December 5th, 2015, 10:40 pm
QuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: outrunQuoteOriginally posted by: CuchulainnQuoteOriginally posted by: outrunA+B=2C!C is not defined.sorry, this then:2C = A+B?but you don't need to know C, it all cancels out:A+B=2CA-B=C(A+B) + (A-B) = 2A = 3C(A+B) - (A-B) = 2B = CA:B = 2A:2B = 3C:C = 3Your C is my x. And it may NOT be 0.YES is may: L'HôpitalUnfortunately1. A = B = 0 in that case, so no (result is undefined).2. A and B are integers, Hospital's is in the limit of real numbers. No derivative possible.You can solve it as a 2X2 matrix systemM y = 0, y = (A,B) and determinant (M) = -2. => y = 0 => A = B = 0.My f(C) = 3C/C is a real valued function whose solution is the solution to the problem. In this case the solution is indeed an integer, 3, -for all C including C=0-. Also, it's nowhere stated that A:B should be integer. If the first line was A+B=77 then it wouldn't have been, right? And you would have given the non integer solution I expect instead of saying that it has no solution.I see. OK, let's call C >x because x is a real. Now we can use Taylor etc.Could you take us through the steps up to f(x) = 3x/x? Then it should be clear.Ok,So I noticed that the numbers after the equal sign had a nice relation, one of them (76) is twice the other (38). So (using your notation) I call one x and the other 2x. I then eliminate thing for the generic solution when one number is twice the other. Turns out that for all those case the solution to A:B is 3. So if the first eq said A+B=6.28 and the 2nd A-B=3.14 then A:B would also be 3 (but that's irrelevant because it's not saying that)What you and I do is first generalize the problem (it has one solution, but we introduce variables) and show that the generalized problem has one solution. If I had said A+B=2x^2 and A-B=|x|^2 then I would have had to do more work to find the only good solution.A=B=0 is not a solution of the original problem, I don't make A and B variables. I think the only assumptions I make about the A,B symbols are that D-D=0 and D+D=2D (substitute D with A or B), so my method would also work if A and B are matrices is seems! It's not a goal, just a property of the chosen generalization. Different generalization can solve different *other* problems too, but that doesn't matter and it's irrelevant.Indeed! A and B are nuisance variables. Gauss Elimination on the general form of the problem (A+B=X; A-B=Y, find A/B) yields a very satisfying answer of (X+Y)/(X-Y) which is 3 for the numbers given.Also, a clever person (which I was not!) would notice the linearity of the system. Thus, one can scale the "76" and "38" by any common constant without affecting the result. From an mental arithmetic standpoint, solving the A+B=2, A-B = 1 lets one avoid two-digit math.
Last edited by
Traden4Alpha on December 5th, 2015, 11:00 pm, edited 1 time in total.