Use Cramer's Rule and the question above to solve the system (write your answer in the lowest term)

Cramer's Rule

$$\begin{array}{rcrcrcr} 1\cdot x &+& 4 \cdot y &-&3 \cdot z &=& -8 \\ 3\cdot x &-& 1 \cdot y &+& 3 \cdot z &=& 12 \\ 1\cdot x &+& 1 \cdot y &+& 6 \cdot z &=& 1 \\ \end{array}$$

$$\small{\text{ $ x= \frac {\left|\begin{array}{rrr} -8 & 4 & -3 \\ 12 & -1 & 3 \\ 1 & 1 & 6 \end{array}\right|} {\left|\begin{array}{rrr} 1 & 4 & -3 \\ 3 & -1 & 3 \\ 1 & 1 & 6 \end{array}\right|} $ }} \quad \small{\text{ $ y= \frac {\left|\begin{array}{rrr} 1 & -8 & -3 \\ 3 & 12 & 3 \\ 1 & 1 & 6 \end{array}\right|} {\left|\begin{array}{rrr} 1 & 4 & -3 \\ 3 & -1 & 3 \\ 1 & 1 & 6 \end{array}\right|} $ }} \quad \small{\text{ $ z= \frac {\left|\begin{array}{rrr} 1 & 4 & -8 \\ 3 & -1 & 12 \\ 1 & 1 & 1 \end{array}\right|} {\left|\begin{array}{rrr} 1 & 4 & -3 \\ 3 & -1 & 3 \\ 1 & 1 & 6 \end{array}\right|} $ }}$$

$$\small{\text{ $ x= \frac { (-8)\cdot(-1)\cdot6+1\cdot4\cdot3+12\cdot 1\cdot(-3) -1\cdot(-1)\cdot(-3)-(-8)\cdot 1\cdot 3 - 12\cdot 4\cdot 6 } { 1\cdot(-1)\cdot 6 + 1\cdot 4 \cdot 3 + 3\cdot 1 \cdot (-3) -1\cdot(-1)\cdot(-3) - 1\cdot 1 \cdot 3- 3\cdot 4 \cdot 6 } =\frac {48+12-36-3+24-288} {-6+12-9-3-3-72} =\frac {-243} {-81} =3 $ }}$$

$$\small{\text{ $ y= \frac { 1\cdot 12 \cdot 6 + 1\cdot(-8)\cdot3 + 3\cdot 1 \cdot (-3) -1\cdot 12 \cdot (-3) - 1\cdot 1 \cdot 3 - 3 \cdot (-8) \cdot 6 } { -81 } =\frac {72-24-9+36-3+144} {-81} =\frac {216} {-81} =-2\frac{2}{3} $ }} \quad$$

$$\small{\text{ $ z= \frac { 1\cdot(-1)\cdot 1 +1\cdot 4 \cdot 12 + 3 \cdot 1 \cdot(-8) -1\cdot(-1)\cdot (-8) -1\cdot 1 \cdot 12 - 3\cdot 4 \cdot 1 } { -81 } =\frac {-1+48-24-8-12 -12} {-81} =\frac {-9} {-81} =\frac{1}{9} $ }} \quad$$