Answer
$\{(1,2,-1)\}$.
Work Step by Step
The given system of equations is
$2x+2y+7z=-1$
$2x+y+2z=2$
$4x+6y+z=15$
The augmented matrix is
$\Rightarrow \left[\begin{array}{ccc|c}
2 & 2 & 7& -1\\
2 &1 & 2& 2 \\
4&6&1&15
\end{array}\right]$
Perform $R_1\Rightarrow R_1/(2)$
$\Rightarrow \left[\begin{array}{ccc|c}
2/(2) & 2/(2) & 7/(2)& -1/(2)\\
2 &1 & 2& 2 \\
4&6&1&15
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 7/2& -1/2\\
2 &1 & 2& 2 \\
4&6&1&15
\end{array}\right]$
Perform $R_2\rightarrow R_2-2 R_1$ and $R_3\rightarrow R_3-4 R_1$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 7/2& -1/2\\
2-2(1) &1-2(1) & 2-2(7/2)& 2-2(-1/2) \\
4-4(1)&6-4(1)&1-4(7/2)&15-4(-1/2)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 7/2& -1/2\\
0 &-1 & -5& 3 \\
0&2&-13&17
\end{array}\right]$
Perform $R_2\Rightarrow R_2/(-1)$
Change the sign of the second row and swap the 3rd and 2nd row.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 7/2& -1/2\\
0/(-1) &-1/(-1) & -5/(-1)& 3/(-1) \\
0&2&-13&17
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 1 & 7/2& -1/2\\
0 &1 & 5& -3 \\
0&2&-13&17
\end{array}\right]$
Perform $R_1\rightarrow R_1- R_2$ and $R_3\rightarrow R_3-2 R_2$.
$\Rightarrow \left[\begin{array}{ccc|c}
1-0 & 1-1 & 7/2-5& -1/2-(-3)\\
0 &1 & 5& -3 \\
0-2(0)&2-2(1)&-13-2(5)&17-2(-3)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -3/2& 5/2\\
0 &1 & 5& -3 \\
0&0&-23&23
\end{array}\right]$
Perform $R_3\rightarrow \frac{R_3}{(-23)}$.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -3/2& 5/2\\
0 &1 & 5& -3 \\
0/(-23)&0/(-23)&-23/(-23)&23/(-23)
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & -3/2& 5/2\\
0 &1 & 5& -3 \\
0&0&1&-1
\end{array}\right]$
Perform $R_1\rightarrow R_1+(3/2)R_3$ and $R_2\rightarrow R_2-5 R_3$.
$\Rightarrow \left[\begin{array}{ccc|c}
1+(3/2)(0) & 0+(3/2)(0) & -3/2+(3/2)(1)& 5/2+(3/2)(-1)\\
0-5(0) &1-5(0) & 5-5(1) & -3-5(-1) \\
0&0&1&-1
\end{array}\right]$
Simplify.
$\Rightarrow \left[\begin{array}{ccc|c}
1 & 0 & 0& 1\\
0&1 & 0 & 2 \\
0&0&1&-1
\end{array}\right]$
Use back substitution to solve the linear system.
$\Rightarrow x=1$.
and
$\Rightarrow y=2$.
and
$\Rightarrow z=-1$.
The solution set is $\{(x,y,z)\}=\{(1,2,-1)\}$.
