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: 16

Answer

$(1,2,-2)$

Work Step by Step

$5x+y+3z=1$ Equation $(1)$ $x-y+3z=-7$ Equation $(2)$ $-x+y=1$ Equation $(3)$ Add Equation $(2)$ and Equation $(3)$ $x-y+3z-x+y=-7+1$ $3z = -6$ $z=-2$ Add Equation $(1)$ and Equation $(2)$ $5x+y+3z+x-y+3z=1-7$ $6x+6z=-6$ $x+z=-1$ Equation $(4)$ Substitute $z$ value in Equation $(4)$ to get $x$ $x+z=-1$ $x-2=-1$ $x=-1+2$ $x=1$ Substitute $x$ value in Equation $(3)$ to get $y$ $-x+y=1$ $-1+y=1$ $y=1+1$ $y=2$ $(1,2,-2)$ satisfies the given equations. Solution is $(1,2,-2)$
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