Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Mid-Chapter Check Point - Page 853: 8

Answer

$(4,- 2,3)$

Work Step by Step

Re-arrange the given equations to get $-6x+3y-6z=-48 $ and $(-6x+3y-6z) +(7x-3y-5z) =-48+19$ $ \implies x-11z =-29$ $ \implies -x+11z =29$ $(x-3z)+(-x+11z)=-5+29$ This gives: $z=3$ Now, $x-3(3)=-5 \implies x=4$ and $2(4)-y+2(3)=16 $ gives: $y=-2$ Our solution is: $(4,- 2,3)$
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