Question 2 (Essay Worth 10 points) ( \( 05.01,05.02,05.03 \mathrm{MC} \) ) An office manager orders one calculator or one calendar for each of the office's 60 employees. Each calculator costs \( \$ 15 \), and each calendar costs \( \$ 10 \). The entire order totaled \( \$ 800 \). Part A: Write the system of equations that models this scenario. ( 5 points) Part B: Use substitution method or elimination method to determine the number of calculators and calendars ordered. Show all necessary steps. ( 5 points)

To tackle this problem effectively, let's break it down! The office manager has ordered two types of items: calculators and calendars. We'll let \( c \) represent the number of calculators and \( d \) represent the number of calendars. From the problem, we can derive two equations based on the information given. To summarize: - The total number of items (calculators and calendars) is 60: \[ c + d = 60 \] - The total cost for calculators (\$15 each) and calendars (\$10 each) is \$800: \[ 15c + 10d = 800 \] Now we have our system of equations: 1. \( c + d = 60 \) 2. \( 15c + 10d = 800 \) For Part B, let’s use the substitution method to solve this. Step 1: Solve the first equation for \( d \): \[ d = 60 - c \] Step 2: Substitute this expression for \( d \) into the second equation: \[ 15c + 10(60 - c) = 800 \] Step 3: Simplify and solve for \( c \): \[ 15c + 600 - 10c = 800 \] \[ 5c + 600 = 800 \] \[ 5c = 800 - 600 \] \[ 5c = 200 \] \[ c = \frac{200}{5} = 40 \] Step 4: Substitute \( c = 40 \) back into the first equation to find \( d \): \[ 40 + d = 60 \] \[ d = 60 - 40 = 20 \] So, the office manager ordered **40 calculators** and **20 calendars**. Remember: to double-check your solution, you can plug these values back into the original equations to ensure they satisfy both conditions!