Chapter 10 - Section 10.3 - Matrices and Systems of Linear Equations - 10.3 Exercises - Page 710: 25

$a. $ $\left\{\begin{array}{l} x-2y+4y=3\\ y+2z=7\\ z=2 \end{array}\right. $ $b.$ $(1,3,2)$

$a. $ $\left\{\begin{array}{l} x-2y+4y=3\\ y+2z=7\\ z=2 \end{array}\right. $ $b.$ Back substitute $z=2 $ into $y+2z=7$ $y+2(2)=7$ $y=7-4$ $y=3$ Back substitute $z=2 ,\ y=3$ into $x-2y+4y=3$ $x-2(3)+4(2)=3$ $x-6+8=3$ $x+2=3$ $x=1$ Solution: $(1,3,2)$