Answer
$x=5$
$y=2$
$z=-6$
Work Step by Step
Write the augmented matrix of the system of linear equations.
$ \begin{bmatrix}
2 & 3 & 1 & 10\\
2 &-3 & -3 & 22\\
4 & -2 & 3 & -2
\end{bmatrix} $
Divide the first row by 2.
$ \begin{bmatrix}
1 & \frac{3}{2} & \frac{1}{2} & 5\\
2 &-3 & -3 & 22\\
4 & -2 & 3 & -2
\end{bmatrix} $
Add -2 times the 1st row to the 2nd row to produce a new 2nd row.
Add -4 times the 1st row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & \frac{3}{2} & \frac{1}{2} & 5\\
0 &-6 & -4 & 12\\
0 & -8 & 1 & -22
\end{bmatrix} $
Divide the second row by -6.
$ \begin{bmatrix}
1 & \frac{3}{2} & \frac{1}{2} & 5\\
0 &1 & \frac{2}{3} & -2\\
0 & -8 & 1 & -22
\end{bmatrix} $
Add 8 times the 2nd row to the 3rd row to produce a new 3rd row.
$ \begin{bmatrix}
1 & \frac{3}{2} & \frac{1}{2} & 5\\
0 &1 & \frac{2}{3} & -2\\
0 & 0 & \frac{19}{3} & -38
\end{bmatrix} $
Multiply the third row by $\frac{3}{19}$
$ \begin{bmatrix}
1 & \frac{3}{2} & \frac{1}{2} & 5\\
0 &1 & \frac{2}{3} & -2\\
0 & 0 & 1 & -6
\end{bmatrix} $
Use back-substitution to find the solution.
$z=-6$
$y=2$
$x=5$
