Take an homogeneous system, Ax = 0. For example in the following equation,
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
The matrix \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} would be called A.
The vector \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} would be called x.
We want to find all x1, x2 such that Ax = 0. Towards that end, expand it out into
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix}.
For any real number x2 that we choose, x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}
This means that \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} so that the value of x2 doesn't affect the value of Ax at all.
So to solve the above equation, we need to solve 1x1 = 0. But whenever we have two numbers a and b such that ab = 0, we know that either a or b must equal 0. Because we know that 1 is not equal to 0, we can conclude that x1 must be.
This means that \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 0 \\ 0 \end{bmatrix} when \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ x_{2} \end{bmatrix} where x2 is any real number that we choose.
We have shown that any solution of Ax = 0 must be of the form \begin{bmatrix} 0 \\ x_{2} \end{bmatrix}.
In other words, there is one solution for every real number x2.
Non-Homogeneous System
Let's now consider a non-homogeneous system Ax = r where r is a non-zero vector. Let's take our example as,
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ r_{1} \end{bmatrix}
Again, we expand this out into,
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix}.
Again, x_{2}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.
So \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = x_{1}\begin{bmatrix} 1 \\ 0 \end{bmatrix}.
Now, 0x1 = 0. This has the consequence that we can only solve the equation if r2 = 0 because that is all we are ever going to get from A. We must also have 1x1 = r1. So we can only solve the equation if
r = \begin{bmatrix} r_{1} \\ 0 \end{bmatrix}
Now, as you know, A(v1 + v2) = Av1 + Av2 so if Av1 = r and Av2 = 0 then A(v1 + v2) = Av1 + Av2 = r + 0 = r
This means that the solution set is all vectors of the form
\begin{bmatrix} r_{1} \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ x_{2} \end{bmatrix}
where r1 is determined by r and x2 is free.
In other words,
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ r_{2} \end{bmatrix}
has solutions only if r2 = 0. In that case, we have
\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} r_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} r_{1} \\ 0 \end{bmatrix} for all real numbers, x2.
Four more good matrices to solve and really look at would be \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} , \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} and \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}
You probably want to use row reduction or some other method for more complicated systems but you should look at those three like we've looked at the one above.
