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: 19

Answer

Solution: $(2,5,\frac{1}{2})$

Work Step by Step

$x+y-4z = 5$ Equation $(1)$ $x-y+2z = -2$ Equation $(2)$ $3x+2y+4z = 18$ Equation $(3)$ Adding Equation $(1)$ and Equation $(2)$ $x+x+y-y-4z+2z = 5-2$ $2x-2z = 3$ Equation $(4)$ Multiplying Equation $(2)$ by $2$ and adding with Equation $(3)$ $2x +3x -2y +2y +4z + 4z = -4+18$ $5x +8z = 14$ Equation $(5)$ Multiplying Equation $(4)$ by $4$ and adding with Equation $(5)$ $8x+5x-8z+8z= 12+14$ $13x = 26$ $x = 2$ Substituting $x$ value in Equation $(4)$ $2(2) -2z = 3$ $4-2z = 3$ $-2z = 3-4$ $-2z = -1$ $z = \frac{-1}{-2}$ $z = \frac{1}{2}$ Substituting $x$ and $z$ values in Equation $(1)$ $x+y-4z = 5$ $2+y-4(\frac{1}{2}) = 5$ $2+y-2 = 5$ $y = 5$ Solution: $(2,5,\frac{1}{2})$
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