Originally Posted by

**Ackbeet** Your conclusion is correct, but not because of the correct reason. The system has infinitely many solutions because in your reduction down to

x1 = 2 - 5(x3)

x2 = 3 - 4(x3),

you can have any value of x3, and x1 and x2 will be determined. The system would have no solutions if you had a system like this after row reduction:

$\displaystyle \left[\begin{matrix}1 &0 &3\\ 0 &1 &4\\ 0 & 0 &0\end{matrix}\left|\begin{matrix}3\\ 5\\ 2\end{matrix}\right]$

The last row there is a contradiction, because if you multiply every variable by zero, you're going to get zero, which is not equal to 2.

In any case, the important thing in your case is that there *is* a solution. Getting back to the original question, what does this tell you?