Intermediate Algebra (6th Edition)

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

Chapter 4 - Sections 4.1-4.3 - Integrated Review - Systems of Linear Equations - Page 234: 16

Answer

Solution: $(1,-3,0)$

Work Step by Step

$y+ 2z = -3$ Equation $(1)$ $x-2y = 7$ Equation $(2)$ $2x-y+z = 5$ Equation $(3)$ Adding Equation $(1)$ and Equation $(2)$ $x+y-2y+2z=-3+7$ $x-y+2z = 4$ Equation $(4)$ Adding Equation $(2)$ and Equation $(3)$ $x+2x -2y -y +z = 7+5$ $3x-3y+z = 12$ Equation $(5)$ Multiplying Equation $(4)$ by $-3$ $-3x+3y-6z=-12$ Equation $(6)$ Adding Equation $(5)$ and Equation $(6)$ $3x-3x-3y+3y+z-6z = 12-12$ $-5z = 0$ $z= 0$ Substituting $z$ value in Equation $(1)$ $y+2(0)= -3$ $y = -3$ Substituting $y$ value in Equation $(2)$ $x-2(-3) = 7$ $x+6 = 7$ $x = 7 - 6$ $x = 1$ Solution: $(1,-3,0)$
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