You havetwoequations inthreeunknowns so, as a purely linear algebra problem, this has no single solutions. But here, of course, a, s, and c must bepositive integersand the problem says "solutions", plural.

However, your row-reduce matrix is wrong- you dropped a sign. (Notice that the top row is equivalent to the equation a+ .83333 c= -10. a and c cannot both be non-negative!)

Your row reduced matrixshouldbe .

That is equivalent to and or and .

Now, since a and s must be non-zero integers, it is clear that c must be a multiple of 6, at least 12, and no larger than 48 ( ). That is, c can be 12, 18, 24, 30, 36, 42, or 48.

If c= 12, a= 10-10= 0 and s= 89- 22= 67.

If c= 18, a= 15- 10= 5 and s= 89- 33= 56, etc.