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: 5

Answer

$(-2,3,-1)$

Work Step by Step

$2x+2y+z=1$ Equation $(1)$ $-x+y+2z=3$ Equation $(2)$ $x+2y+4z=0$ Equation $(3)$ Add Equation $(2)$ and Equation $(3)$ $-x+y+2z+x+2y+4z=3+0$ $3y+6z = 3$ $y+2z = 1$ Equation $(4)$ Multiply Equation $(2)$ by $2$ and add with Equation $(1)$ $2(-x+y+2z)+2x+2y+z=2(3)+1$ $-2x+2y+4z+2x+2y+z=6+1$ $4y+5z = 7$ Equation $(5)$ Multiply Equation $(4)$ by $4$ and subtract from Equation $(5)$ $4y+5z-4(y+2z) = 7-4(1)$ $4y+5z-4y-8z=7-4$ $-3z=3$ $z=-1$ Substitute $z$ value in Equation $(4)$ $y+2z = 1$ $y+2(-1) = 1$ $y-2=1$ $y=3$ Substitute $z$ and $y$ values in Equation $(3)$ $x+2y+4z=0$ $x+2(3)+4(-1)=0$ $x+6-4=0$ $x+2=0$ $x=-2$ The solution is $(-2,3,-1)$
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