Answer
$\displaystyle \{\ (3,-\frac{8}{3},\frac{1}{9})\ \}$
Work Step by Step
$\left\{\begin{array}{lllll}
x & +4y & -3z & =-8 & (1)\\
3x & -y & +3z & =12 & (2)\\
x & +y & +6z & =1 & (3)
\end{array}\right.$
Add the first two equations ... eliminates z.,
Add 2$\times$(eq.$1$)+(eq.$3$) ... eliminates z.
$\left\{\begin{array}{lllll}
4x & +3y & & =4 & (4)\\
3x & +9y & & =-15 & (5)
\end{array}\right.$
Add $(-3)\times$(eq.$4$)+(eq.$5$) ... eliminates y.
$-9x=-27$
$x=3$
Back-substitute into (4)
$12+3y=4$
$3y=-8$
$y=-\displaystyle \frac{8}{3}$
Back-substitute into ($3$)
$3-\displaystyle \frac{8}{3}+6z=1$
$6z=1-3+\displaystyle \frac{8}{3}$
$6z=\displaystyle \frac{2}{3}\qquad/\div 6$
$z=\displaystyle \frac{1}{9}$
Solution set = $\displaystyle \{\ (3,-\frac{8}{3},\frac{1}{9})\ \}$
