Question 1156104
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There is an endless number of ways to do this....<br>
{{{matrix(2,3,3,-2,2,5,-5,10)}}}<br>
First objective: get a 1 in row 1, column 1.  I choose to do that by dividing the second row by 5 and switching rows.<br>
{{{matrix(2,3,1,-1,2,3,-2,2)}}}<br>
Next objective: 0 in row 2, column 1.  Only one way to do that and keep the 1 in row 1 column 1.  Replace row 2 with (row 2) minus (3 times row 1).<br>
{{{matrix(2,3,1,-2,2,0,1,-4)}}}<br>
Next objective: 1 in row 2 column 2.  No work needed -- it's already there.<br>
Last objective: Again only 1 way to do it without changing the entries we have already fixed -- replace row 2 with (row 2) plus (row 1).<br>
{{{matrix(2,3,1,0,-2,0,1,-4)}}}<br>
The solution is (-2,-4)<br>