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: 6

Answer

$(-3,-2,5)$

Work Step by Step

$2x-3y+z=5$ Equation $(1)$ $x+y+z=0$ Equation $(2)$ $4x+2y+4z=4$ Equation $(3)$ Multiply Equation $(2)$ by $3$ and add with Equation $(1)$ $3(x+y+z)+2x-3y+z=3(0)+5$ $3x+3y+3z+2x-3y+z=5$ Combine like terms. $5x+4z=5$ Equation $(4)$ Multiply Equation $(2)$ by $-2$ and add with Equation $(3)$ $-2(x+y+z)+4x+2y+4z=-2(0)+4$ $-2x-2y-2z+4x+2y+4z=4$ Combine like terms. $2x+2z=4$ Equation $(5)$ Multiply Equation $(4)$ by $2$, Equation $(5)$ by $-4$ and add the resultant equations. $2(5x+4z)-4(2x+2z)=2(5)-4(4)$ $10x+8z-8x-8z=10-16$ Combine like terms. $2x=-6$ $x=-3$ Substitute $x$ value in Equation $(5)$ $2x+2z=4$ $2(-3)+2z=4$ $-6+2z=4$ $2z=10$ $z=5$ Substitute $x$ and $z$ values in Equation $(2)$ $x+y+z=0$ $-3+y+5=0$ $y+2=0$ $y=-2$ $(-3,-2,5)$ satisfies the given equations. Solution: $(-3,-2,5)$
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