Reduced row-echelon form

The "simple" matrix you want to end up with when you apply elementary row operations to an augmented matrix is a matrix in a very specific form. Some terminology: the first non-zero number in any row of your augmented matrix is called the leading entry, or pivot entry for that row.

A matrix is in reduced row-echelon form when

any rows of all zeroes (called zero rows) are at the bottom

the leading entry in any other row is a 1 (a leading 1, or pivot 1)

each leading 1 is further to the right than any leading 1's above it

any column with a leading 1 has zeroes in the rest of that column.

Once your augmented matrix has been transformed into reduced row-echelon form, there's a systematic way to solve the linear system. Each row of your new matrix corresponds to an equation, and the new system of equations has the same solutions as the old one.

If your reduced matrix contains a "bad row" – a string of zeroes followed by a leading 1 in the very last column – that bad row translates into the impossible equation 0 = 1, so your system has no solution, no matter what the other equations are. Stop.
Any zero row in the reduced matrix just translates into the equation 0 = 0, which is true but contributes no information – ignore that row/equation.
For the remaining rows:
- Identify the leading variables (those corresponding to a leading 1) and the free variables (the others).
- The free variables are called free because they can take on any value; none of the equations relates any of them to each other. Assign arbitrary parameter variables to these free variables to indicate this freedom.
- Solve each equation for its leading variable in terms of the parameters.

Here's how that works for the augmented matrix above.

There's no bad row.
The leading variables are x1, x3 and x4. The free variables are then x2 and x5, so set x2 = s and x5 = t, where s and t are parameters.
The non-zero rows of the matrix give the equations

x1 – 2x2 + 3x5	= 2	or	x1 =	2 + 2s – 3t
x3 – 5x5	= –3	or	x3 =	–3 +5t
x4 + x5	= 7	or	x4=	7 – t

The full solution is then

x1 =	2 + 2s – 3t	where s and t are parameters.
x2 =	s
x3 =	–3 + 5t
x4 =	7 – t
x5 =	t

To summarize the process:

To solve a linear system when its augmented matrix is in reduced row-echelon form

If there's a leading 1 in the last column, stop: there is no solution.

Otherwise, identify the leading variables and the free variables. Assign parameter values to the free variables.

Translate the non-zero rows of the matrix back into equations.

Solve each equation for its leading variable in terms of the parameters.