Ticket Sales: A theater sells adult tickets for $15 and child tickets for $10. If they sell 500 tickets for $6000, how many of each kind were sold? (Represent with equations and solve.)

Imagine the spotlight shining on a packed theater, buzzing with excitement. But behind the scenes, the box office is facing a mathematical mystery: How many adult and child tickets were sold to reach a total of $6000?

This isn’t just a drama for aspiring actors, it’s a prime opportunity to flex your algebra muscles! Grab your pencils and join us on a journey of equations and solutions.

**The Clues:**

- Adult tickets cost $15, while child tickets are a sweet deal at $10.
- The grand total for 500 tickets sold is a whopping $6000.

**The Equation Detective:**

Our mission is to find two unknowns: the number of adult tickets (let’s call them **a**) and the number of child tickets (let’s call them **c**). To do this, we need to craft two equations based on the clues:

**Total Tickets:**This is a simple addition problem. a + c = 500 (adults + children = total tickets).**Total Revenue:**Multiplying the number of tickets by their price gives us the revenue. 15a + 10c = 6000 (adult price * adult tickets + child price * child tickets = total revenue).

**The Grand Unveiling:**

Now, we have two equations and two unknowns, but how do we find the missing pieces? There are two paths to uncover the solution:

**Path 1: The Substitution Sleuth:**

- We can solve the first equation for c (the child tickets) to get c = 500 – a.
- Substituting this expression for c in the second equation, we get a tricky math puzzle: 15a + 10(500 – a) = 6000.
- Solving this equation for a reveals our first clue: 200 adult tickets were sold!
- Finally, plugging 200 back into the first equation, we solve for c: 300 child tickets!

**Path 2: The Elimination Expert:**

- We can multiply the first equation by -10 to get -10a – 10c = -5000.
- Adding this equation to the second one, voila! The c terms vanish, leaving us with a neat 5a = 1000.
- Solving for a again, we find the same answer: 200 adult tickets.
- The path to find child tickets remains the same: using 200 in the first equation, we get c = 300.

**Curtain Call:**

The stage is set, the mystery is solved! The theater sold 200 adult tickets and 300 child tickets, generating a $6000 jackpot. This journey through equations proves that math plays a starring role in even the most exciting of ticket sales dramas.

But the show doesn’t end here! Remember, this method of cracking the code with equations can be applied to countless similar scenarios. So, the next time you face a real-world puzzle involving quantities and costs, don’t forget your algebra! With a little bit of equation magic, you too can become a champion problem solver!