Certainly! Let's go through the step-by-step process to solve this problem.
### Problem Statement:
In 2005, College A has 13,300 students, and the student enrollment is projected to increase by 1,000 students per year. College B has 26,800 students, with a projected enrollment decline of 500 students per year. We need to find the year when both colleges will have the same number of students and determine what that enrollment number will be.
### Step-by-Step Solution:
Step 1: Define the variables and expressions
Let [tex]\( x \)[/tex] be the number of years after 2005.
- Enrollment in College A after [tex]\( x \)[/tex] years:
[tex]\[
13300 + 1000x
\][/tex]
- Enrollment in College B after [tex]\( x \)[/tex] years:
[tex]\[
26800 - 500x
\][/tex]
Step 2: Set up the equation for equal enrollment
We want to find the value of [tex]\( x \)[/tex] when the enrollment in both colleges is the same. Thus, we set the expressions equal to each other:
[tex]\[
13300 + 1000x = 26800 - 500x
\][/tex]
Step 3: Solve for [tex]\( x \)[/tex]
1. Combine like terms:
[tex]\[
1000x + 500x = 26800 - 13300
\][/tex]
2. Simplify:
[tex]\[
1500x = 13500
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{13500}{1500} = 9
\][/tex]
So, the number of years after 2005 when the enrollments are equal is 9 years.
Step 4: Find the specific year
To find the specific year when this occurs:
[tex]\[
2005 + 9 = 2014
\][/tex]
Step 5: Calculate the enrollment at that time
Now, we can substitute [tex]\( x = 9 \)[/tex] back into either expression to find the enrollment number.
For College A:
[tex]\[
13300 + 1000 \times 9 = 13300 + 9000 = 22300
\][/tex]
For College B:
[tex]\[
26800 - 500 \times 9 = 26800 - 4500 = 22300
\][/tex]
Results:
- The colleges will have the same enrollment in the year 2014.
- The enrollment in each college at that time will be 22,300 students.
### Part (b): Use your table feature on your calculator to numerically check your work
You can use your calculator to create tables for both colleges' enrollments over the years to confirm these results.
- College A:
[tex]\[
13300 + 1000x
\][/tex]
- College B:
[tex]\[
26800 - 500x
\][/tex]
When you create the table and look for the year when both enrollments are equal, you should see that in the year 2014, both colleges have 22,300 students enrolled.
This confirms our work and shows that the solution is correct.
