College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 6, Matrices and Determinants - Section 6.1 - Matrices and Systems of Linear Equations - 6.1 Exercises - Page 501: 26


$a.\qquad\left\{\begin{aligned} x+y-3z&=8\\ y-3z&=5\\ z&=-1\end{aligned}\right.$ $b.\qquad (3,2,-1)$

Work Step by Step

$a.\left\{\begin{aligned} x+y-3z&=8\\ y-3z&=5\\ z&=-1\end{aligned}\right.$ $b.$ The third equation gives $z=-1$. Back-substituting $z=-1$ into the second equation, $y-3(-1)=5\quad \Rightarrow\quad y=2.$ Back-substituting $z=-1 $and $y=2$ into the first equation, $x+2-3(-1)=8\Rightarrow x=3.$ The solution is $(3,2,-1)$
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