Answer
The solution of the equations is, $\left\{ \left( -2z+2,\ 2z+\frac{1}{2},\ z \right) \right\}$.
Work Step by Step
The given system of equations can be written in matrix form as below: $\left[ \left. \begin{matrix}
5 & 8 & -6 \\
3 & 4 & -2 \\
1 & 2 & -2 \\
\end{matrix} \right|\begin{matrix}
14 \\
8 \\
3 \\
\end{matrix} \right]$
Solve this matrix as below:
$\left[ \left. \begin{matrix}
5 & 8 & -6 \\
0 & 2 & -4 \\
1 & 2 & -2 \\
\end{matrix} \right|\begin{matrix}
14 \\
1 \\
3 \\
\end{matrix} \right]$ $ By,\ 3{{R}_{3}}-{{R}_{2}}\to {{R}_{2}}$
$\left[ \left. \begin{matrix}
5 & 8 & -6 \\
0 & 2 & -4 \\
0 & 2 & -4 \\
\end{matrix} \right|\begin{matrix}
14 \\
1 \\
1 \\
\end{matrix} \right]$ $ By,\ 5{{R}_{3}}-{{R}_{1}}\to {{R}_{3}}$
$\left[ \left. \begin{matrix}
5 & 8 & -6 \\
0 & 2 & -4 \\
0 & 0 & 0 \\
\end{matrix} \right|\begin{matrix}
14 \\
1 \\
0 \\
\end{matrix} \right]$ $ By,\ {{R}_{2}}-{{R}_{3}}\to {{R}_{3}}$
Convert the last row into equation form, $\begin{align}
0x+0y+0z=0 & \\
0=0 & \\
\end{align}$
The last row does not add any information about the variables; therefore we will drop it.
$\left[ \left. \begin{align}
& \begin{matrix}
5 & 8 & -6 \\
\end{matrix} \\
& \begin{matrix}
0 & 2 & -4 \\
\end{matrix} \\
\end{align} \right|\begin{matrix}
14 \\
1 \\
\end{matrix} \right]$
Convert this system into linear equations.
$\begin{align}
5x+8y-6z=14 & \\
2y-4z=1 & \\
\end{align}$
Write the values of x and y in terms of z and solve the equation,
$\begin{align}
& x=-2z+2 \\
& y=2z+\frac{1}{2} \\
\end{align}$
Hence the numbers satisfy the set of solutions $\left\{ \left( -2z+2,\ 2z+\frac{1}{2},\ z \right) \right\}$
