College Algebra 7th Edition

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

Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 16

Answer

$\begin{cases} x-3y+2z=-1\\ 0x+y+z=-1\\ 0x+0y-3z=3 \end{cases}$

Work Step by Step

$\begin{cases} x-3y+2z=-1\\ 0x+y+z=-1\\ 0x+2y-z=1 \end{cases}$ Multiplying Equation 2 by -2 and adding it to Equation 3. results in a new Equation 3. $\begin{cases} -2y-2z=2\\ 2y-z=1\\ -- -- -\\ -3z=3 \end{cases}$ Therefore, The new equivalent system is... $\begin{cases} x-3y+2z=-1\\ 0x+y+z=-1\\ 0x+0y-3z=3 \end{cases}$
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