Formulate a system of equations for the situation below and solve.

For the opening night at the Opera House, a total of 1000 tickets were sold. Front orchestra seats cost $80 apiece, rear orchestra seats cost $60 apiece, and front balcony seats cost $50 apiece. The combined number of tickets sold for the front orchestra and rear orchestra exceeded twice the number of front balcony tickets sold by 400. The total receipts for the performance were $63,600. Determine how many tickets of each type were sold.

To solve this problem, let's define the variables for the number of tickets sold for each section:

- Let $ x $ represent the number of front orchestra seats sold.
- Let $ y $ represent the number of rear orchestra seats sold.
- Let $ z $ represent the number of front balcony seats sold.

Now, based on the information provided in the problem, we can formulate the following system of equations:

### Equation 1: Total number of tickets sold
The total number of tickets sold is 1000, so:
\[
x + y + z = 1000
\]

### Equation 2: Relationship between front orchestra, rear orchestra, and front balcony tickets
The combined number of front and rear orchestra tickets exceeds twice the number of front balcony tickets by 400. This gives:
\[
x + y = 2z + 400
\]

### Equation 3: Total receipts
The total receipts for the performance were $63,600. Each ticket price for the different sections is provided, so:
\[
80x + 60y + 50z = 63,600
\]

### System of equations:
1. $ x + y + z = 1000 $
2. $ x + y = 2z + 400 $
3. $ 80x + 60y + 50z = 63,600 $

Now, let's solve this system step by step.

#### Step 1: Solve for $ x + y $ using equation (2)
From equation (2):
\[
x + y = 2z + 400
\]
We'll substitute this expression for $ x + y $ into equations (1) and (3).

#### Step 2: Substitute $ x + y = 2z + 400 $ into equation (1)
Substitute into equation (1):
\[
(2z + 400) + z = 1000
\]
Simplify:
\[
3z + 400 = 1000
\]
\[
3z = 600
\]
\[
z = 200
\]
So, 200 front balcony tickets were sold.

#### Step 3: Substitute $ z = 200 $ into equation (2)
Substitute $ z = 200 $ into equation (2):
\[
x + y = 2(200) + 400 = 800
\]
So, $ x + y = 800 $.

#### Step 4: Substitute $ x + y = 800 $ and $ z = 200 $ into equation (3)
Substitute into equation (3):
\[
80x + 60y + 50(200) = 63,600
\]
Simplify:
\[
80x + 60y + 10,000 = 63,600
\]
\[
80x + 60y = 53,600
\]

#### Step 5: Solve the system of two equations
We now have the system:
1. $ x + y = 800 $
2. $ 80x + 60y = 53,600 $

Solve the first equation for $ y $:
\[
y = 800 - x
\]

Substitute $ y = 800 - x $ into the second equation:
\[
80x + 60(800 - x) = 53,600
\]
Simplify:
\[
80x + 48,000 - 60x = 53,600
\]
\[
20x = 5,600
\]
\[
x = 280
\]
So, 280 front orchestra tickets were sold.

#### Step 6: Solve for $ y $
Substitute $ x = 280 $ into $ y = 800 - x $:
\[
y = 800 - 280 = 520
\]
So, 520 rear orchestra tickets were sold.

### Final Answer:
- Front orchestra tickets sold: $ x = 280 $
- Rear orchestra tickets sold: $ y = 520 $
- Front balcony tickets sold: $ z = 200 $

Would you like more details on the solution process or further clarification?

### 5 Related Questions:
1. How can we solve this system of equations using matrices?
2. How would the solution change if the total receipts were higher or lower?
3. What if the number of front orchestra tickets was capped at 300?
4. How could we verify the solution by plugging it back into all equations?
5. What would the system look like if there were an additional ticket type?

### Tip:
Always double-check your system of equations by interpreting the problem step-by-step before solving.

Learn how to solve a system of equations to determine ticket sales for an opera house. This problem involves algebraic concepts, including systems of linear equations, substitution method, and handling real-world scenarios. Suitable for students in Grades 9-12.

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