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: 20

Answer

$(1,1,\frac{1}{3})$

Work Step by Step

$2x-y+3z = 2$ Equation $(1)$ $x+y -6z=0$ Equation $(2)$ $3x+4y-3z=6$ Equation $(3)$ Adding Equation $(1)$ and Equation $(2)$ $2x+x-y+y+3z-6z=2+0$ $3x -3z = 2$ Equation $(4)$ Multiplying Equation $(1)$ by $4$ and adding with Equation $(3)$ $8x+3x-4y+4y+12z-3z = 8 +6$ $11x +9z = 14$ Equation $(5)$ Multiplying Equation $(4)$ by $3$ and adding with Equation $(5)$ $9x+11x-9z+9z = 6+14$ $20x = 20$ $x= 1$ Substituting $x$ value in Equation $(4)$ $3(1)-3z=2$ $3-3z = 2$ $-3z=2-3$ $-3z=-1$ $3z=1$ $z=\frac{1}{3}$ Substituting $x$ and $z$ values in Equation $(2)$ $x+y -6z=0$ $1+y -6(\frac{1}{3})=0$ $1+y-2=0$ $y-1=0$ $y=1$ Solution: $(1,1,\frac{1}{3})$
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