SOLUTION: In ​2011, there were 12500 students at college​ A, with a projected enrollment increase of 800 students per year. In the same​ year, there were 24200 students at college​ B

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Question 1166699: In ​2011, there were 12500 students at college​ A, with a projected enrollment increase of 800 students per year. In the same​ year, there were 24200 students at college​ B, with a projected enrollment decline of 500 students per year. Complete parts a and b.
a. According to these​ projections, when will the colleges have the same​ enrollment? What will be the enrollment in each college at that​ time?
According to these​ projections, in the year ____ the colleges will have the same enrollment and at that time ____ students were enrolled.
b. To obtain the​ table, the entered equations are Y1= __​, and Y2= __
​(Type expressions using x as the​ variable.)
TABLE:
X Y1 Y2
---------------
7 18100 20700
---------------
8 18900 20200
---------------
9 19700 19700
---------------
10 20500 19200
---------------
11 21300 18700
---------------
12 22100 18200
---------------
13 22900 17700
---------------
14 23700 17200
---------------
15 24500 16700
---------------
16 25300 16200
---------------
17 26100 15700
---------------
X = 7
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.

In my post, I will response part (a) only. 

Enrollment at college A    = 12500 + 800*(n-2011).


Enrollment at college B    = 24200 - 500*(n-2011).


We want to know when it will happen   = .


We write the equation


    12500 + 800*(n-2011) = 24200 - 500*(n-2011)


We simplify it


    800*(n-2011) + 500*(n-2011) = 24200 - 12500

    (800+500)*(n-2011) = 11700

     1300*(n-2011) = 11700

          n - 2011 = 1170/1300 = 9.


The long-waited event will happen at the year  9 + 2011 = 2020.                                     ANSWER


The number of enrolled students will be  12500 + 800*9 = 19700 = 24200 - 500*9  in each college.    ANSWER

Solved.

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What is written next in your post - is a riddle to me . . .



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