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I obtained a list of Y list after using Table[NSolve[expression,X],{var,varmin,varmax,step}]. Then I created a list of X with Range. I concatenated both of these lists and took the transpose to get x,y points. I need to plot the two lists. I think I am getting issue because of the X-> obtained in the list. Can anyone tell how to get rid of X-> in the list. It seems something trivial but I don't seem to know it.

Some points of the list {{0.1,X->0.333333}, {0.2,X->0.5}, {0.3,X->0.6}, {0.4,X->0.666667}} .

Code

expression = k1 *kinase *(Xtot - X) - k11 *phase* X == 0;
phase = 0.2; k1 = 1; k11 = 1; Xtot = 1;
Yvalues = Table[NSolve[expression, X], {kinase , 0.1, 2, 0.1}] // Flatten;
Xvalues = Range[0.1, 2, 0.1];
Finallist = Transpose[{Xvalues,Yvalues}]
ListPlot[Finallist]

If anyone knows a better way to plot it would be much better.

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  • 2
    $\begingroup$ Change the definition of Yvalues to Yvalues = Table[X /. NSolve[expression, X], {kinase, 0.1, 2, 0.1}] // Flatten; $\endgroup$ – Bob Hanlon Mar 6 at 19:05
  • $\begingroup$ See also Values $\endgroup$ – OkkesDulgerci Mar 7 at 5:47
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This will do it for you

    expression = k1*kinase*(Xtot - X) - k11*phase*X == 0;
    phase = 0.2; k1 = 1; k11 = 1; Xtot = 1;
    Yvalues = 
      Table[X /. NSolve[expression, X], {kinase, 0.1, 2, 0.1}] // Flatten;
    Xvalues = Range[0.1, 2, 0.1];
    Finallist = Transpose[{Xvalues, Yvalues}];
    ListPlot[Finallist]
    ListLinePlot[Finallist]

Output plot

With line

You need to look up ReplaceAll to get rid of the X-> The output is in this form to prevent X being set to a numerical value. Also note that it is best to use lower case variables since these won't clash with Mathematica variables.

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  • $\begingroup$ Thanks. Do you know how to get a line connecting all these points? $\endgroup$ – A Q Mar 6 at 19:07
  • $\begingroup$ See edit....... $\endgroup$ – Hugh Mar 6 at 19:10
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Try this:

expression = k1*kinase*(Xtot - X) - k11*phase*X == 0;
phase = 0.2; k1 = 1; k11 = 1; Xtot = 1;
Yvalues = 
  Table[Solve[expression, X], {kinase, 0.1, 2, 0.1}] // Flatten;
Xvalues = Range[0.1, 2, 0.1];
finallist = Transpose[{Xvalues, Yvalues}] /. {a_, X -> b_} -> {a, b}

(*
{{0.1, 0.333333}, {0.2, 0.5}, {0.3, 0.6}, {0.4, 0.666667}, {0.5, 
  0.714286}, {0.6, 0.75}, {0.7, 0.777778}, {0.8, 0.8}, {0.9, 
  0.818182}, {1., 0.833333}, {1.1, 0.846154}, {1.2, 0.857143}, {1.3, 
  0.866667}, {1.4, 0.875}, {1.5, 0.882353}, {1.6, 0.888889}, {1.7, 
  0.894737}, {1.8, 0.9}, {1.9, 0.904762}, {2., 0.909091}}
*)

Then

ListPlot[finallist]

yielding this:

enter image description here

Have fun!

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  • $\begingroup$ By the way, you could make your final list in a more concise way: finallist = Table[{kinase, Solve[expression, X][[1, 1, 2]]}, {kinase, 0.1, 2, 0.1}]. $\endgroup$ – Alexei Boulbitch Mar 6 at 19:48
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Your problem can be solved exactly.

expression = k1*kinase*(Xtot - X) - k11*phase*X == 0;
phase = 1/5; k1 = 1; k11 = 1; Xtot = 1;

As with the other solutions, using ReplaceAll

x[kinase_] = X /. Solve[expression, X][[1]]

(* (5 kinase)/(1 + 5 kinase) *)

Plot[x[kinase], {kinase, 0, 2},
 AxesLabel -> {"kinase", "X"},
 PlotRange->All] (* EDIT: Added PlotRange -> All *)

enter image description here

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We can directly use ContourPlot to plot the solution of the equation k1*kinase*(Xtot - X) - k11*phase*X.

phase = 0.2; k1 = 1; k11 = 1; Xtot = 1;
ContourPlot[
 k1*kinase*(Xtot - X) - k11*phase*X == 0, {kinase, 0.1, 2}, {X, 0.1, 
  2}]

enter image description here

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