Answer
a) after 9 years; that is, year 2017; Number of students for both colleges = 22300
b) $Y_1=13300+1000x$ and $Y_2=26800-500x$
Work Step by Step
Number of years = x
Need to solve for the number of students at college A. In order to find this as per the given statement in the question, we have an equation as $A=13300+1000x$ for college A
and
Need to solve for the number of students at college B. In order to find this as per the given statement in the question, we have an equation as $B=26800-500x$ for college B.
a) As per condition when A = B
Thus, or, $13300+1000x=26800-500x$
or, $1000x+500x=26800-13300$
or, $x=9$
Now, when x= 9 then $A=13300+1000(9)=22300$ and when x= 9 then $B=26800-500(9)=22300$
Thus, Number of students for both colleges = 22300
Hence, we can conclude it will be after 9 years when the both colleges will have same students.This means that in the year of $2017$
b) Two conditions will be:
$Y_1=13300+1000x$ and $Y_2=26800-500x$
