#### 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)
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(1.) Perform elimination by addition method (eliminate y) by
$Eq.2+Eq1 \Rightarrow$
$(3+1)x+(1-1)y+(-2+3)z=-2+8$
$Eq.3+4\times Eq1 \Rightarrow$
$(2+4)x+(4-4)y+(1+12)z=0+24$
$\left[\begin{array}{ll}
4x+z=6 & I\\
6x+13z=24 & II
\end{array}\right.$ ...,
(2.) ...$ (II-13\times I)$ eliminates z:
$(6-52)x=24-78$
$-46x=-46$
$x=1$
Back-substitute into $\ \ 4x+z=6$
$4(1)+z=6$
$z=2$
(3.) Back-substitute into one of the initial equations:
x-y+3z=8
$1-y+3(2)=8$
$7-8=y$
$y=-1$
Solution set: : $\{ (1,-1,2) \}$