Intermediate Algebra (6th Edition)

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

Chapter 4 - Review - Page 250: 16

Answer

The numbers are $63$ and $21$

Work Step by Step

Let the first number is $x$ and second number is $y$ Given, One number is three times a second number, $x=3y$ Equation $(1)$ Twice the sum of the numbers is $168$ $2(x+y)= 168$ $x+y=84$ Equation $(2)$ From Equation $(1)$ and Equation $(2)$ $x+y=84$ $3y+y=84$ $4y=84$ $y=21$ Substituting $y$ value in Equation $(1)$, we get, $x=3y$ $x=3(21)$ $x=63$ The numbers are $63$ and $21$
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