Solving system of linear equations ( 4 variables, 3 equations) Finding the solution(s) for:
$$
\begin{align*}
\begin{cases}
2x_1+x_2+3x_3+2x_4 &=5\\
 x_1+x_2+x_3+2x_4&=3\\
-x_2+x_3+6x_4&=3
\end{cases}
\end{align*}
$$
I tried using elimination to rewrite the system in row echelon form, then back substituting. I have no idea what I did wrong, or if I was doing the elimination process correctly.. but I kept getting stuck. I am confused and not really sure how to proceed, I must be awful because I've been staring at this problem for hours. 
 A: You can call $x_1=p$ and get the system:
$$
\begin{align*}
\begin{cases}
x_2+3x_3+2x_4 &=5-2p\\
x_2+x_3+2x_4&=3-p\\
-x_2+x_3+6x_4&=3
\end{cases}
\end{align*}
$$
If you sum the last equation at the first and second you get
$$
\begin{align*}
\begin{cases}
4x_3+8x_4 &=8-2p\\
2x_3+8x_4&=6-p\\
-x_2+x_3+6x_4&=3
\end{cases}
\end{align*}
$$
Now subtract the second at the first
$$
\begin{align*}
\begin{cases}
2x_3 &=2-p\\
2x_3+8x_4&=6-p\\
-x_2+x_3+6x_4&=3
\end{cases}
\end{align*}
$$
And solve as usual. Your solution will be a function of $p$.
Can you finish?
A: $\begin{matrix}
2& 1 &3 &2 &|&5\\
1 &1 &1 &2 &|&3\\
0 &-1& 1& -2&|&-1\\
0&-1& 1& 6& |&3\\
0& 0& 0& -8& |&-4\end{matrix}\\
x_4 = \frac 12\\
-x_2+x_3 =0\\
x_2 = x_3 = t\\
x_1 + 2t + 1 = 3\\
x_1 = 2-2t\\$
$(x_1,x_2,x_3,x_4) = (2,0,0,\frac 12) + (-2,1,1,0)t$
A: Adding the third row to the first one and adding the third one to the second one you obtain
\begin{array}{lcl}
2x_1+4x_3+8x_4&=&8\\
x_1+2x_3+8x_4&=&6\\
-x_2+x_3+6x_4&=&3
\end{array}
Now multiply the second equation by $2$ and subtract it from the first one
\begin{array}{lcl}
-8x_4&=&-4\\
x_1+2x_3+8x_4&=&6\\
-x_2+x_3+6x_4&=&3
\end{array}
Now you know that $x_4=2$ and hence
\begin{array}{lcl}
x_4&=&2\\
x_1+2x_3&=&-10\\
x_3&=&-9+x_2
\end{array}
Finally subtract the last one multiplied by two from the second equation
\begin{array}{lcl}
x_4&=&2\\
x_1&=&8-2x_2\\
x_3&=&-9+x_2
\end{array}
Now it's clear that your system has an infinite number of solutions (one per each value given to $x_2$). This is not surprising since you have four variables and just three equations.
