## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 3 Linear Systems and Matrices - 3.4 Solve Systems of Linear Equations in Three Variables - 3.4 Exercises - Skill Practice - Page 183: 37

#### Answer

$a=12,b=-4,c=10$

#### Work Step by Step

Plugging in $x=-1,y=2,z=-3$ into the equations we can easily obtain the constants, since there must be an equality between the two sides. Hence: $1(-1)+2(2)-3(-3)=-1+4+9=12=a$ $-(-1)-2+1(-3)=1-2-3=-4=b$ $2(-1)+3(2)-2(-3)=-2+6+6=10=c$

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