Chapter 2 - Systems of Linear Equations - 2.1 Introduction to Systems of Linear Equations - Exercises 2.1 - Page 63: 21

$(\dfrac{2}{3},\dfrac{1}{3},\dfrac{-1}{3})$

Consider $3z=-1 \implies z=\dfrac{-1}{3}$ and $2y-z=1 \implies 2y-(\dfrac{-1}{3})=1$ or, $y=\dfrac{1}{3}$ Now, $x-y+z=0 \implies x-\dfrac{1}{3}+(\dfrac{-1}{3})=0$ or, $x=\dfrac{2}{3}$ Our result is : $(x,y,z)=(\dfrac{2}{3},\dfrac{1}{3},\dfrac{-1}{3})$