College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 5 - Mid-Chapter Check Point - Page 561: 1

Answer

$ x = \frac{9}{5}$ $y = -\frac{26 }{15}$

Work Step by Step

$x - 3y -7 $ $4x +3y -2$ Rearrange $x - 3y = 7 $ -> equation 1 $4x +3y = 2$ -> equation 2 Add $ x - 3y = 7 $ + $4x +3y = 2$ $= 5x + 0 = 9$ $= x = \frac{9}{5}$ Replace x in equation 1 $ \frac{9}{5} - 3y = 7 $ $= - 3y = 7 - \frac{9}{5}$ $= -3y = \frac{35 -9 }{5} $ $ -3y = \frac{26 }{5}$ $y = -\frac{26 }{15}$
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