Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 3 - Review - Review Exercises - Page 219: 8

Answer

$x = \frac 5 3$ and $y = \frac 1 3$

Work Step by Step

1. Notice that, if we add the two equations, neither $x$ or $y$ will be eliminated. But, if we multiply the first equation by 2, the $y$ unknown will be eliminated: $2(0.2x - 0.1y) = 2(0.3)$ $0.2x + 0.2y = 0.4$ $0.4x - 0.2y = 0.6$ $0.2x + 0.2y = 0.4$ Adding the equations: $0.4x + 0.2x - 0.2y + 0.2y = 0.6 + 0.4$ $0.6x = 1$ $\frac{0.6x}{0.6} = \frac{1}{0.6}$ $x = \frac{1}{0.6} = \frac{10}{6} = \frac 5 3$ 2. Calculate the $y$ value by substituting the value for $x$ on the first equation: $0.2 (\frac{5}{3}) - 0.1y = 0.3$ $\frac{1}{3} - 0.1y = 0.3$ $\frac{1}{3} - 0.3 = 0.1y$ $\frac 1 3 - \frac {0.9} {3} = 0.1y$ $\frac {0.1} 3 = 0.1y$ $\frac 1 3 = y$
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