Question 814179
<pre>
 5     3     7     4
 3    26     2     9
 7     2    10     5
Multiply row 1 by 1/5:
 1   3/5   7/5   4/5
 3    26     2     9
 7     2    10     5
Add -3 times row 1 to row 2:
 1    3/5   7/5   4/5
 0  121/5 -11/5  33/5
 7     2    10     5
Add -7 times row 1 to row 3:
 1    3/5   7/5   4/5
 0  121/5 -11/5  33/5
 0  -11/5   1/5  -3/5
Multiply row 2 by 5/121:
 1    3/5   7/5   4/5
 0      1 -1/11  3/11
 0  -11/5   1/5  -3/5
Add -3/5 times row 2 to row 1:
 1      0  16/11  7/11
 0      1  -1/11  3/11
 0  -11/5   1/5  -3/5
Add 11/5 times row 2 to row 3:
 1      0  16/11  7/11
 0      1  -1/11  3/11
 0      0    0     0
</pre>This is as far as we can go. This translates into:
 x + (16/11)z = 7/11 or x = (-16/11)z + 7/11
 y - (1/11)z = 3/11 or y = (1/11)z + 3/11
So there are an infinite number of solutions. One way to express the solution would be:
((-16/11)z + 7/11, (1/11)z + 3/11, z) where z is any real number.