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 221: 42

Answer

Solution:$(5,1,-2,1)$

Work Step by Step

$5x+4y=29$ Equation $(1)$ $y+z-w = -2$ Equation $(2)$ $5x+z = 23$ Equation $(3)$ $y-z+w = 4$ Equation $(4)$ Adding Equation $(2)$ and Equation $(4)$ $y+y+z-z-w+w = -2+4$ $2y = 2$ $y = 1$ Substituting $y$ value in Equation $(1)$ $5x+4y=29$ $5x+4(1)=29$ $5x+4 = 29$ $5x= 29-4$ $5x= 25$ $x=5$ Substituting $x$ value in Equation $(3)$ $5x+z = 23$ $5(5)+z = 23$ $25+z = 23$ $z=23-25$ $z= -2$ Substituting $z$ and $y$ values in Equation $(4)$ $y-z+w = 4$ $1-(-2)+w = 4$ $1+2+w = 4$ $3+w = 4$ $w = 1$ Solution:$(x,y,z,w) = (5,1,-2,1)$
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