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when you try to remove the 4 below the 1 you take row two and subtract 4 times row 1. Doing this gets rid of the 4, but it also gets rid of the 12. Leaving you with a matrix that looks like this
If you ever have row, such as the second row, where the last entry is not zero but all the other entries are zero, then there is no solution, which means the lines do not cross. However if you have the same number of equations as variables and row reduction yield a row of all zeros , then there are infinitely many solutions. So basically a long story short, you are correct there is no solution to this problem.
Hi Lisa. Another way to tell that this is insoluble is to remember that a solution means an ordered pair (x,y) such that both equations are true. This means the intersection of the two lines. Remember back to when you first learned about line equations. Standard form for a line is: Ax + By = C. and from this, the slope is A/B. Since your two equations have the same A/B, they are parallel and hence, don't intersect, so there is no solution. This isn't proving by augmented matrices, but it is a perfectly valid way to check your answer. Stephen La Rocque.  


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