To solve the problem, we can define the equations for the students in both high schools over time \( t \), where \( t \) is measured in years.
High School A:
- Current number of students: 900
- Growth rate: 50 students per year
So the equation for the number of students in High School A after \( t \) years, \( A(t) \), can be written as: \[ A(t) = 900 + 50t \]
High School B:
- Current number of students: 700
- Growth rate: 100 students per year
So the equation for the number of students in High School B after \( t \) years, \( B(t) \), is: \[ B(t) = 700 + 100t \]
Determine the Number of Students After 2 Years
Now we will calculate the number of students in each school after 2 years.
For High School A: \[ A(2) = 900 + 50 \times 2 = 900 + 100 = 1000 \]
For High School B: \[ B(2) = 700 + 100 \times 2 = 700 + 200 = 900 \]
Conclusion After 2 Years:
- High School A will have 1000 students.
- High School B will have 900 students.
Therefore, High School A is projected to have more students than High School B after 2 years.
Finding When the Two Schools Intercept
To find when the two schools will have the same number of students, we set \( A(t) \) equal to \( B(t) \):
\[ 900 + 50t = 700 + 100t \]
Now, we can solve for \( t \):
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Subtract \( 50t \) from both sides: \[ 900 = 700 + 50t \]
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Subtract 700 from both sides: \[ 200 = 50t \]
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Divide both sides by 50: \[ t = \frac{200}{50} = 4 \]
Conclusion
The two high schools will intercept in 4 years.
Summary:
- After 2 years, High School A will have 1000 students, and High School B will have 900 students.
- High School A is larger at that time.
- The two schools will have the same number of students in 4 years.