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: 19

Answer

$(3, 2, 2)$

Work Step by Step

$2x-3y+z=2$ Equation $(1)$ $x-5y+5z=3$ Equation $(2)$ $3x+y-3z=5$ Equation $(3)$ $2\times$Equation $(2)$ $-$ Equation $(1)$ $(2x-10y+10z)-(2x-3y+z)=(6-2)$ $-7y+9z=4$ Equation $(4)$ $3\times$Equation $(2)$ $-$ Equation $(3)$ $(3x-15y+15z)-(3x+y-3z)=(9-5)$ $-16y+18z=4$ Equation $(5)$ $2\times$Equation $(4)$ $-$ Equation $(5)$ $(-14y+18z)-(-16y+18z)=(8-4)$ $2y=4$ $y=2$ Substitute $y=2$ into Equation $(4)$ to get $z$ $-7(2)+9z=4$ $-14+9z=4$ $9z=18$ $z=2$ Substitute known values for $y$ and $z$ into Equation $(1)$ to get $x$ $2x-3(2)+2=2$ $2x-4=2$ $2x=6$ $x=3$ $(3, 2, 2)$ satisfies the given equations. Solution is $(3, 2, 2)$
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