Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations in Three Variables - Exercise Set - Page 221: 44

Answer

$(1,2,3,4)$

Work Step by Step

We multiply the second equation by two and add it to the third equation to find: $3y+2z=12$ We multiply the first equation by 2 and add to the new equation above: $4x+3y=10$ We now substitute $ x = 3-y$, which can be found using equation 4, to find: $4(3-y)+3y=10$ $12-y=10$ $y=2$ We substitute $y=2$ into the last equation to find: $x+2=3$ $x=1$ Now, we substitute x into the first equation to have: $2(1)-z=-1$ $z=3$ Now, use the third equation and solve for z: $ (2)-2w=-6$ We solve to find: $w=4$ Thus, we have a solution: $(1,2,3,4)$
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