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Solve [(((750/(k*v*2))*(1 - e^(-2 k)))/(1 - e^(-k*8)))*e^(-6 k) == 
    50 && (((750/(k*v*2))*(1 - e^(-2 k)))/(1 - e^(-k*6)))*e^(-4 k) == 
    70  , {v, k}
    ] /. e -> E 

It said "Solve::nsmet: This system cannot be solved with the methods available to Solve." However, if I put another variable to Solve (even if it does not exist) like this: Solve [.... , {v,k,s}] it can actually solve the equations. What just happened? is it just a bug or there are anything explanation?


Im using Wolfram Mathematica 8.0 and

Solve [(((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*8)))*E^(-6 k) == 
   50 && (((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*6)))*E^(-4 k) == 
   70, {k, v}, Method -> Reduce]

does not work for me it keeps running all day and returns

No more memory available. Mathematica kernel has shut down. Try quitting other applications and then retry.

However, if i let 50 equal variable S and solve for v,k,S it somehow gives

Solve [(((750/(k*v*2))*(1 - e^(-2 k)))/(1 - e^(-k*8)))*e^(-6 k) == 
    S && (((750/(k*v*2))*(1 - e^(-2 k)))/(1 - e^(-k*6)))*e^(-4 k) == 
    70, {v, k, S}] /. e -> E 

v -> 75/(14 (1 + E^(2 k) + E^(4 k)) Log[E^k]), S -> (
 70 (1 + E^(2 k) + E^(4 k)))/(1 + E^(2 k) + E^(4 k) + E^(6 k))

which i know S==50 then I can solve for k and v later on

Solve [S == (70 (1 + E^(2 k) + E^(4 k)))/(
    1 + E^(2 k) + E^(4 k) + E^(6 k)), k] /. S -> 50 // N

How come putting exact number is more complicated than defining a variable as it could be any numbers and 50 is just one of them

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    $\begingroup$ Per the text on bugs, it should only be used once the community or WRI has confirmed it is a bug. $\endgroup$ – rcollyer Aug 11 '16 at 15:39
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When you run your code, you get the warning:

Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information.

Why not try using Reduce?

Reduce[(((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*8)))*E^(-6 k) == 
   50 && (((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*6)))*E^(-4 k) == 
   70, {v, k}]

This gives a clear answer.

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    $\begingroup$ Now, why when using a system, would one want to pay attention to the advice the system gives its users? ;) (+1) $\endgroup$ – Michael E2 Aug 11 '16 at 15:45
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This is just to remind people that (1) you may access Reduce in Solve through the Method option and (2) sometimes the order of the variables matter when constructing a CAD.

This returns a solution in a few seconds

Solve[(((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*8)))*E^(-6 k) == 
   50 && (((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*6)))*E^(-4 k) == 
   70, {k, v}, Method -> Reduce]

but this does not (it takes longer I'm willing to wait, several minutes):

Solve[(((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*8)))*E^(-6 k) == 
   50 && (((750/(k*v*2))*(1 - E^(-2 k)))/(1 - E^(-k*6)))*E^(-4 k) == 
   70, {v, k}, Method -> Reduce]
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