How to solve system of three equations I got these equations as a system 
$2x+3y-2z=2$
$3x-y+2z=-1$
$7x+16y-12z=11$
And I cannot figure out how to solve this. I have tried addition, tried to get rid of z and only solve for x and y but always end up with $0=0$
 A: It's no wonder you're having problem solving the system, since the matrix
$$\begin{pmatrix}
2 & 3 & -2 \\
3 & -1 & 2 \\
7 & 16 & -12 \\
\end{pmatrix}$$
is singular, which means there is no unique solution.
A: We add six times the second equation to the third equation and get
$$
\begin{align}
2x&+3y&-2z&=2\\
3x&-y&+2z&=-1\\
25x&+10y&&=5
\end{align}
$$
Then we add the first equation to the second, and we get
$$
\begin{align}
2x&+3y&-2z&=2\\
5x&+2y&&=1\\
25x&+10y&&=5
\end{align}
$$
Finally, we divide the third equation by $5$ to get
$$
\begin{align}
2x&+3y&-2z&=2\\
5x&+2y&&=1\\
5x&+2y&&=1
\end{align}
$$
and we see that the second and third equations say the same thing. Thus the third equation gives us only information that the first and second already contain. So while what you had looked like three equations in three unknowns, it was really just two equations in three unknowns. Thus there is no unique solution to this equation set.
A: $$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
3 & -1 & 2 & -1 \\
7 & 16 & -12 & 11\\
\end{array}\right)$$
Add $-\frac{3}{2}$ times the first row to the second, and $-\frac{7}{2}$ times the first row to the third:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
0 & \frac{11}{2} & -5 & 4\\
\end{array}\right)$$
Now add the second row to the third:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
0 & 0 & 0 & 0\\
\end{array}\right)$$
Remove the third row:
$$\left(\begin{array}{ccc|c}
2 & 3 & -2 & 2 \\
0 & \frac{-11}{2} & 5 & -4 \\
\end{array}\right)$$
Move the third column to the right side (all of the numbers there will be multipled by $-1$), and name it as $p$ (so $z=p$)
$$\left(\begin{array}{cc|cc}
2 & 3 & 2 & 2 \\
0 & \frac{-11}{2} & -4 & -5 \\
\end{array}\right)$$
Now add $\frac{6}{11}$ times the second row to the first:
$$\left(\begin{array}{cc|cc}
2 & 0 & \frac{-2}{11} & \frac{-8}{11} \\
0 & \frac{-11}{2} & -4 & -5 \\
\end{array}\right)$$
Now the equations:
$$2x=-\frac{2}{11}-\frac{8}{11}p$$
$$-\frac{11}{2}y=-4-5p$$
$$z=p$$
So:
$$x=-\frac{1}{11}-\frac{4}{11}p$$
$$y=-\frac{8}{11}-\frac{10}{11}p$$
$$z=p$$
In vector form:
$$\begin{pmatrix} x \\ y \\ z\end{pmatrix}=\frac{1}{11}\begin{pmatrix} -1 \\ -8 \\ 0\end{pmatrix} + \frac{p}{11} \begin{pmatrix} -4 \\ -10 \\ 11\end{pmatrix}$$
