Intermediate Algebra for College Students (7th Edition)

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

Chapter 3 - Section 3.2 - Problem Solving and Business Applications Using Systems of Equations - Exercise Set - Page 207: 47



Work Step by Step

Here, $C(x)=18,000+20x; R(x)=80x$ when the revenue breaks with the costs then $C(x)=R(x)$ Thus, $18,000+20x=80x$ or, $18,000=80x-20x$ or, $x=300$
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