x + 132x = 1040\)
\(260x = 1040\)

Step4: Solve for \(x\)

Calculate student ticket cost:
\(x = \frac{1040}{260} = 4\)

Step5: Solve for \(y\)

Use \(y=2x\) to find adult cost:
\(y = 2(4) = 8\)

Create linear equations where substituting \(x=1, y=6\) makes the equation true. Two simple equations can be formed using basic operations like addition and multiplication.

Use elimination (add equations to eliminate \(x\)).

Step1: Add the two equations

Eliminate \(x\) to solve for \(y\):
\((y - x) + (x + y) = 4 + 6\)
\(2y = 10\)

Step2: Solve for \(y\)

Calculate \(y\) value:
\(y = \frac{10}{2} = 5\)

Step3: Substitute \(y=5\) into \(x+y=6\)

Solve for \(x\):
\(x + 5 = 6\)
\(x = 6 - 5 = 1\)

Answer:

a. System of equations:
\(64y + 132x = 1040\)
\(y = 2x\)
b. Student ticket: \$4, Adult ticket: \$8

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Problem 18