Math Problem Statement

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.

Solution

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

  • Let xx represent the number of front orchestra seats sold.
  • Let yy represent the number of rear orchestra seats sold.
  • Let zz 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=1000x + 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+400x + 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=63,60080x + 60y + 50z = 63,600

System of equations:

  1. x+y+z=1000x + y + z = 1000
  2. x+y=2z+400x + y = 2z + 400
  3. 80x+60y+50z=63,60080x + 60y + 50z = 63,600

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

Step 1: Solve for x+yx + y using equation (2)

From equation (2): x+y=2z+400x + y = 2z + 400 We'll substitute this expression for x+yx + y into equations (1) and (3).

Step 2: Substitute x+y=2z+400x + y = 2z + 400 into equation (1)

Substitute into equation (1): (2z+400)+z=1000(2z + 400) + z = 1000 Simplify: 3z+400=10003z + 400 = 1000 3z=6003z = 600 z=200z = 200 So, 200 front balcony tickets were sold.

Step 3: Substitute z=200z = 200 into equation (2)

Substitute z=200z = 200 into equation (2): x+y=2(200)+400=800x + y = 2(200) + 400 = 800 So, x+y=800x + y = 800.

Step 4: Substitute x+y=800x + y = 800 and z=200z = 200 into equation (3)

Substitute into equation (3): 80x+60y+50(200)=63,60080x + 60y + 50(200) = 63,600 Simplify: 80x+60y+10,000=63,60080x + 60y + 10,000 = 63,600 80x+60y=53,60080x + 60y = 53,600

Step 5: Solve the system of two equations

We now have the system:

  1. x+y=800x + y = 800
  2. 80x+60y=53,60080x + 60y = 53,600

Solve the first equation for yy: y=800xy = 800 - x

Substitute y=800xy = 800 - x into the second equation: 80x+60(800x)=53,60080x + 60(800 - x) = 53,600 Simplify: 80x+48,00060x=53,60080x + 48,000 - 60x = 53,600 20x=5,60020x = 5,600 x=280x = 280 So, 280 front orchestra tickets were sold.

Step 6: Solve for yy

Substitute x=280x = 280 into y=800xy = 800 - x: y=800280=520y = 800 - 280 = 520 So, 520 rear orchestra tickets were sold.

Final Answer:

  • Front orchestra tickets sold: x=280x = 280
  • Rear orchestra tickets sold: y=520y = 520
  • Front balcony tickets sold: z=200z = 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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Equations

Formulas

x + y + z = 1000
x + y = 2z + 400
80x + 60y + 50z = 63,600

Theorems

Linear System Solving
Substitution Method

Suitable Grade Level

Grades 9-12