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: 7

Answer

$x = \frac 6 5$ and $y = \frac 7 5$

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 second equation by 2, the $y$ unknown will be eliminated: $x + 2y = 4$ $2(2x - y) = 2(1)$ $x + 2y = 4$ $4x - 2y = 2$ Adding the equations: $x + 4x + 2y - 2y = 4 + 2$ $5x = 6$ $\frac{5x}{5} = \frac{6}{5}$ $x = \frac{6}{5}$ 2. Calculate the $y$ value by substituting the value for $x$ on the first equation: $(\frac 6 5) + 2y = 4$ $2y = 4 - \frac 6 5$ $2y = \frac{20}{5} - \frac 6 5$ $2y = \frac {14} 5$ $y = \frac 7 5$
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