Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 9 - Section 9.1 - Solving Systems of Linear Equations in Three Variables and Problem Solving - Exercise Set - Page 638: 27

Answer

$[12, 6, 4]$

Work Step by Step

$3/4*x-1/3*y+1/2*z=9$ $1/6*x+1/3*y-1/2*z=2$ $1/2*x-y+1/2*z=2$ $3/4*x-1/3*y+1/2*z=9$ $12*(3/4*x-1/3*y+1/2*z)=12*9$ $9x-4y+6z=108$ $1/6*x+1/3*y-1/2*z=2$ $6*(1/6*x+1/3*y-1/2*z)=6*2$ $x+2y-3z=12$ $1/2*x-y+1/2*z=2$ $2*(1/2*x-y+1/2*z)=2*2$ $x-2y+z=4$ $x+2y-3z=12$ $x+2y-3z-(x-2y+z)=12-4$ $x+2y-3z-x+2y-z=8$ $4y-4z=8$ $4y-4z+4z=8+4z$ $4y=8+4z$ $4y/4=(8+4z)/4$ $y=2+z$ $9x-4y+6z=108$ $9x-4y+6z-(9)(x-2y+z)=108-(9)(4)$ $9x-4y+6z-9x+18y-9z=108-36$ $14y-3z=72$ $14(2+z)-3z=72$ $28+14z-3z=72$ $28+11z=72$ $28+11z-28=72-28$ $11z=44$ $11z/11 =44/11$ $z=4$ $y=2+z$ $y=2+4$ $y=6$ $x-2y+z=4$ $x-2*6+4=4$ $x-12+4=4$ $x-12+4-4=4-4$ $x-12=0$ $x-12+12=0+12$ $x=12$
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