Mathematica solve eliminate If your solutions are always numbers, you can produce all solutions with Reduce and then filter them - it may be faster in some cases. 12. The sizes of the Eliminate intermediate I want to find the positive root of the following function F[x], but using Solve does not work. You should have Eliminating complex numbers from solutions. In[1]:= Solve[x + y + z == 5 && y == 3, {x}, MaxExtraConditions -> All] Out[1]= {{x -> ConditionalExpression[2 - z, y == 3]}} Given a expression (polynomial, or any equation in general) such as a s^2+b = 0 I want to solve for s^2, to get s^2 = -b/a. Sanjay Manohar New to Mathematica, trying to solve a simple economics problem but having troubles with using Solve and Eliminate. Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. However Mathematica can't Solve [expr, vars] attempts to solve the system expr of equations or inequalities for the variables vars. That syntax makes Mathematica assume that y can be anything, and if y is different from 13/2, then there's no x, so it can't solve it for the general case. 057; n0 = 0; n1 = 1; gE0 = 1/g0 Mathematica. My code is: ClearAll[a, b, x, y, a12, a22] Assuming[ {a > 0, b > 0, a12 > 0, a22 > 0}, Solve[ Skip to main content. e. solutions that are always valid independently of the values of the parameters. I have to perform a really simple operation, but I'm really struggling with that. This page titled 5. 7 & 8 I'm going to suggest how to deal with it in earlier versions. " No result is shown, however. Is there an easier way to do all this? For example, can I somehow specify in a single step the variables I want to eliminate and the variable for which I want to solve the equation and then get the output in InputForm? Eliminate seems to do the job, but dosent work well when the resulting constraint should be an inequality (for example y==x*x, eliminating x, I would like to get y>=0. Ask Question Asked 10 years, 10 months ago. It is something from Simplifying polynomials with conditions on variables , that intend to minimize the number of variables. 1. In the Wolfram Language, rules associate symbols with values. I would like to eliminate all of the variables except for {en, s, s1000} and then obtain the solution in the form en[s, s1000]. Maple gives solution but in terms of unresolved integrals. , {VOUT}. py This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. Follow answered Apr 13, 2022 at 13:22. ; What are Root objects?. Solve[{loop1, loop2, loop3}, {i1, i2, i3}] returns the solution. Also, let's define Av as Vo/Vi. Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information. The variables to be eliminated must be in list brackets, e. Follow Why Mathematica can't solve this? NSolve or Solve. = iNt[x] /. I have a set of 4 equations, for example (they can also be more complex): $ AA = If expr consists only of integers and Root and AlgebraicNumber objects combined using algebraic operations, then the result from RootReduce [expr] will always be a single Root object. The functions use a collection of algorithms applicable to classes of problems satisfying particular properties, I have a series of equations where I use eliminate to get a new equation with desired variables. That adds one equation to the original equation, so that you now have the freedom to eliminate one of the original unknowns. An example of my expression is: K=a*b*t/((t+f)c*d); (the actual expr There is nothing wrong with the 1/x term in the ODEs. I basically want to solve these simyltaneous equiations: NSolve[ { - That syntax makes Mathematica assume that y can be anything, and if y is different from 13/2, then there's no x, so it can't solve it for the general case. You could also explicitly ask solve to eliminate y. The equation x==y tests whether x is equal to y. This is good Mathematica practice. 3: Solve Systems of Equations by When using the Solve function with Mathematica, you can specify for what variables you want Solve to specify the solutions. Solve[expr, vars, dom] solves over the domain dom. 1. Also, according to the docs, Assuming only affects those functions that have an Assumptions option. For linear functions in any number of variables, GroebnerBasis is equivalent to Gaussian elimination. but while there are some results for equation pairs, for the whole system Mathematica just runs for ever Solve and other functions such as FindInstance, NSolve, and NDSolve return a list of rules. 2. I have used Eliminate[] to reduce a system of equations as found here, I then proceeded to verify that indeed my system has zeros using this code. there are many many complex solutions so the restriction of the domain to Reals is important to get $\begingroup$ @JandR: As a point of note, Solve changed between v. For example: exp1 = Expand[(1 + x)^9] Take[exp1, 5] 1 + 9 x + 36 x^2 + 84 x^3 + 126 x^4 + 126 x^5 + 84 x^6 + 36 Looking in the documentation to the Solve command, I don't find any examples of solving functional equations. Off the top $\begingroup$ In Mathematica highlight the unknown symbol (e. My question is, what is the best way to retrieve those zeros for x1 x 1, x2 x 2, This message is generated when and related functions use inverse functions to represent a solution. Replacing x by 1/2, the solution you're looking for, doesn't make the equalities be true. Use ContourPlot and Graphics`MeshFindIntersections. Asking for help, clarification, or responding to other The set of polynomials in a Gr ö bner basis have the same collection of roots as the original polynomials. 8 it is for the domain. We can stuff all of the equations into Eliminate, asking it to eliminate Vo, Vi, VL, VR, and vd: Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. The current situation is (1) Eliminate and Solve are both frequently used i. A single variable or a list of variables can be specified. Python odeint clearly returning wrong solution. 8. I'm having an issue with my code. Improve this answer. Note the underscores; they matter (in a way that will be explained later). Stack Exchange Network. To do that, we can multiply both sides of an equation by the same factor. $\endgroup$ – eyorble. > Solve::ratnz: Solve was unable to solve the system with inexact > coefficients. ContourPlot[{ -1 + 4/(5 x^(3/10)) + 6/(5 y^(3/10)) == 0, -1 + (2 x^(7/10))/5 You can rewrite expressions in this way using a bare symbol, Eliminate, and Solve: ClearAll[f]; expr = (I Sqrt[Subscript[\[Sigma], 2]/Subscript[\[Sigma], 1 You can also eliminate variables using Mathematica. Here we discuss equations, which test equality. ; Simple Root objects may in turn automatically evaluate to rational expressions or combinations of radicals. I think this should be easy to numerically solve in Mathematica, but for some reason I'm not finding the correct way to do it. Substitute the solution from Step 4 into one of the original equations. So you should write . march march. Then solve for the other variable. For them to be true, you need to solve for y also $\begingroup$ @JandR: As a point of note, Solve changed between v. I want to eliminate variables from an equation using MATLAB. In v. Thanks! wolfram-mathematica; Share. $\endgroup$ – Using Mathematica 8. Modified 3 years, 11 months ago. $\endgroup$ – These are very hard for computers to solve. . If you need a math solver, MathGPT is the AI I am trying to get an equation in terms of variable z by eliminating variables g0 and g1 but Running below code is taking time forever In[10]:= y = 0. @@@ muSolutions but this is This answer is to summarize the most important points about working with Root objects. I tend to think that novice users might come up to speed a little easier and be a little less confused if we show them Map and Apply and ReplaceAll with [ and ] neatly containing their arguments than if we just give them strings of /@ and @@ and /. It takes conditions restricting those variables as part of its 1st argument. NSolve gives {} if there are no solutions to the equations. ; Inverse functions normally give one branch of a general inverse, so equations that are solved in terms of inverse functions may omit solutions associated with other branches. In the simplest example, I am hoping the following Asking Mathematica to Eliminate two of the three variables I'd like to remove and then simplifying the result gets me to the answer fairly quickly: How can I make Eliminates life easier, or are there other tools I could try to solve this system of equations? I would like to eliminate p and generate either a single polynomial in c or complex numerical solutions to c. This equation is then solved for a specific variable and then the result is presented in a specific Perform Eliminate, Solve and Format at once in Mathematica. $\endgroup$ – Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information. 5, 0. I am trying to solve a system of non-linear equations using some easy assumptions. Please read the documentation of Solve to see that the only valid third argument of this function is a domain. And then use FindRoot again. Start by defining Mathematica functions for g and h. >> I get the warning, "Solve::ratnz: Solve was unable to solve the system with inexact coefficients. How can I solve a set of equations with variables and eliminate a When I run *Mathematica* to get the solution of a differential equation it gives the result in terms of inverse function. The question: Now suppose that this agent makes two choices, expressed by x and y and that these You do get a warning in mathematica 11. I'm stuck on how to write the code for this problem: a) Eliminate the parameter to find a Cartesian equation for a parametric curve. Incorporating methods that span from antiquity to the latest cutting-edge research at Wolfram Research, the Wolfram Language has the world's broadest and deepest integrated web of polynomial algorithms. python odeint gives strange results. If you were to enter just VOUT then Mma would first try to interpreted it as a domain resulting in the warning: If your polynomial is actually as simple as shown, with a term for every power of x and none others, you can simply use Take or Part to extract only those terms that you want because of the automatic ordering (in Plus) that Mathematica uses. Since it involves algebraic rewriting of a rational expression, a standard method from computational algebra involves Groebner bases. DSolve[y'[t] == ε/Sqrt[1 + k/y[t]^2], y[t], t] My question is how Solve's 2nd argument must be a variable or list of variables to solve for. I tried Solve, Reduce, Eliminate, tried taking them pairwise etc. Follow edited Sep 11, 2014 at 23:52. To review, open the file in an editor that reveals hidden Unicode characters. Per Alexandersson There are no conditions given to solve for all the constants. We all know that one can't just write Solve[eq==0,s^2] because s^2 i Using Mathematica 8. When I use Solve I get: Solve::nsmet: This system cannot be solved with the methods available to Solve. >>. Assumptions that contain solve variables are considered to be a part of the system to solve: SolveValues does not eliminate solutions that are neither generically correct nor generically incorrect: Mathematica complains when I am including the nonlinear part. Solve Equations from Equations Mathematica. 7 the third param was for variables to eliminate, and in v. $\begingroup$ Dear @RenéG, you even do not need to use Solve. Improve this question. I understand that Mathematica evaluates symbols as early as possible, and so I have tried playing with Hold but it didn't help. Since you do not want the B variables present in the solution for A, solve for A1 and all of the B variables and then use only the A1 solution. However, the constraints are enforced in the s plot domain specification, so I thought it There is no solution to since there is a leading 1 in the last column; confirm using Solve: Find all solutions of the following system of equations: First, write the coefficient matrix , variable vector and constant vector : This message is generated when Solve and related functions use inverse functions to represent a solution. y == a x (x + 1) Exp[b (1 - x/85)^(1/2)] + c x^2 + d x for x and y values x={16,18,20,22}, Mathematica does have bugs and limitations, but doubting about its capability should always be the last thing to do. Provide details and share your research! But avoid . Commented Dec 10, 2020 at 21:14 $\begingroup$ Maybe I should use LinearSolve The OP indicates that the solution in my comment is preferred, so in keeping with SE philosophy, I will include in my answer. After failing to get any output after several minutes, I eventually have to abort the evaluation. Share. Eliminate[%, {c}] leaves no equations, and Eliminate returns (* True *) (Why it does not return Null, instead, I do not know. Besides, Normal@Solve[. Solving ode with python getting wrong solution. Follow edited Jun 15, 2015 at 3:30. Ask Question Asked 7 months ago. b) To solve this system by elimination, we first need to rewrite one or both equations so that one variable can be eliminated. The answer was obtained by solving a corresponding exact system and numericizing the result. For example, lets consider the following equations: p = (m + n) q = (m - n) r = (m^3 - n^3) Now, r could be expressed in terms of p and q by entirely eliminating m and n like this: r = (3*p^2*q + q^3)/4. 24. What is going on, and how can I enter the equations normally, and solve for them in terms of specific variables? equation-solving; Share. Simultaneous equations can be combined either in a list or with &&. From a contour plot of your equations I believe this result is correct and there are no real valued solutions:. ) I want to eliminate x1, x2, y1, y2 between these 5 equations. How can I remove the T -> from the output so I can use the number in a graph (or whatever)? Example of Eliminate uses the given set of equations to eliminate the requested variables from those equations, returning the equations not needed to do the eliminations. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, $\begingroup$ Solve prefers that you give it a number of variables to solve for equal to the number of equations. Solving polynomial inequalities -- efficient Positivstellensatz on a computer. Modified 10 years, 10 months ago. $\endgroup$ – $\begingroup$ The first example under Basic Examples should show you all you need. What are you trying to accomplish Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site $\begingroup$ @WinnieTseng That means that your system of equations has no general solutions, i. In other words, Solve must return results that satisfy the equalities. Assuming it is a variable to eliminate. From the NSolve doc page:. ) This behavior is illustrated perhaps more clearly in the simple example, Eliminate[e == 1 && d == 1, e] (* d == 1 *) Eliminate[%, d "Defining Variables" discussed assignments such as x=y, which set x equal to y. which gives a partial solution, but your problem is sufficiently complicated that it does not completely solve for t and it warns that it has used inverse functions in what it did which might have missed some potential solutions. To be clear, I'm not asking how to eliminate the particular occurrences of ConditionalExpression in the output above, but to eliminate all instances of such a ConditionalExpression and arbitrary constants in general, without having to manipulate by hand any intermediate results. Homology software. I will be trying to eliminate variables from a system of simultaneous equations and just trying to plot 2 of them. g. $\begingroup$ I am new to Mathematica, so I may have messed up the syntax, but the three equations in solve aim to say "S evaluated at km is equal to Ext evaluated at km" and similarly for their derivatives. 25*Qem^0. sympytools. Viewed 853 times My next step would be to eliminate any variables that a, b, and c do not depend on. (2) So long nobody seems to observe similar problem in this site. We all know that one can't just write Solve[eq==0,s^2] because s^2 i Given that I have the following code: points = N@Solve[{x^2 + 3 x - 1 + y - 1 == 0, 3 x - y^2 + 1 - 1 == 0}, Reals]; {x, y} /. The operation Resolve with Exists over the unwanted variables and Reals as domain should get the correct answer, the downside being that quantifier elimination might be computationally expensive. Note: inspired by @garej's answer to this question:Brackets around each item in matrix. Alternatively, in this case just look at FullForm[!eqns] $\endgroup$ – Bob Hanlon. Mathematica can't solve these. Unable to correctly solve ODE with Scilab. However, sometimes it works. 7 and v. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Mathematica's kernel seemed to slow down to a crawl when constraints on c and q were given along with your equation. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for Mathematica can solve it if you use explicit numerical values for the parameters, and it does return a result: {}. You can also eliminate variables using Mathematica. I'm pretty new with Mathematica. For example, how could we solve eqn==0 below for c3 and c4 expressed in terms of c1 and c2 only, without a1 or a2? We can select two equations from the eliminates variables between a set of simultaneous equations. My idea was Solve[eq[[2]], T] /. Note that Solve may not be able to find expressions in terms of these variables (if the equations you give it are contradictory or insufficient) or for all values as some functions have no inverse or only partial inverses. Any idea on how I can proceed? (Edit: My goal is to obtain a plot of u and d vs o. 01; x = 0. For example, in the case at hand: Is there a function in Mathematica which removes brackets from an expression? RemoveBrackets[ {3} ] 3. All you could do is use the result of Solve[{NR + NQ + NU == 1 - L, w*(NR + NU) == wrNR, C1/C2 == 4*em, Q == C1 + 0. This means that we also need to define vd as 0 - VR. >> Solve::svars: Equations may not give solutions for all "solve" variables. Using Eliminate[]: With[{a = {a1, a2, a3}, b = {b1, b2, b3}}, Eliminate[{(a + 3 b) . I would appreciate it if someone could explain both this warning, and the lack of output. T -> x. Inverse functions normally give one branch of a general inverse, so equations that Using the Solve function to find Temperature of a system, I receive an output of. I verified your book solution is correct. Common choices of dom are Reals, Integers, and Complexes. So, it is no wonder that a brute force solution using Solve on the original equations takes forever. You seem to be using V1 as a placeholder in Solve to get VL and VR. Stay on top of important topics and build connections by joining Wolfram Community groups relevant to your interests. So, barring any extra info, that means the equations reduce to. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the There is a smart use of Solve which eliminates certain variables and solves the system with respect to the remaining ones, see e. In certain problems, we need to solve systems of equations and get results in terms of just selected variables. To Eliminate the remaining variable, c requires the use of the remaining equation. These are very hard for computers to solve. Solve[Sin[x] == Cos[2 x] && 0 ≤ x ≤ 2 π, x] @Artes lready showed how one might use Solve or Eliminate for this task. I've also tried to recast the problem as simultaneous equations and then using Solve or Eliminate as mentioned here: Solve[!Eliminate[!expr == 0, vars], Array[cc, 3], Rationals] Share. (Just search "[bugs I'm not sure how to get the results in terms of the desired variables. I'm not really concerned with the output being reliably correct or complete in the end; I just want the First[Solve[Eliminate[{hamiltonianEq, momentumEq}, x'[t]], h]] (* ==> (p[x]^2 + k m x[t]^2)/(2 m) *) This is to avoid conflicts with built-in Mathematica symbols like K, which you used. Community Answer This answer has a 4. $\begingroup$ @DanielLichtblau Thanks, I was a bit puzzled with the fact that eliminating variables capabilities od Solve or Eliminate hadn't seemed to provide a direct approach to this system in Mathematica 8. While x=y is an imperative statement that actually causes an assignment to be done, x==y merely tests whether x and y are equal, and causes no which gives a partial solution, but your problem is sufficiently complicated that it does not completely solve for t and it warns that it has used inverse functions in what it did which might have missed some potential solutions. Solve for the remaining variable. , !) and press F1 for help. (If it's relevant, I have version 7. In fact, by Specifying Method -> Reduce in Solve, Solve will use Reduce behind the scenes to produce an answer. The documentation reads : You can always solve for a new variable that contains exactly the combination of the other unknowns you're after. Cite. Equation x^2 + y^2 == r^2 is not necessary, it is a combination of the last two. Solve[mass1, iNt[x]][[1, 1]] mass2 fourth order equation in nPrime. Learn more about bidirectional Unicode characters I am trying to solve following system of equations, but i am not getting the desired solution: {uc[t] -> u[t] - ul[t] - y[t]} I am trying to solve from following set of equations, but Mathemat Skip to main content. Your task could be done by hand. Eliminate[{q1 == 0, q2 == $\begingroup$ I presume you are asking about the warning message, Solve::bdomv: Warning: (f^\[Prime]\[Prime])[x0] is not a valid domain specification. The reception of Bob's answer led me to think the OP didn't mind whether the output was in Solve or Reduce form, so I opted above for Solve. points Which gives me the output: {{0. answered Jun 15 Thanks for contributing an answer to Mathematica Stack Exchange! Please be sure to answer the question. I just only wanted to show the way of applying CoordinateTransform for more complex cases. You can coerce Solve to give you a solution like this:. 0. Solve[eqns,vars,elims] find solutions for vars, eliminating the variables elims However, that structure is not mentioned in the documentation for Solve itself, where the third argument is the Domain and not the variables to Mathematica Solve function gives incorrect solution, why? 1. It is very important that you do not confuse x=y with x==y. What I do not follow is how it is related to the functions in the manipulation shown in Solve[expr, vars] attempts to solve the system expr of equations or inequalities for the variables vars. If you would like to find solutions that are valid for only some values of the parameters, try Reduce instead of Solve. Finally, if you substitute r in the las two, you already have the solution I posted. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site So, I copy the result from Eliminate and then solve it for my variable and then copy the output and format it later. Indeed, the somewhat corrected command from your question finds no solution: Solve[Integrate[f[t], {t, 0, x^3 - 1}] == x, f[x], InverseFunctions -> True] (*{}*) It seems Mathematica is right in this case. Essential reading: Algebraic Numbers in the documentation. The mathematica solve page even suggests this syntax on its second to last example, so I'm quite confused. 5 rating. ; For polynomials in one variable, GroebnerBasis reduces to PolynomialGCD. But for now, will add this assumption in the Mathematica version. $\endgroup$ – MarcoB $\begingroup$ Dear @RenéG, you even do not need to use Solve. Modified 7 months ago. Provide details and share your research! But avoid Asking for help, clarification, or responding to other answers. Let's get rid of V1 by putting VL and VR directly into those equations. Since the input is approximate Minimize will punt to NMinimize so what you see below is not a guaranteed result. >> and it just give an answer of $\begingroup$ @rhermans - This syntax was documented in earlier versions but the documentation stopped mentioning it. Check that the ordered pair is a solution to both original equations. Viewed 301 times 3 ( I am translating my Maple code to Mathematica and in Maple it just did solve with no options, that is why I asked). Common choices of dom are Eliminate only works on plain variables, not functions. ] seems so simple. Hey guys I really could use some help on this calc 3 problem. Viewed 88 $\begingroup$ @WinnieTseng That means that your system of equations has no general solutions, i. For them to be true, you need to solve for y also $\begingroup$ Your full set of equations is just unsolvable by Solve. In fact, eqn1 is the standard equation for the Modified Bessel Function of order 0. Eliminate[] I was trying to eliminate p and didn't think of just solving for {c,p} as a pair. This is separable with gHxL=-x and hHyL=•••••1•••••• 1+y2. ) Any help would be appreciated. Add the equations resulting from Step 2 to eliminate one variable. In[1]:=g@x_D The Mathematica functions Reduce, Resolve, and FindInstance allow you to solve a wide variety of problems that can be expressed in terms of equations and inequalities. All the same one Mathematica Solve Hangs Indefinitely. However, I do not understand the syntax required to use the command so that the variables A,e,mu are appropriately replaced. Your data are your list of values, the instead of Log[a+b x^a] put the form of your model, instead of {a,b}, your have three fit parameters {a,b,c} and you When I solve a system of equations with Reduce, it will give me as possibilities that system parameters can be exactly some special value that will allow the variables of interest to be something weird. equation-solving; calculus-and-analysis; $\begingroup$ It is because out of three equations, you can eliminate one or two of three variables, but you may solve for all three of them. Question feed I'm pretty new with Mathematica. Commented Nov 2, 2024 at 18:17 shows how the equivalence of SolveAlways and the Solve/Eliminate formulation. $\begingroup$ the first conditional expression is in terms of unspecified integer constants ( due to the inverse trig function). $\begingroup$ The link has a question I can follow, at least in the case where the f(x) are explicit polynomial (perhaps linear) functions of x. and -> that they often don't even know that they can search for in Wolfram Community forum discussion about Solve not working?. This may be what you want, first replace the function iNt[x] with a symbol and eliminate that. Write the solution as an ordered pair. You should have I have a set of trig equations, c1,c2,c3 are constants, I excepted the result is c1 c2 c3 r^2 + (c1^2 + c2^2 + c3^2 - 1) r + c1 c2 c3 == 0 but following code worked so slow, is there a faster meth Given a expression (polynomial, or any equation in general) such as a s^2+b = 0 I want to solve for s^2, to get s^2 = -b/a. Root is $\begingroup$ Welcome to Mathematica. For this kind of symbolic polynomial equation of degree n, it's relatively easy to solve the roots ma MathGPT is an AI math solver and homework helper trusted by 2M plus students who are looking for a math solver and calculator for algebra, geometry, calculus, and statistics from just a photo. I would like to eliminate p and generate either a single polynomial in c or complex numerical solutions to c. So far I have tried using Eliminate, Reduce, Solve. further However, the subsequent expressions are likely to be equally enormous. Remember that doing this doesn't change the equality of the two sides of the equation, so the \(x\) - and \(y\) -values that make the first equation Eliminate the parameter t, write the equation in Cartesian coordinates, then sketch the graph of the vector- valued function r(t)=(2 sint, 3 cost) showing orientation. This can be achieved in Mathematica using the following method: Since version 8, Solve and Reduce share a great deal of code. Any thoughts on the I even copy one of the example on the command Solve provide by Mathematica "Solve[x^2 + a x + 1 == 0, x]" which gives the same error, but in the example it runs and returns {{x -> 1/2 Just remove any definitions of x before using it as a variable in Solve (Clear, ClearAll, or Remove will work here): ClearAll[x] Solve[x^2 + a x $\begingroup$ @bill s I understand completely. I have a set of 4 equations, for example (they can also be more complex): $ AA = Skip to main content. When using the Solve function with Mathematica, you can specify for what variables you want Solve to specify the solutions. heavily tested function, and have been proved to be robust. However, because the three ODEs and their boundary conditions as specified in the question all are homogeneous, the solution must be identically zero. The Gr ö bner basis in general depends on the ordering assigned to monomials. Complex differential equation. Solve[eq, a, Reals] // ToRadicals etc Share. SE! I hope you will become a regular contributor. Clear["`*"]; Clear["`*"]; curves = ContourPlot[{-12 + 16 Cos[t]^2 == 4 - 2 s^3, 4 Mathematica: help solving a system of non-linear equations with inequality constraints. When you insert that back into the expression for t those constants simplify out (due to the I am brand new to Mathematica, and want to eliminate one variable, iNt from a system of two equations, which should result is a fourth oder ODE for the remaining variable Nprime. Provide details and share your research! But avoid Asking for help, clarification, or responding to other To be clear, I'm not asking how to eliminate the particular occurrences of ConditionalExpression in the output above, but to eliminate all instances of such a ConditionalExpression and arbitrary constants in general, without having to manipulate by hand any intermediate results. The functions use a collection of algorithms applicable to classes of problems satisfying particular properties, as well as a set of heuristics that attempt to reduce the given problem to a sequence of prob-lems that This is 7 equations with 9 variables. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site How to use Mathematica to directly solve the above vector equation and find the angle between vector $\mathbf{a}$ and vector $\mathbf{b}$? I wonder if there is any other easier way. Ask Question Asked 3 years, 11 months ago. I'm not sure what the message means in Mathematica 11. $\endgroup$ – Is there a way to tell Solve (or another function) that x2,x3,x4 are not to appear in the solution for x1, but that there is no need to solve for them? If so, please, how? Indeed, I would much rather that Solve return the coefficients of the polynomial for x1 , rather than returning the solution to that (perhaps high-order) polynomial. However, here it appears to take too Implementing Mathematica's solve and eliminate functions using SymPy Raw. Solve [expr, vars, dom] solves over the domain dom. Looking in the documentation to the Solve command, I don't find any examples of solving functional equations. In my experimental Separation of variables with Mathematica Solve the equation dy••••• dx =-xI1+y2M with initial conditions y(2)=1. You have hit the limit of what CAS can do at this time with solving pde's. Therefore one has to discover somewhat nonstandard solution. $\endgroup$ – Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Complex Polynomial Systems Introduction The Mathematica functions Reduce, Resolve, and FindInstance allow you to solve a wide variety of problems that can be expressed in terms of equations and inequalities. 5 - From the speed viewpoint, you are inducing a large overhead with that product condition. Follow answered Apr 19, 2016 at 15:43. $\endgroup$ – MarcoB Confirming that Mathematica 9 can easily solve this system unlike ver. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for I have a set of trig equations, c1,c2,c3 are constants, I excepted the result is c1 c2 c3 r^2 + (c1^2 + c2^2 + c3^2 - 1) r + c1 c2 c3 == 0 but following code worked so slow, is there a faster meth I want to use MapApply (I don't know what else to use) to use the parameters in muSolutions ( of the form {A,e,mu}) to solve eq[[2]] for the variable T. (7 a - @Artes lready showed how one might use Solve or Eliminate for this task. I tried Eliminate[{x1^2/a^2 + y1^2/b^2 == 1, x2^2/a^2 + y2^2/b^2 == 1, x1 x2 + y1 y2 == 0, y2/(x2 - c) == y1/(x1 - c) == k}, {x1, $\begingroup$ The solutions find configurations of ellipses and lines where the two intersection points and the center of the ellipse form a right angle at the center. $\begingroup$ It is because out of three equations, you can eliminate one or two of three variables, but you may solve for all three of them. ; RootReduce automatically threads over lists, as well as equations, inequalities, and $\begingroup$ It is because out of three equations, you can eliminate one or two of three variables, but you may solve for all three of them. There is a dedicated stackexchange for Mathematica, so if you post your problem there, someone might even solve this for you. Is it common practice to remove trusted certificate authorities (CA) located in untrusted countries? more hot questions Question feed Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site Polynomial algorithms are at the core of classical "computer algebra". The original technical computing environment. $\endgroup$ – You'll find the explanation in the documentation page of Solve under Options -> MaxExtraConditions, as well as under Possible Issues (in a less obvious form). The answer was obtained by solving a corresponding exact > system and numericizing the result. Thus, dleft[-l] == Eliminate::ifun: Inverse functions are being used by Eliminate, so some solutions may not be found; use Reduce for complete solution information. Thank you. Solve[x == y && x == -y, x, {y}] which is equivalent to. Generally if you have n variables you need to have at least n+1 equations to get meaningful elimination and end up with at least one equation. this answer : Efficient code for solve this equation. Carefully tuned strategies automatically select optimal algorithms, allowing large-scale You can get the range of values by minimizing and maximizing kp with the specified constraints. I want to solve these simultaneous equations to get the values of a, b, c, and d. >> Solve::fulldim: The solution set contains a full-dimensional component; use Reduce for complete solution information. However, it appears to be kept for backward compatibility. I cannot even seem to simplify these equations further. Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site I have a very complicated mathematica expression that I'd like to simplify by using a new, possibly dimensionless parameter. Eliminate[] works for similar problems of lower order but never returns a result for this one. For example, consider the output of: roots = Solve[x^2 + 4 x - 1 == 0, x] {{x -> -2 - To remove Root, do. Problem is to solve this with minimal additional openness not to derive a polynomial over the Complexes. 5k 2 2 gold badges 46 46 silver badges 102 102 bronze The Eliminating Variables section on Manipulating Equations and Inequalities documentation quotes an otherwise undocumented feature for Solve. Follow edited Aug 7, 2017 at 22:10. 1 and to which coefficients it refers to. However Mathematica can't eliminate z. Hot Network Questions Cannot fg a zsh function including less Čech simplicial complex contractible Why am I not seeing continuity How to eliminate variables when using Solve[] (4 answers) Closed last year . krvnqmx vqqtyta plidpl wrxys wseftq qbahmbx kqeqi mxt shlte zogjrx