A teacher is choosing between two options for a class field trip to an orchard.

Question

At orchard A, admission costs $9 per person, and 3 chaperones are required.
At Orchard B, the cost is $10 per person, but only 1 chaperone is required
At each orchard, the same price applies to both chaperones and students

To help her compare the cost of her two options, the teacher first writes the equation 9 (n + 3) = 10 (n + 1), and then she writes the inequality 9 (n + 3) < 10 (n + 1)

Answer 1

To analyze the teacher's equations, let's break down the problem and solve the equations and inequality provided.

Given:

1. Orchard A: Admission is $9 per person, with 3 required chaperones.

   • Total cost for Orchard A: \( 9(n + 3) \)
   • Where \( n \) is the number of students.

2. Orchard B: Admission is $10 per person, with 1 required chaperone.

   • Total cost for Orchard B: \( 10(n + 1) \)

Exploring the Equation:

The equation given by the teacher: \[ 9(n + 3) = 10(n + 1) \]

Let's solve it step by step.

1. Expand both sides: \[ 9n + 27 = 10n + 10 \]

2. Rearranging the equation: \[ 27 - 10 = 10n - 9n \] \[ 17 = n \]

So, at \( n = 17 \), the costs of both orchards are equal.

Exploring the Inequality:

Next, let's consider the inequality: \[ 9(n + 3) < 10(n + 1) \]

1. Expand both sides: \[ 9n + 27 < 10n + 10 \]

2. Rearranging the inequality: \[ 27 - 10 < 10n - 9n \] \[ 17 < n \]

This means that for \( n > 17 \), the cost of Orchard A (with \( n + 3 \) chaperones) will be less than the cost of Orchard B (with \( n + 1 \) chaperone).

Summary:

• The costs for both orchards are equal when \( n = 17 \).
• For more than 17 students (\( n > 17 \)), Orchard A becomes the cheaper option.
• For exactly 17 students, the costs are the same.
• For fewer than 17 students, Orchard B is the cheaper option.

This analysis can help the teacher make a more informed decision based on how many students are participating in the field trip.