Let's denote the number of students in High School A after \( t \) years as \( A(t) \) and the number of students in High School B after \( t \) years as \( B(t) \).
Step 1: Define the functions
High School A: \[ A(t) = 900 + 50t \] High School B: \[ B(t) = 700 + 100t \]
Step 2: Calculate the number of students in 2 years
To find out how many students each high school is projected to have in 2 years, we can substitute \( t = 2 \) into both functions.
For High School A: \[ A(2) = 900 + 50 \times 2 = 900 + 100 = 1000 \]
For High School B: \[ B(2) = 700 + 100 \times 2 = 700 + 200 = 900 \]
In 2 years:
- High School A will have 1000 students.
- High School B will have 900 students.
Conclusion at \( t = 2 \):
High School A is projected to have more students than High School B in 2 years.
Step 3: Determine when the number of students will be equal (intercepts)
To find out when the number of students in each school will be equal, set \( A(t) \) equal to \( B(t) \): \[ 900 + 50t = 700 + 100t \]
Step 4: Solve for \( t \)
Rearranging the equation: \[ 900 - 700 = 100t - 50t \] \[ 200 = 50t \] \[ t = \frac{200}{50} = 4 \]
Step 5: Calculate the number of students when they intersect
Now we can find out how many students there will be at that time. Substitute \( t = 4 \) into either function: \[ A(4) = 900 + 50 \times 4 = 900 + 200 = 1100 \]
And for High School B: \[ B(4) = 700 + 100 \times 4 = 700 + 400 = 1100 \]
Conclusion for intercept coordinates:
The two high schools are projected to have the same number of students (1100 students) in 4 years.
Summary
- In 2 years, High School A has 1000 students, and High School B has 900 students.
- The intercept occurs at \( t = 4 \) years, where both high schools will have 1100 students.
The coordinates of the intercept are \( (4, 1100) \).
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