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: 15

Answer

$(6,22,-20)$

Work Step by Step

$x+y+z=8$ Equation $(1)$ $2x-y-z=10$ Equation $(2)$ $x-2y-3z=22$ Equation $(3)$ Add Equation $(1)$ and Equation $(2)$ $x+y+z+2x-y-z=8+10$ Combine like terms. $3x=18$ $x= 6$ Multiply Equation $(1)$ by $2$ and add with Equation $(3)$ $2(x+y+z)+x-2y-3z=2(8)+22$ $2x+2y+2z+x-2y-3z=16+22$ Combine like terms. $3x-z=38$ Equation $(4)$ Substitute $x$ value in Equation $(4)$ to get $z$ value. $3x-z=38$ $3(6)-z=38$ $18-z=38$ $-z=38-18$ $-z=20$ $z=-20$ Substitute $x$ and $z$ values in Equation $(3)$ to get $y$ value. $x-2y-3z=22$ $6-2y-3(-20)=22$ $6-2y+60=22$ $-2y+66=22$ $-2y=22-66$ $-2y=-44$ $y=22$ $(6,22,-20)$ satisfies the given equations. Solution:$(6,22,-20)$
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