question about Solve

*To*: mathgroup at smc.vnet.net*Subject*: [mg91690] question about Solve*From*: "Jaccard Florian" <Florian.Jaccard at he-arc.ch>*Date*: Fri, 5 Sep 2008 07:14:18 -0400 (EDT)

Dear All, I don't understand the following behaviour of Solve. Consider the following system : Solve[{x*y==(a+2*b)/(c+2*d),1/Sqrt[2]==Sqrt[(e*y)/(z*f*g*h)],2*Pi*i==0.9/(z*f)},{x,y,z}] Everything fine, I obtain : {{z -> 0.1432394487827058/(f*i), x -> (1.419827298426263*^-9*(9.83403688*^9*a + 1.966807376*^10*b)*e*i)/((c + 2.*d)*g*h), y -> (0.0716197243913529*g*h)/(e*i)}} But if I ask for the answer of almost the same (only a 4 in the denominator of the second equation), Solve isn't abble anymore to manage without using inverse functions... why? Solve[{x*y == (a + 2*b)/(c + 2*d), 1/Sqrt[2] == Sqrt[(e*y)/(4*z*f*g*h)], 2*Pi*i == 0.9/(z*f)}, {x, y, z}] Worse: If I have numerical values for a, b, c, d, e, f, g, h, i: a = 65/10^6; b = 1/10^3; c = 1.9; d = 0.19; e = 1/(2.5/10^3); v = 18; w = 8; g = (2*v)/((c + 2*d)*w); i = 3000; h = 0.2; Then Solve isn't able anymore! Mathematica thinks there is no solution. But there is one. I have to use Reduce or give the numerical values as rules after the Solve to find them. In[247]:= Solve[{x*y == (a + 2*b)/(c + 2*d), 1/Sqrt[2] == Sqrt[(e*y)/(z*f*g*h)], 2*Pi*i == 0.9/(z*f)}, {x, y, z}] Out[247]= {} Can somebody explain this? Thanks in advance, Regards F.Jaccard

**Follow-Ups**:**Re: question about Solve***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>