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: 35

Answer

\begin{cases} x+y+z=-3 \\ 2x+y+z=-4 \\ x+2y+z=-1 \end{cases}

Work Step by Step

To find a system that has $( -1,2,-4 )$ as a solution, make equations using the given points. Since $ -1+2+(-4)=-3 ,$ then $ x+y+z=-3 $ has the given point as a solution. Since $ 2(-1)+2+(-4)=-4 ,$ then $ 2x+y+z=-4 $ has the given point as a solution. Since $ -1+2(2)+(-4)=-1 ,$ then $ x+2y+z=-1 $ has the given point as a solution. Hence, a system that has the given point as a solution is \begin{cases} x+y+z=-3 \\ 2x+y+z=-4 \\ x+2y+z=-1 \end{cases}
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