Once a linear system is in echelon form, its entire solution set can be read from the pivots, free variables and any contradictory rows that appear.

No solution

A row such as

$$\left[\begin{array}{ccc|c}0&0&0&5\end{array}\right]$$

represents the impossible equation

$$0=5.$$

If an echelon form contains such a row, the system is inconsistent and has no solution.

One solution

A consistent system has a unique solution when every variable is determined by a pivot. For example,

$$\left[\begin{array}{cc|c} 1&0&2\ 0&1&3 \end{array}\right]$$

gives directly

$$x=2,\qquad y=3.$$

Infinitely many solutions

A consistent system with one or more free variables has infinitely many solutions. Consider

$$x+2y=3.$$

If $y$ is free, write

$$y=t.$$

Then

$$x=3-2t,$$

so the complete solution set is

$$(x,y)=(3-2t,t),\qquad t\in\mathbb R.$$

Equivalently,

$$\begin{pmatrix}x\y\end{pmatrix}

\begin{pmatrix}3\0\end{pmatrix} +t\begin{pmatrix}-2\1\end{pmatrix}.$$

The free parameter describes every solution rather than selecting only one.

Thus a linear system has exactly three possibilities: no solution, one solution or infinitely many solutions.