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Bob Hanlon
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Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *)   

In this next example one of the definitions (= vice ==) is embedded in the Solve since it cannot be included with the other definitions since it depends on them.

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[dbtot = Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) 

Or it can be pulled out

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, dbtot = Sqrt[x^2 + y^2 + z^2]; Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *)   With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[dbtot = Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) 

Or

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, dbtot = Sqrt[x^2 + y^2 + z^2]; Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) 

In this next example one of the definitions (= vice ==) is embedded in the Solve since it cannot be included with the other definitions since it depends on them.

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[dbtot = Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) 

Or it can be pulled out

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, dbtot = Sqrt[x^2 + y^2 + z^2]; Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 
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Bob Hanlon
  • 167.4k
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  • 82
  • 213

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtotSolve[dbtot === Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) 

Or

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, dbtot = Sqrt[x^2 + y^2 + z^2]; Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *)   Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r)}, dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[dbtot = Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) 

Or

With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, dbtot = Sqrt[x^2 + y^2 + z^2]; Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *)   Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 
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David G. Stork
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Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[dbtotSolve[{dbtot === Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r)}, dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[dbtot = Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 

Solve seems to work better if you separate definitions from equations.

Off[Solve::ratnz] With[{r = .4, dbtot = 2.06155}, Solve[1/dib == -2/r - 1/(dbtot - r), dib]] (* {{dib -> -0.178513}} *) With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2]; 1/dib == -2/r - 1/(dbtot - r)}, dib]] (* {{dib -> -0.178513}} *) Clear[dbtot] With[{x = 2, y = 0.5, z = 0.0, r = 0.4}, Solve[{dbtot == Sqrt[x^2 + y^2 + z^2], 1/dib == -2/r - 1/(dbtot - r)}, {dbtot, dib}]] (* {{dbtot -> 2.06155, dib -> -0.178513}} *) 
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Bob Hanlon
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