Intermediate Algebra (6th Edition)

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

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations in Three Variables - Exercise Set - Page 221: 34

Answer

$z = \frac{-27}{14}$

Work Step by Step

$z-3(z+7) = 6(2z+1)$ Using distributive property, $z-3z-21 = 12z + 6$ $-2z-21 = 12z + 6$ Adding $21$ to both sides. $-2z-21+21 = 12z + 6 +21$ $-2z = 12z + 27$ Adding $-12z$ to both sides. $-2z-12z = 12z + 27-12z$ $-14z = 27$ Divide both sides by $-14$ $z = \frac{27}{-14}$ $z = - \frac{27}{14}$
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