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: 41

Answer

$(1,1,0,2)$

Work Step by Step

$x+y-w=0$ Equation $(1)$ $y+2z+w=3$ Equation $(2)$ $x-z=1$ Equation $(3)$ $2x-y-w=-1$ Equation $(4)$ Adding Equation $(2)$ and Equation $(4)$ we get, $y+2z+w+2x-y-w=3-1$ $2x+2z = 2$ $x+z = 1$ Equation $(5)$ Adding Equation $(3)$ and Equation $(5)$ we get, $x-z+x+z=1+1$ $2x=2$ $x=1$ Substituting $x$ value in Equation $(3)$ $x-z=1$ $1-z=1$ $-z=1-1$ $-z=0$ $z=0$ Adding Equation $(1)$ and Equation $(2)$ we get, $x+y-w+y+2z+w=0+3$ $x+2y+2z=3$ Equation $(6)$ Substituting $x$ and $z $ values in Equation $(6)$ $x+2y+2z=3$ $1+2y+2(0)=3$ $1+2y+0=3$ $2y=2$ $y=1$ Substituting $x$ and $y$ values in Equation $(1)$ $x+y-w=0$ $1+1-w=0$ $2-w=0$ $-w=-2$ $w=2$ $(1,1,0,2)$ satisfies the given equations. Solution: $(1,1,0,2)$
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.