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 206: 26



Work Step by Step

As per the statement: $x+y=15$ and $0.10x+0.25y=3.30$ After plugging the variable $x$ form the first equation in the second equation, we get $0.10(15-y)+0.25y=3.30 \implies 0.15y=1.80$ or, $y=12$ Now, $x+12=15 \implies x=3$ Thus, $x=3,y=12$
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