What is the general method for solving $Ax = b$ when A is rectangular Suppose I am given that
$A = \begin{bmatrix} 2 &4& 6 &-2 \\ 1& 0& 1 &1 \\ 0& 2 &2& -2 \end{bmatrix}$
and 
$b = \begin{bmatrix} 4 \\ 2 \\ 1 \end{bmatrix}$
What should I do to solve for $x$?
Should I attempt solving the simultaneous equation:
$2x_1 + 4x_2 + 6x_3 - 2x_4 = 4$
$x_1 + x_3 + x_4 = 2$
$2x_2 + 2x_3 -2x_4 = 1$
Is this the only way to approach this problem?
 A: 
Is this the only way to approach this problem?

In fact, $Ax=b$ is equivalent to the given system of linear equations (just a different way of writing it).
But it would be easier to solve if we convert the augmented matrix to row echelon form first:
\begin{align*}
\left[\begin{array}{cccc|c} 2 & 4 & 6 & -2 & 4 \\ 1 & 0 & 1 & 1 & 2 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right]
& \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 2 & 4 & 6 & -2 & 4 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right] \\
& \xrightarrow{R_2 \gets R_2-2R_1}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 0 & 4 & 4 & -4 & 0 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right] \\
& \xrightarrow{R_2 \gets \tfrac{1}{4} R_2}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 2 & 2 & -2 & 1 \\ \end{array}\right] \\
& \xrightarrow{R_3 \gets R_3-2R_2}
\left[\begin{array}{cccc|c} 1 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{array}\right].
\end{align*}
Since row operations preserve the set of solutions, the system of equations $Ax=b$ will have the same set of solutions as the system of equations
$$
\begin{bmatrix} 1 & 0 & 1 & 1 \\ 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 
0 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}.
$$
A: If there are solutions, they can be found using the Moore-Penrose pseudoinverse.
https://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse#Obtaining_all_solutions_of_a_linear_system
They will be of the form
$$A^+b + (I - A^+A)w$$
for pseudoinverse $A^+$ and any $w$.
