Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.8 - Modeling with Equations - Exercise Set - Page 123: 13


The same enrollment ( $22,300$ ) will be achieved in 2019.

Work Step by Step

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 $13,300+1000(9)=22,300$
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