Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Cumulative Review - Page 253: 44

Answer

$(-1,2,5)$

Work Step by Step

Given Equations, $x-2y+z=0$ Equation $(1)$ $3x-y-2z=-15$ Equation $(2)$ $2x-3y+3z=7$ Equation $(3)$ Multiply Equation $(1)$ by $2$ and add with Equation $(2)$ $2(x-2y+z)+3x-y-2z=2(0)-15$ $2x-4y+2z+3x-y-2z=-15$ $5x-5y=-15$ $x-y=-3$ Equation $(4)$ Multiply Equation $(2)$ by $3$ $3(3x-y-2z)=3(-15)$ $9x-3y-6z=-45$ Equation $(5)$ Multiply Equation $(3)$ by $2$ $2(2x-3y+3z)=2(7)$ $4x-6y+6z=14$ Equation $(6)$ Add Equation $(5)$ and Equation $(6)$ $9x-3y-6z+4x-6y+6z=-45+14$ $13x-9y=-31$ Equation $(7)$ Multiply Equation $(4)$ by $9$ and subtract from Equation $(7)$ $13x-9y-9(x-y)=-31-9(-3)$ $13x-9y-9x+9y=-31+27$ $4x=-4$ $x=-1$ Substituting $x$ value in Equation $(4)$ $x-y=-3$ $-1-y=-3$ $-y=-3+1$ $-y=-2$ $y=2$ Substituting $x$ and $y$ value in Equation $(1)$ $x-2y+z=0$ $-1-2(2)+z=0$ $-1-4+z=0$ $-5+z=0$ $z=5$ Solution: $(-1,2,5)$
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