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

Cramer's Rule

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

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

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

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

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