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: 13

Answer

$(-3,-35,-7)$

Work Step by Step

$6x-5z=17$ Equation $(1)$ $5x-y+3z=-1$ Equation $(2)$ $2x+y=-41$ Equation $(3)$ Add Equation $(2)$ and Equation $(3)$ $5x-y+3z+2x+y=-1-41$ $7x+3z = -42$ Equation $(4)$ Multiply Equation $(1)$ by $3$ $3(6x-5z)=3(17)$ $18x-15z=51$ Equation $(5)$ Multiply Equation $(4)$ by $5$ $5(7x+3z) =5( -42)$ $35x+15z = -210$ Equation $(6)$ Add Equation $(5)$ and Equation $(6)$ to find $x$ $18x-15z+35x+15z=51-210$ $53x=-159$ $x=-3$ Substitute $x$ value in Equation $(3)$ to find $y$ $2x+y=-41$ $2(-3)+y=-41$ $-6+y=-41$ $y=-41+6$ $y=-35$ Substitute $x$ value in Equation $(1)$ to find $z$ $6x-5z=17$ $6(-3)-5z=17$ $-18-5z=17$ $-5z=17+18 $ $-5z=35$ $z=-7$ Solution: $(-3,-35,-7)$
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