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: 7

Answer

Infinitely many solutions. Solution set: $\{ (x,y,z)|x-2y+z=-5\}$

Work Step by Step

$x-2y+z=-5$ Equation $(1)$ $-3x+6y-3z=15$ Equation $(2)$ $2x-4y+2z=-10$ Equation $(3)$ Divide both sides of Equation $(2)$ by $-3$ and Equation $(3)$ by $2$, the resulting system is $x-2y+z=-5$ $x-2y+z=-5$ $x-2y+z=-5$ All three equations are identical and equivalent. The equations are dependent equations. So, the system has infinitely many solutions. Solution set: $\{ (x,y,z)|x-2y+z=-5\}$
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