College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 3 - Review Exercises - Page 380: 87

Answer

$y=35$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use $ x=ky $ and solve for the value of $k$ with the given $ x $ and $ y $ values. Then use the equation of variation to solve for the value of the unknown variable. $\bf{\text{Solution Details:}}$ Since $x$ varies directly as $y,$ then $ x=ky .$ Substituting the given values, $ x=20 $ and $ y=14 ,$ then the value of $k$ is \begin{array}{l}\require{cancel} x=ky \\\\ 20=k(14) \\\\ \dfrac{20}{14}=k \\\\ k=\dfrac{10}{7} .\end{array} Hence, the equation of variation is given by \begin{array}{l}\require{cancel} x=ky \\\\ x=\dfrac{10}{7}y .\end{array} If $x=50,$ then \begin{array}{l}\require{cancel} x=\dfrac{10}{7}y \\\\ 50=\dfrac{10}{7}y \\\\ \dfrac{7}{10}(50)=\left( \dfrac{10}{7}y \right) \dfrac{7}{10} \\\\ \dfrac{7}{\cancel{10}}(\cancel{50}^5)=y \\\\ y=7(5) \\\\ y=35 .\end{array}
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