Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.5 - Problem Solving and Using Formulas - Exercise Set - Page 67: 33

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 $Y_2=26800-500x$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.