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 220: 25

Answer

$(0, 0.5, -4)$

Work Step by Step

$x+2y-z=5$ Equation $(1)$ $-3x-2y-3z=11$ Equation $(2)$ $4x+4y+5z=-18$ Equation $(3)$ Equation $(1)$ + Equation $(2)$ $(x+2y-z)+(-3x-2y-3z)=5+11$ $-2x-4z=16$ Equation $(4)$ $2\times$ Equation $(2)$ + Equation $(3)$ $2(-3x-2y-3z)+(4x+4y+5z)=2(11)+(-18)$ $(-6x-4y-6z)+(4x+4y+5z)=22-18$ $-2x-z=4$ Equation $(5)$ Equation $(4)$ $-$ Equation $(5)$ $(-2x-4z)-(-2x-z)=16-4$ $-3z=12$ $z=-4$ Substitute $z=-4$ into Equation $(5)$. $-2x+4=4$ $-2x=0$ $x=0$ Substitute known values for $x$ and $z$ into Equation $(1)$. $0+2y+4=5$ $2y=1$ $y=0.5$ $(0, 0.5, -4)$ satisfies the given equations. Solution is $(0, 0.5, -4)$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.