Answer
$\{ (1,1,2) \}$
Work Step by Step
Plan (see p.533):
1. Eliminate one variable and arrive at a system of two equations in two variables.
2. Solve the system of two equations in two variables.
3. Back-substitute the solutions of (2) to find the eliminated varable in (1)
------------
(1.) Perform elimination by addition method (eliminate $x$) by
$Eq.1-Eq.1 \Rightarrow$
$(1-1)x+(2-0)y+(-1-1)z=1-3$
$Eq.3-2\times Eq.1 \Rightarrow$
$(2-2)x+(-1-0)y+(1-2)z=3-6$
$\left[\begin{array}{ll}
2y-2z=-2 & /\div 2\\
-y-z=-3 & /\times(-1)
\end{array}\right.$ ,
$\left[\begin{array}{ll}
y-z=-1 & I\\
y+z=3 & II
\end{array}\right.$ ...,
(2.) ... $I+II$ eliminates $z$:
$2y=2$
$y=1$
Back-substitute into $y+z=3$
$1+z=3$
$z=2$
(3.) Back-substitute into one of the initial equations:
$x+z=3$
$x+2=3$
$x=1$
Solution set: : $\{ (1,1,2) \}$
