Answer
Using Cramer's rule to solve
$\left\{ \begin{align}
& 3x+y+4z=-8 \\
& 2x+3y-2z=11 \\
& x-3y-2z=4 \\
\end{align} \right.$ for $y$,
we obtain
$y=\frac{\left| \begin{matrix}
\underline{3} & \underline{-8} & \underline{4} \\
\underline{2} & \underline{11} & \underline{-2} \\
\underline{1} & \underline{4} & \underline{-2} \\
\end{matrix} \right|}{\left| \begin{matrix}
\underline{3} & \underline{1} & \underline{4} \\
\underline{2} & \underline{3} & \underline{-2} \\
\underline{1} & \underline{-3} & \underline{-2} \\
\end{matrix} \right|}$
Work Step by Step
Cramer's rule states that for any linear equations:
$\begin{align}
& {{a}_{1}}x+{{b}_{1}}y={{c}_{1}} \\
& {{a}_{2}}x+{{b}_{2}}y={{c}_{2}}
\end{align}$
The value for $x,y$ can be calculated as here:
$x=\frac{\left| \begin{matrix}
{{c}_{1}} & {{b}_{1}} \\
{{c}_{2}} & {{b}_{2}} \\
\end{matrix} \right|}{\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} \\
{{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right|}$
$y=\frac{\left| \begin{matrix}
{{a}_{1}} & {{c}_{1}} \\
{{a}_{2}} & {{c}_{2}} \\
\end{matrix} \right|}{\left| \begin{matrix}
{{a}_{1}} & {{b}_{1}} \\
{{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right|}$
Therefore, the values of $y$ for the given linear equations are:
$y=\frac{\left| \begin{matrix}
3 & -8 & 4 \\
2 & 11 & -2 \\
1 & 4 & -2 \\
\end{matrix} \right|}{\left| \begin{matrix}
3 & 1 & 4 \\
2 & 3 & -2 \\
1 & -3 & -2 \\
\end{matrix} \right|}$
