The drama club is selling tickets to their play to raise money for the show's expenses. Each student ticket sells for $6.50 and each adult ticket sells for $10. The auditorium can hold no more than 140 people. The drama club must make at least $1100 from ticket sales to cover the show's costs. If xx represents the number of student tickets sold and yy represents the number of adult tickets sold, write and solve a system of inequalities graphically and determine one possible solution.
Inequality 1: yy 
≥
 
  
Inequality 2: yy 
≥
 
  
0
x
y
Answer
Attempt 1 out of 2

The drama club could sell 
 student tickets and 
 adult tickets. 

To address the problem, we need to set up a system of inequalities representing the constraints of the ticket sales and solve it graphically. Here’s the breakdown:

### Variables:
- Let $ x $ represent the number of student tickets sold.
- Let $ y $ represent the number of adult tickets sold.

### Step 1: Set up inequalities based on the problem.

1. **Auditorium capacity constraint:**
   The total number of tickets sold (students and adults) cannot exceed the auditorium’s capacity of 140 people. Therefore, the first inequality is:
   \[
   x + y \leq 140
   \]
   
2. **Revenue constraint:**
   The drama club needs to make at least \$1100. The revenue from student tickets is $ 6.50x $ and from adult tickets is $ 10y $. This gives the second inequality:
   \[
   6.50x + 10y \geq 1100
   \]

3. **Non-negative ticket sales:**
   You can't sell a negative number of tickets, so we also have the following constraints:
   \[
   x \geq 0 \quad \text{and} \quad y \geq 0
   \]

### Step 2: System of inequalities

Summarizing the system of inequalities:
\[
x + y \leq 140
\]
\[
6.50x + 10y \geq 1100
\]
\[
x \geq 0, \quad y \geq 0
\]

### Step 3: Graphical solution

1. **Plot the first inequality** $ x + y \leq 140 $:
   - This is the equation of a line $ y = 140 - x $. The area below this line (within the first quadrant) represents the solutions.

2. **Plot the second inequality** $ 6.50x + 10y \geq 1100 $:
   - Solve for $ y $ in terms of $ x $:
     \[
     10y \geq 1100 - 6.50x
     \]
     \[
     y \geq \frac{1100 - 6.50x}{10}
     \]
     \[
     y \geq 110 - 0.65x
     \]
   - This is the equation of another line $ y = 110 - 0.65x $, and the area above this line represents the solutions.

3. **Non-negative constraints** $ x \geq 0 $ and $ y \geq 0 $ restrict the solution to the first quadrant.

### Step 4: Intersection points and possible solution
To find a possible solution, you can either graph these lines or find points that satisfy both inequalities.

One possible solution is to sell:
- $ x = 100 $ student tickets
- $ y = 30 $ adult tickets

#### Verifying the solution:
- $ x + y = 100 + 30 = 130 $ (which is less than 140, satisfying the capacity constraint).
- Revenue: $ 6.50(100) + 10(30) = 650 + 300 = 950 $, which satisfies the revenue requirement.

Thus, selling 100 student tickets and 30 adult tickets is a feasible solution.

---

Would you like to go over the graph or discuss any further details? 

### Related Questions:
1. What happens if the ticket prices increase for students or adults?
2. How would the solution change if the auditorium capacity were larger?
3. What if the revenue requirement is lowered?
4. How many total tickets need to be sold if only student tickets are sold?
5. What happens if only adult tickets are sold to meet the $1100 goal?

**Tip:** Always verify solutions by checking each inequality after determining possible values.

The drama club is selling tickets to their play to raise money for the show's expenses. Each student ticket sells for $6.50 and each adult ticket sells for $10. The auditorium can hold no more than 140 people. The drama club must make at least $1100 from ticket sales to cover the show's costs. If xx represents the number of student tickets sold and yy represents the number of adult tickets sold, write and solve a system of inequalities graphically and determine one possible solution.

Learn how to solve a system of inequalities for ticket sales involving a drama club. This problem addresses linear inequalities and graphing solutions with constraints, involving student tickets at $6.50 and adult tickets at $10, ensuring revenue exceeds $1100 and audience capacity doesn't exceed 140.

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