quote:
Originally posted by -dt-
quote:
Originally posted by Adeptus
quote:
Originally posted by -dt-
i have notes on how to do them if they are >= something but (Smilie) I've never encountered a <= one so I'm unsure of how to solve it.
It seems to me if you know how to solve it for x1 >= 5 (which you are implying you do), then you could subtract the number of those solutions from the total number of solutions and get the answer you want....
oh 
thanks I never thought of that
Will that work? Remember that the procedure you are aware of also includes the = solutions so, those will be common. By Adeptus method, you'll obtain the < solutions. Or am I looking at this in a wrong way?
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