Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.4 Solve Systems of Linear Equations in Three Variables - 3.4 Exercises - Skill Practice - Page 182: 18

Answer

No solution

Work Step by Step

The first equation: $x+3y-z=12$ Solve for x: $\rightarrow x=-3y+z+12$ Substitute for z in the second equation: $2x+4y-2z=6$ $2(-3y+z+12)+4y-2z=6$ $-9y+3z+36+4y-2z=6$ $-2y+z=-30$ (1) Substitute for z in the third equation: $-x - 2y + z = -6$ $-(-3y+z+12)- 2y + z = -6$ $y=6$ It is impossible for $y$ to have two different values, so the system is inconsistent and has no solution.
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