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 - Vocabulary, Readiness & Video Check - Page 219: 3

Answer

Yes. Ordered triple $(-1,3,1)$ satisfies all three equations, $x+y+z = 3$ $-x+y+z = 5$ $x+2y-3z =2$ So, $(-1,3,1)$ is a solution.

Work Step by Step

Substituting $x=-1$, $y=3$, $z= 1$ in $x+y+z = 3$ $-x+y+z = 5$ $x+2y-3z =2$, $x+y+z = 3$ $-1+3+1=3$ $3 = 3$ $-x+y+z = 5$ $-(-1)+3+1= 5$ $1+3+1= 5$ $5=5$ $x+2y-3z =2$ $-1+2(3)-3(1)= 2$ $-1+6-3 = 2$ $2 = 2 $ Ordered triple $(-1,3,1)$ satisfies all three equations. So, $(-1,3,1)$ is a solution.
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