# The reduced row echelon form of the augmented matrix of a system of linear equations is given. Tell whether the system has one solution, no solution,

The reduced row echelon form of the augmented matrix of a system of linear equations is given. Tell whether the system has one solution, no solution, or infinitely many solutions. Write the solutions or, if there is no solution, say the system is inconsistent. $\left[\begin{array}{cccccc}1& 2& 0& 0& |& -4\\ 0& 0& 1& 0& |& -3\\ 0& 0& 0& 1& |& 2\\ 0& 0& 0& 0& |& 0\end{array}\right]$

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There are no impossible equations, such as $\left(0=1\right)$, the system is consistent,
One variables does not have a leading 1 in its corresponding column, so we take it as a parameter ... the system is consistent and has infinitely many solutions.
Parameters: $x2\in R$
Interpreting row by row as equations.