Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.2 - An Introduction to Problem Solving - Exercise Set: 13


45, 225, and 145

Work Step by Step

We set the first number equal to $x$. We are given that a second number is 5 times the first number. Therefore, the second number can be represented by $5x$. A third number is 100 more than the first number, so the third number can be represented by $x+100$. The sum of the three numbers is 415, so $x+5x+(x+100)=415$. Group like terms on the left side. $7x+100=415$ Subtract 100 from both sides. $7x=315$ Divide both sides by 7. $x=45$ Therefore, number one is $45$, number 2 is $5\times45=225$, and number 3 is $45+100=145$
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