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# Why am I not getting a result for the following computation?

Posted 9 years ago
 It says that WA is not able to calculate a solution for the following system of equations. I assume it should be simple enough xa=.10, xb+1-x=.89, xc=.01, xa+(1-x)d=.1, xb+(1-x)e=0, xc+(1-x)f=.9,(1-x)d=0, x+(1-x)e,(1-x)f=.89
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Posted 9 years ago
 As I said, what I had done still does not solve your problem with WolframAlpha.If I put in your first six equations it does give me a solution for those.If you see the same then is it possible for you to then take that solution along with your last two equations and ask WolframAlpha to solve that as a separate problem and get you a solution to your entire problem?Or can you modify this idea, to take some subset of your equations, find a solution and use that with the rest of your equations?
Posted 9 years ago
 Thanks for the help so far. @Bill Simpson, I see that in the link you sent me, the last equation, shown below was omitted (1-x)g+x=.11This is the one that you mentioned I inputted incorrectly. Now that I put it in, it still does not give me definite values for my variables. This is what I would expect, given 8 equations and 7 unknowns. Is this due to lack of computing power in WA? How can I rectify this?
Posted 9 years ago
 If he is using WolframAlpha, where xa is the product of two variables, and, based on some experimentation, I suspect he may have meant to write x + (1 - x) e= .11 instead of just x + (1 - x) e and I translate this calculation into Mathematica just to see the result then it seems In:= eqns = {x a == 1/10, x b + 1 - x == 89/100, x c == 1/100, x a + (1 - x) d == 1/10, x b + (1 - x) e == 0, x c + (1 - x) f == 9/10, (1 - x) d == 0, x + (1 - x) e == 11/100, (1 - x) f == 89/100}; sol = Reduce[eqns, {a, b, c, d, e, f, x}, Backsubstitution -> True] Out= b == 1/10 (10 - 11 a) && c == a/10 && d == 0 && -1 + 10 a != 0 && e == (-10 + 11 a)/(10 (-1 + 10 a)) && f == (89 a)/(10 (-1 + 10 a)) && a != 0 && x == 1/(10 a) In:= Simplify[eqns //. ToRules[sol]] Out= {True, True, True, True, True, True, True, True, True} But this still doesn't address why WolframAlpha doesn't seem to provide this, even when you replace 'e' with 'g' or another name.
Posted 9 years ago
 Oh, right, my mistake... All good points.
Posted 9 years ago
 The set of equations is inconsistent. here are your equations:  In:= eqns = {xa == .10, 1 - x == .89 - xb, xc == .01, xa + (1 - x) d == .1, xb + (1 - x) e == 0, xc + (1 - x) f == .9, (1 - x) d == 0, x + (1 - x) e, (1 - x) f == .89} Now substitute the explicitly set values that are already in them In:= eqns /. {xa -> .1, xc -> .01} Out= {True, 1 - x == 0.89 - xb, True, 0.1 + d (1 - x) == 0.1, e (1 - x) + xb == 0, 0.01 + f (1 - x) == 0.9, d (1 - x) == 0, e (1 - x) + x, f (1 - x) == 0.89} Ok, from d (1 - x) == 0, one sees that x==0 so let's use that too: In:= eqns /. {xa -> .1, xc -> .01, x -> 1} Out= {True, 0 == 0.89 - xb, True, True, xb == 0, False, True, 1, False} So the set of equations is inconsistent and there are therefore no solutions.
Posted 9 years ago
 WolframAlpha is interpreting, without giving you any hint it has done this or any option to change this, that e=Euler's Constant=2.71828...I assume that is not what you intended.It would be really nice for novice users if WolframAlpha made this very clear. It has provided information and options for other interpretations when solving other problems so this feature might not be out of character to incorporate.If you use perhaps g instead of e you can get around this.Your next to last item lacks an = and is not an equation. None of the values I can guess give a consistent set of equations or allow WolframAlpha to find a solution.If I remove the last two items WolframAlpha is able to quickly provide a solution to the partial problem.Here