## Precalculus (6th Edition) Blitzer

The same enrollment ( $22,300$ ) will be achieved in 2019.
Let $x=$ the number of years after 2010. Enrolled at college A, after x years = $13,300+1000x$ Enrolled at college B, after x years = $26,800-500x$ We want to find x when these two are equal. $13,300+1000x=26,800-500x \qquad$ ... add $500x-13,300$ $1000x+500x=26,800-13,500\qquad$ $1500x=13,500\qquad$ ... divide with 1500 $x=9$ The two colleges will have the same enrollment $9$ years after 2010, in 2019. To calculate the enrollment, substitute $x=9$ into one of the expressions $13,300+1000(9)=22,300$