College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.3 - Page 134: 31

Answer

a) 2019; 22,300 students b) $Y_{1}=13,000+1000x$ and $Y_{2}=28,600-500x$

Work Step by Step

a) A system of linear equations is needed to solve this problem. Let $x$ represent the number of years since 2010 (the starting year). Remember that a linear equation consists of a slope and y-intercept, so, since 13,300 students are currently enrolled at college A (initial value) and 1000 students $per$ $year$ are projected to enroll, the linear equation can be written as $A=13,300+1,000x$. The same can be done for college B using the starting enrollment of 26,800 students and projected enrollment decline of 500 students per year to get $B=28,600-500x$. Now, solve for $x$ by setting the two linear equations equal to each other: $13,300+1,000x=28,600-500x$. Combine like terms by adding the 500 to the left side and subtract 13,300 from the right side. This results in $1,500x=13,500$. $x=13,500\div1,500=9$. So, in 2019 (9 years after 2010), the colleges will have the same enrollment. Plug 9 into one of the linear equations to find the enrollments for both colleges in 2019: 13,300+1000(9)=22,300 or 26,800-500(9)=22,300. b) The table verifies that when x=9 (2019), the enrollment at both colleges will be 22,300 students. $Y_{1}=13,300+1000x$ and $Y_{2}=28,600-500x$.
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