I understand Mathematica can't assign the results of a Solve to the unknowns because there may be more than 1 solution. How can I assign the 4 values of following result to variables?

enter image description here

Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}]
  • $\begingroup$ Could you elaborate on what you are trying to achieve? In principle you can use Replace (or Part) to assign the values to variables. $\endgroup$ – sebhofer Jun 11 '12 at 13:36
  • $\begingroup$ @sebhofer - I want to assign the first x value to a variable X1, then the y value to Y1, the other two to X2 and Y2. I just don't seem to get the hang of references yet. $\endgroup$ – stevenvh Jun 11 '12 at 13:40

You can do this :

s = Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}];
xx = s[[All, 1, 2]];
yy = s[[All, 2, 2]];

Now you can access solutions, this way xx[[1]], yy[[2]].

If you prefer to collect solutions in Array, there is another way :

X = Array[ x, {Length@s}];
Y = Array[ y, {Length@s}];
x[k_] /; MemberQ[ Range[ Length @ s], k] := s[[k, 1, 2]]
y[k_] /; MemberQ[ Range[ Length @ s], k] := s[[k, 2, 2]]

now X is equivalent to s[[All, 1, 2]], while Y to s[[All, 2, 2]], e.g. :

X[[1]] == x[1]
Y == s[[All, 2, 2]]

You do not have to use or even to define X and Y arrays, e.g.

{x[1], y[1]}
{(-11181 - Sqrt[2242057])/74498, 1/386 (13 - Sqrt[2242057])}

We've used Condition i.e. /; to assure definitions of x[i], y[i] only for i in an appropriate range determined by Length @ s, i.e. number of solutions.

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    $\begingroup$ I think the OP is a complete beginner, and all he's looking for is ReplaceAll. This might be a bit too advanced for someone new to Mma. $\endgroup$ – Szabolcs Jun 11 '12 at 13:57
  • $\begingroup$ @Szabolcs Literally there is the assignment tag, so he is rather looking for Set or SetDelayed applications. $\endgroup$ – Artes Jun 11 '12 at 16:07
  • $\begingroup$ I think I'll use your first solution for now, until I get the hang of ReplaceAll. @Szabolcs: Yes, Mma virgin. Thanks, all. $\endgroup$ – stevenvh Jun 12 '12 at 7:09
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    $\begingroup$ @anhnha That's all information you can get, unless you've defined s in a different way, however one can use FullForm[s]. $\endgroup$ – Artes Sep 21 '17 at 5:06
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    $\begingroup$ @anhnha Read the doccumentation pages of Part then you'll understand what such a notation as [[i,j,k]] mean. $\endgroup$ – Artes Sep 21 '17 at 6:55

Usually you don't want to actually assign values to x and y, and you would use replacement rules instead:

sols = Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}];

{x, y} /. sols[[1]]

or for the second solution:

{x, y} /. sols[[2]]

If you really want to assign values to x and y globally, you could use:

Set @@@ sols[[1]]

but you must clear x and y before using another set:

Clear[x, y]
Set @@@ sols[[2]]

If you want to assign values to x and y within a Block you could do something like this:

Hold @@ {sols[[2]]} /. Rule -> Set /. _[vars_] :>
   Sin[x] + Sqrt[y] // N

This uses what I am calling the injector pattern to get the values into Block in the right syntax without it prematurely evaluating.

Related questions:

Getting rid of the “x ->” in FindInstance results

Using the output of Solve

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  • 1
    $\begingroup$ +1.Something related to your last comment and the injector pattern. $\endgroup$ – Leonid Shifrin Jun 11 '12 at 14:05
  • $\begingroup$ I agree that often you just want to use rules, but in my case I'm trying to set a global variable before going on to solve for other things (where Solve will work better if it has this global value instead of trying to remain general for all possible values). I tried the equivalent of Set[sols[[1]]], and when that didn't work. Of course the "help" file is no help because it insists on explaining the symbol = and not the function Set, so I ended up here. What is Set @@@ sols[[1]] doing? $\endgroup$ – Travis Bemrose Feb 21 '16 at 22:39
  • $\begingroup$ @Travis Assuming you want this in my own words please reference (46248). @@@ is shorthand for Apply at levelspec {1}. So foo @@@ {x -> val1, y -> val2} becomes {foo[x, val1], foo[y, val2]}, or in the code in the question {Set[x, val1], Set[y, val2]}, which then assigns the values with Set. (= is the shorthand for Set.) $\endgroup$ – Mr.Wizard Feb 22 '16 at 4:38
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    $\begingroup$ @Travis Okay. You need to become familiar with the long form of all these "shorthand" operators. You can see the long form by wrapping code in HoldForm[FullForm[ code ]] -- for example HoldForm[FullForm[ {x -> val1, y = val2} ]] will reveal List[Rule[x, val1], Set[y, val2]]. This "FullForm" expression is what you need to visualize when you think about Mathematica manipulating code. Virtually all Replace, Map, Apply, Part, etc. operations effectively "see" this structure, not the short form you type in. See e.g. HoldForm[FullForm[ a - b ]] $\endgroup$ – Mr.Wizard Feb 22 '16 at 16:45
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    $\begingroup$ @Travis the canonical Q&A here is (15567), and I suggest you also look at the links in my answer to (29750). If these do not answer all your questions please let me know. $\endgroup$ – Mr.Wizard Mar 8 '16 at 12:45

Update: Version 10 built-in function Values does value extraction conveniently for rules appearing in lists of arbitrary lengths and depths:

{{x1, y1}, {x2, y2}} = Values[Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}]]
(* {{(-11181-Sqrt[2242057])/74498,1/386 (13-Sqrt[2242057])}, 
    {(-11181+Sqrt[2242057])/74498,1/386 (13+Sqrt[2242057])}} *)

Another example:

(* {{1,2},{3},{{4}},{5,{6,{7}}}} *)

Original post:

{{x1, y1}, {x2, y2}} = Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}][[All, All, -1]]
(* {{(-11181 - Sqrt[2242057])/74498, 1/386 (13 - Sqrt[2242057])}, 
    {(-11181 + Sqrt[2242057])/74498, 1/386 (13 + Sqrt[2242057])}} *)

{x1, y2}
(* {(-11181- Sqrt[2242057]) / 74498, 1 / 386 (13 + Sqrt[2242057])} *)
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  • $\begingroup$ I'm sorry you've deleted your previous answer nevertheless suppresing duplicates we reduce increasing overall entropy of this site, +1. $\endgroup$ – Artes Sep 19 '14 at 21:07
  • $\begingroup$ @artes, thank you for the upvote. I agree with your concern over excessive duplicates. The other Q/A is indeed a special case of this one. However, because of its special structure, some tricks that work there do not work here, e.g Last@@@Solve[...]. $\endgroup$ – kglr Sep 19 '14 at 21:16

If you really wish to assign solutions to variables, you can do something like this:

In[1]:= ClearAll[Subscript]


Subscript[y,2]=1/386 (13+Sqrt[2242057])

Then you can use the solutions for demonstration purposes:

enter image description here

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Here's a painless solution:

s = Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}];

Assign the four results of the solution s above to a variable each, in sequence

{X1, Y1, X2, Y2} = s // Values // Flatten;


{X1, Y1, X2, Y2}
{(-11181 - Sqrt[2242057])/74498, 1/386 (13 - Sqrt[2242057]), 
 (-11181 + Sqrt[2242057])/74498, 1/386 (13 + Sqrt[2242057])}
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Ah, they finally implemented it in version 10, then! Here's a procedure I've been using since version 5, it might provide similar features in versions prior to the introduction of Value. (I'm not sure, but maybe I posted it on the MathGroup... so forgive me if this is not news)

I had called it "ToValues". I gave it two options:

Options[ToValues] = {
      Flattening -> Automatic, IndexedFunction -> False};

The help message is hopefully self-explicating:

ToValues::usage =
    suppresses the Rule wrapper in every part of list li.\n ToValues[li,F] \
applies the function F to every rhs of Rule, turning var->value into \
F[value]. If the function F has a parametrized head, then it is possible to \
pass the lhs of Rule to it by setting the option IndexedFunction->True. It will \
turn var->value into F[var][value].\n When the option Flattening is set to \
Automatic, ToValues flattens li to yield a simplified structure (the \
flattening is tuned to get the simplest list of values for the solution of a \
system of several equation in several variables). With Flattening set to None \
the original structure is left intact.";

The code is really short.

ToValues[li_, opts___Rule] := Module[
    {newli, vars, sols, fl},
    fl = Flattening /. {opts} /. Options[ToValues];
    sols = First[Dimensions[li]]; vars = Last[Dimensions[li]];
    newli = li /. Rule[_, v_] -> v;
    If[fl == Automatic && vars == 1, newli = Flatten[newli]];
    If[fl == Automatic && sols == 1, First[newli], newli]

ToValues[li_, fun_, opts___Rule] := 
    {newli, vars, sols, foo, fl, mi},
    mi = IndexedFunction /. {opts} /. Options[ToValues];
    fl = Flattening /. {opts} /. Options[ToValues];
    If[mi == True,
          newli = li /. (x_ -> v_) -> foo[x][v],
          newli = li /. (_ -> v_) -> foo[v]
    sols = First[Dimensions[li]]; vars = Last[Dimensions[li]];
    If[fl == Automatic && vars == 1, newli = Flatten[newli]];
    If[fl == Automatic && sols == 1, First[newli], newli] //. foo -> fun

Example data:

sols = {{x -> 1}, {y -> 2}, {z -> 3}};

Application of ToValues to lists of rules produces a list of values

ToValues[sols] // InputForm

{1, 2, 3}

Of course assignment is immediate, here

{x1,x2,x3} = ToValues[sols]

By default, ToValues returns the simplest list of values possible, suppressing nested list. If you want to preserve the original nesting, this is what the Flattening option does:

ToValues[sols, Flattening -> None] // InputForm

{{1}, {2}, {3}}

We can specify a function to be applied to the returned values.

ToValues[sols, F] // InputForm

{F[1], F[2], F[3]}

We can also use function that make use of the right hand sides of the rules. Just define them as functions with a parametrized Head and set IndexedFunction->True to instruct ToValues to make use of that. In this case we want to return lists in the form {variable name, variable value}:

F[var_][value_] := {var, value}

ToValues[sols, F, IndexedFunction -> True] // InputForm

{{x, 1}, {y, 2}, {z, 3}}

ToValues[sols, F, IndexedFunction -> True, Flattening -> None] // InputForm

{{{x, 1}}, {{y, 2}}, {{z, 3}}}

Real world applications

ToValues is normally used to extract solutions from the lists returned by Solve, NSolve, DSolve, etc. In its default form it can be applied in this way:

Solve[{x + y == 1, x - y == 2}] // ToValues

{3/2, 1/2}

This gives a list of the complex solutions

Solve[x^5 == 1] // ToValues

This uses the optional function to compute the real and imaginary part of each solution

coords = ToValues[Solve[x^5 == 1, x], {Re[#], Im[#]} &] // N;
ListPlot[coords, AspectRatio -> Automatic, Frame -> True, 
   PlotStyle -> PointSize[.018]];

And this pushes the function to create graphics objects (Points) based on those values inside ToValues:

pts = ToValues[Solve[x^9 == 1, x], Point[{Re[#], Im[#]}] &];
Show[Graphics[{PointSize[.018], pts}],
    AspectRatio -> 1, Frame -> True, Axes -> True];

Someone who has patience enough might want to add the plots.

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Here is an easy solution

s = Solve[y^2 == 13 x + 17 && y == 193 x + 29, {x, y}];
x1 = x /. s[[1, 1]];
y1 = y /. s[[1, 2]];
x2 = x /. s[[2, 1]];
y2 = y /. s[[2, 2]];

We can verify with

MatrixForm[{{x1, y1}, {x2, y2}}]

The output looks like


For reference, please look at assign.

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  • 1
    $\begingroup$ Correct but seems to duplicate parts of earlier answers. $\endgroup$ – bbgodfrey Jun 8 '18 at 21:44
  • $\begingroup$ Also requires knowing the structure of the solution set. The outcome should not depend on whether Solveorders y results before or after x. Here it does though, and that's not good at all. $\endgroup$ – Daniel Lichtblau Jun 8 '18 at 23:26
  • $\begingroup$ It is different, I used replacement with variables instead of direct assignment. $\endgroup$ – zyy Jun 9 '18 at 17:03
  • $\begingroup$ As for requirement of knowing the structure, you have to know the structure before you do the assignment, as Mathematica is not showing the solutions in a particular order, which is a pain for me as well. $\endgroup$ – zyy Jun 9 '18 at 17:05

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