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: 1

Answer

$(-1,5,2)$

Work Step by Step

$x-y+z = -4$ Equation $(1)$ $3x+2y-z = 5$ Equation $ (2)$ $-2x+3y-z = 15$ Equation $(3)$ Add Equation $(1)$ and $(2)$ to eliminate $z$ $4x+y = 1$ Equation $(4)$ Add Equation $(1)$ and $(3)$ to eliminate $z$ $-x+2y = 11 $ Equation $(5)$ Add $(4)$ and $ 4 \times (5)$ to eliminate $x$ and get $y$ $4x -(4)x +y +(4)2y = 1+ (4)11$ $4x -4x +y +8y = 1+ 44$ $9y = 45$ $y=5$ Substituting $y$ value in Equation $(4)$ to get $x$ $4x+y = 1$ $4x+5= 1$ $4x = 1-5$ $4x = -4$ $x = -1$ Substituting $x $ and $y$ values in Equation $(1)$ to get $z$ $x-y+z = -4$ $-1-5+z = -4$ $-6+z = -4$ $z = -4+6$ $z= 2$ $x = -1$ $y=5$ $z= 2$ Solution is $(-1,5,2)$
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