Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 3 - Section 3.3 - Systems of Linear Equations in Three Variables - Exercise Set - Page 215: 6

Answer

$(1,-1,1)$

Work Step by Step

To solve the given problem we will have to plug the simplest equation into the other equation and then solve for the other variable. $2x+y-2z=-1$ and $-6x+6y+2z=-10$ This gives, $-4x+7y=-11$ Now, $9x-9y-3z=15$ and $x-2y+3z=6$ This gives $10x-11y=21$ or, $y=-1$ or, $x=2$ Also, $-4x+7y=-11$ or, $-4x+7(-1)=-11$ or, $x=1$ Thus, $9x-9y-3z=15$ this gives $9(1)-9(-1)-3z=15$ or, $z=1$ Hence, the desired result is $(1,-1,1)$
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