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: 2

Answer

$(3,-3,1)$

Work Step by Step

$x+y-z=-1$ Equation $(1)$ $-4x-y+2z=-7$ Equation $(2)$ $2x-2y-5z=7$ Equation $(3)$ Add Equation $(1)$ and Equation $(2)$ $x+y-z-4x-y+2z=-1-7$ Combine like terms. $-3x+z=-8$ Equation $(4)$ Multiply Equation $(2)$ by $-2$ and add with Equation $(3)$ $-2(-4x-y+2z)+2x-2y-5z = -2(-7)+7$ $8x+2y-4z+2x-2y-5z=14+7$ Combine like terms. $10x-9z=21$ Equation $(5)$ Multiply Equation $(4)$ by $9$ and add with Equation $(5)$ $9(-3x+z)+10x-9z=9(-8)+21$ $-27x+9z+10x-9z=-72+21$ Combine like terms. $-17x=-51$ $x = 3 $ Substitute $x$ value in Equation $(4)$ $-3x+z=-8$ $-3(3)+z=-8$ $-9+z=-8$ $z=-8+9$ $z=1$ Substitute $x$ and $z$ values in Equation $(1)$ $x+y-z=-1$ $3+y-1=-1$ $2+y=-1$ $y=-3$ $(3,-3,1)$ satisfies the given three equations. The solution is $(3,-3,1)$
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.