Help solve 2 equations with 4 unknowns

For my class, we have to solve 2 systems with these equations:

(a*c)/(b*d) = 22.27 +/- 0.01

a+b = c+d

the numbers must be integers and greater than 15.

How would i go about solving this?

I have already tried for a bit and reached a dead end.

I used

syms a b c d;
solve([(a*c)/(b*d) == 22.27, a+b == c+d], [a,b,c,d])
Warning: 6 equations in 4 variables. 
> In C:\Program Files\MATLAB\R2014a\toolbox\symbolic\symbolic\symengine.p>symengine at 56
  In mupadengine.mupadengine>mupadengine.evalin at 97
  In mupadengine.mupadengine>mupadengine.feval at 150
  In solve at 170 
Warning: Explicit solution could not be found. 
> In solve at 179

and get the above error

Any help would be very appreciated.

Answer 1

[Roger Stafford]

I don't think the approach you describe or anything like it would succeed, David. The 'solve' function is not smart enough for that job. If I were doing your problem I would reduce the number of unknowns by taking the second equation into consideration. Call s = a+b = c+d and replace b and d by the corresponding s, a, and c. Then to avoid round-off problems that can cause you to miss a solution I would rewrite your first equation as

   100*a*c-2228*(s-a)*(s-c) = 0 (or 100*a*c-2226*(s-a)*(s-c) = 0)

Finally, I would devise three nested for-loops which go through all possible combinations of integers a, b, and s whose magnitudes do not exceed some large number like N = 100 or whatever you think it would take to find a solution. Set up a test to detect when the above expression is equal to zero and produce a printout of the integers.

[Roger Stafford]

I tried out my idea above and realized, especially with the 22.26 case, that you would need to use a smarter method to avoid an excessive amount of computation. I would suggest just two nested for-loops looking like this:

   N = some pretty big number
   for s = 32:N
     for a = 16:s-16

Then instead of searching for c in another inside for-loop, just solve for c in the above equation and determine if the result is an integer greater than 15 with d also greater than 15. That would be much, much more efficient. I got lots of solutions that way.