allsolutions = truefalse

Return more solutions for non-algebraic equations (default is false).

conditionalsolutions = truefalse

Return piecewise conditional solutions (default is true).

dropmultiplicity = truefalse

Do not return multiple identical solutions of polynomials (default is false).

explicit = {posint, truefalse}

Express RootOfs in solution as radicals when possible (default is false).

maxsols = {posint, pos_infinity}

Limit the number of solutions returned (default is 100).

tryhard = truefalse

Use radical form for quadratic solutions (default is false).

useassumptions = truefalse

Convert assumptions into additional inequalities (default is false).

useunits = {truefalse, list(variable::unit)}

Do preprocessing to handle units in the solve input. By default solve does this automatically if it detects units. Set this option to true to force preprocessing if detection fails.  Set this option to false to have units treated as symbols. The default behavior is that solve analyzes the input to determine what units are associated with the symbols in the input.  To aid that analysis or to override the units it finds by other units, units can be specified in this option as a list of variable::unit pairs.

If the solve command cannot confirm that the solution set returned is complete, it sets the global variable _SolutionsMayBeLost to true and issues a warning.

The solve command can return conditional solutions to many types of parametric equations and inequalities. Systems of linear inequalities in one or more variables are fully supported, as well as relations involving some types of simple functions of linear expressions. Conditional solutions are returned in the form of piecewise functions whose conditions depend upon the values of the parameters. Each output of the piecewise function will be a list of lists or sets, identifying all of the solutions that are valid under the corresponding condition. If a branch of the piecewise function contains no solutions, then its output will be an empty list.

To prevent solve from returning conditional solutions, use the solve option conditionalsolutions=false. In this case, if solve encounters a conditional solution, it will be discarded, a warning message will be displayed, and only solutions that hold unconditionally, if any, will be returned.  You can also set the environment variable _EnvConditionalSolutions := false.

A single quadratic equation with constant coefficients in one variable is solved directly by substitution into the quadratic formula. To attempt to express solutions in the common radical form, which may take a longer time to process, use the option tryhard=true. You can also set the environment variable _EnvTryHard := true.

In general, explicit solutions in terms of radicals do not exist for polynomial equations of degree greater than 4. The solve command returns explicit solutions for low degree polynomial equations (by default, degree less than 4). For higher degree equations, implicit solutions are given in terms of RootOf.

To control the form of solutions returned, use the option explicit. The behavior is described in the following table.

By default, solve may return multiple identical solutions to a polynomial (i.e. the roots of multiplicity greater than one). To have solve return only one of each of these multiple roots use the option dropmultiplicity=true or set the environment variable _EnvDropMultiplicity := true.

To control the total number of solutions returned, use the option maxsols. This integer specifies the maximum number of solutions generated inside the solve engine. The default value is 100. You can set the environment variable _MaxSols instead of using the option to the solve command, but take case, because other commands may also be effected by environment variables.

The solve command returns all solutions for polynomial equations. In general for transcendental equations, the solve command returns only one solution, but does not set _SolutionsMayBeLost to true. To force the solve command to return the entire set of solutions for all inverse transcendental functions, provide the solve option allsolutions = true or set the environment variable _EnvAllSolutions := true.

In the solution, Maple may generate variables that take numeric values. Normally such variables are named with the prefix _Z for integer values, _NN for non-negative integer values, and _B for binary values (0 and 1).  If other conditions on these values apply, then the conditions will be converted to assumptions and simplified, if possible. Note that if the solution includes one of these variables, you can use about to find out information on the variable. If such conditions are sufficient to allow Maple to determine that the variable may only take on a finite number of values, then the full list of solutions can be requested by using the option explicit=true.

In most cases, solve ignores assumptions on the variables for which it is solving.  However, the useassumptions option may be supplied to force solve to inspect any assumptions on the variables, and extract inequalities that it will add to the input.  For example:

solve(x^2-1,{x}, useassumptions) assuming x>0;

$\left\{x=1\right\}$

solve({x^2-1, x>0},{x});

$\left\{x=1\right\}$

solve(x^2+1,{x}, useassumptions) assuming x::real;

Warning, solve may not respect assumed property 'real' on 'x'.

$\left\{x=\mathrm{I}\right\},\left\{x=\mathrm{-I}\right\}$

$\mathrm{solve}\left(\left\{x^2=4\mathrm{Unit}\left(m^2\right)\right\}\,\left\{x\right\}\right)$

$\left\{x=2⟦m⟧\right\},\left\{x=-2⟦m⟧\right\}$

$\mathrm{solve}\left(\left\{x^2=4\mathrm{Unit}\left(m^2\right)\right\}\,\left\{x\right\}\,\mathrm{useunits}=\mathrm{false}\right)$

$\left\{x=2\sqrt{⟦m^2⟧}\right\},\left\{x=-2\sqrt{⟦m^2⟧}\right\}$

$\mathrm{solve}\left(\left\{x-y\,x^2=4\mathrm{Unit}\left(m^2\right)\right\}\right)$

$\left\{x=2⟦m⟧\,y=2⟦m⟧\right\},\left\{x=-2⟦m⟧\,y=-2⟦m⟧\right\}$

$\mathrm{solve}\left(\left\{x-y\,x^2=4\mathrm{Unit}\left(m^2\right)\right\}\,\mathrm{useunits}=\left[x::\mathrm{Unit}\left(\mathrm{ft}\right)\,y::\mathrm{Unit}\left(\mathrm{cm}\right)\right]\right)$

$\left\{x=\frac{2500}{381}⟦\mathrm{ft}⟧\,y=200⟦\mathrm{cm}⟧\right\},\left\{x=-\frac{2500}{381}⟦\mathrm{ft}⟧\,y=-200⟦\mathrm{cm}⟧\right\}$

Units can be specified for parameters

$\mathrm{solve}\left(\left\{x-y\,x^2=a^2\mathrm{Unit}\left(m^2\right)\right\}\,\left\{x\,y\right\}\right)$

$\left\{x=a⟦m⟧\,y=a⟦m⟧\right\},\left\{x=-a⟦m⟧\,y=-a⟦m⟧\right\}$

$\mathrm{solve}\left(\left\{x-y\,x^2=a^2\mathrm{Unit}\left(m^2\right)\right\}\,\left\{x\,y\right\}\,\mathrm{useunits}=\left\{a::\left(\mathrm{Unit}\left(\frac{1}{m}\right)\right)\right\}\right)$

$\left\{x=\frac{a}{⟦\frac{1}{m}⟧}\,y=\frac{a}{⟦\frac{1}{m}⟧}\right\},\left\{x=-\frac{a}{⟦\frac{1}{m}⟧}\,y=-\frac{a}{⟦\frac{1}{m}⟧}\right\}$

$\mathrm{solve}\left(x^4\,x\right)$

$0,0,0,0$

$\mathrm{solve}\left(x^4\,x\,\mathrm{dropmultiplicity}\right)$

$0$

$\mathrm{eq}≔x^4-5x^2+6x=2$

$\mathrm{eq}≔x^4-5x^2+6x=2$

$\left[\mathrm{solve}\left(\mathrm{eq}\,x\right)\right]$

$\left[\sqrt{3}-1\,-1-\sqrt{3}\,1\,1\right]$

$\mathrm{solutions}≔\left[\mathrm{solve}\left(\mathrm{eq}\,x\right)\right]$

$\mathrm{solutions}≔\left[\sqrt{3}-1\,-1-\sqrt{3}\,1\,1\right]$

$\mathrm{solutions}\left[1\right]$

$\sqrt{3}-1$

To convert to floating-point values, use the evalf command.

$\mathrm{evalf}\left(\mathrm{solutions}\right)$

$\left[0.732050808\,\mathrm{-2.732050808}\,1.\,1.\right]$

$\mathrm{solutions}≔\left\{\mathrm{solve}\left(\mathrm{eq}\,x\right)\right\}$

$\mathrm{solutions}≔\left\{1\,-1-\sqrt{3}\,\sqrt{3}-1\right\}$

$\mathrm{solutions}\left[1\right]$

$1$

To verify the solution, use the eval command.

$\mathrm{eval}\left(\mathrm{eq}\,x=\mathrm{solutions}\left[1\right]\right)$

$2=2$

$\mathrm{solve}\left(x^4+x+1\,x\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^4+\mathrm{\_Z}+1\,\mathrm{index}=1\right),\mathrm{RootOf}\left({\mathrm{\_Z}}^4+\mathrm{\_Z}+1\,\mathrm{index}=2\right),\mathrm{RootOf}\left({\mathrm{\_Z}}^4+\mathrm{\_Z}+1\,\mathrm{index}=3\right),\mathrm{RootOf}\left({\mathrm{\_Z}}^4+\mathrm{\_Z}+1\,\mathrm{index}=4\right)$

To convert to floating-point values, use the evalf command.

$\mathrm{evalf}\left(\left\{\right\}\right)$

$\left\{\mathrm{-0.7271360845}-0.4300142883\mathrm{I}\,\mathrm{-0.7271360845}+0.4300142883\mathrm{I}\,0.7271360845-0.9340992895\mathrm{I}\,0.7271360845+0.9340992895\mathrm{I}\right\}$

To find an explicit symbolic representation for the roots, use map with the allvalues command.  This applies allvalues to each solution and is generally better than applying allvalues directly to all the solutions at once.

$\mathrm{solve}\left(x^4-2x^3+2\,x\right)$

$\mathrm{RootOf}\left({\mathrm{\_Z}}^4-2{\mathrm{\_Z}}^3+2\,\mathrm{index}=1\right),\mathrm{RootOf}\left({\mathrm{\_Z}}^4-2{\mathrm{\_Z}}^3+2\,\mathrm{index}=2\right),\mathrm{RootOf}\left({\mathrm{\_Z}}^4-2{\mathrm{\_Z}}^3+2\,\mathrm{index}=3\right),\mathrm{RootOf}\left({\mathrm{\_Z}}^4-2{\mathrm{\_Z}}^3+2\,\mathrm{index}=4\right)$

$\mathrm{map}\left(\mathrm{allvalues}\,\left\{\right\}\right)$

$\left\{\frac{1}{2}-\frac{\mathrm{I}}{2}-\frac{\sqrt{4+2\mathrm{I}}}{2}\,\frac{1}{2}-\frac{\mathrm{I}}{2}+\frac{\sqrt{4+2\mathrm{I}}}{2}\,\frac{1}{2}+\frac{\mathrm{I}}{2}-\frac{\sqrt{4-2\mathrm{I}}}{2}\,\frac{1}{2}+\frac{\mathrm{I}}{2}+\frac{\sqrt{4-2\mathrm{I}}}{2}\right\}$

$\mathrm{solve}\left(\left\{x^2{y}^2-b\,x^2-y^2-a\right\}\,\left[x\,y\right]\right)$

$\left[\left[x=\mathrm{RootOf}\left({\mathrm{\_Z}}^2-{\mathrm{RootOf}\left({\mathrm{\_Z}}^4+a{\mathrm{\_Z}}^2-b\right)}^2-a\right)\,y=\mathrm{RootOf}\left({\mathrm{\_Z}}^4+a{\mathrm{\_Z}}^2-b\right)\right]\right]$

$\mathrm{map}\left(\mathrm{allvalues}\,\right)$

$\left[\left[x=\sqrt{\frac{a}{2}+\frac{\sqrt{a^2+4b}}{2}}\,y=\frac{\sqrt{2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=-\sqrt{\frac{a}{2}+\frac{\sqrt{a^2+4b}}{2}}\,y=\frac{\sqrt{2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=\sqrt{\frac{a}{2}+\frac{\sqrt{a^2+4b}}{2}}\,y=-\frac{\sqrt{2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=-\sqrt{\frac{a}{2}+\frac{\sqrt{a^2+4b}}{2}}\,y=-\frac{\sqrt{2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=\sqrt{\frac{a}{2}-\frac{\sqrt{a^2+4b}}{2}}\,y=\frac{\sqrt{-2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=-\sqrt{\frac{a}{2}-\frac{\sqrt{a^2+4b}}{2}}\,y=\frac{\sqrt{-2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=\sqrt{\frac{a}{2}-\frac{\sqrt{a^2+4b}}{2}}\,y=-\frac{\sqrt{-2\sqrt{a^2+4b}-2a}}{2}\right]\,\left[x=-\sqrt{\frac{a}{2}-\frac{\sqrt{a^2+4b}}{2}}\,y=-\frac{\sqrt{-2\sqrt{a^2+4b}-2a}}{2}\right]\right]$

$\mathrm{equations}≔\left\{3u+v=3\,u+v+w=1\,u-2v-w=0\right\}$

$\mathrm{equations}≔\left\{3u+v=3\,u-2v-w=0\,u+v+w=1\right\}$

$\mathrm{solutions}≔\mathrm{solve}\left(\mathrm{equations}\right)$

$\mathrm{solutions}≔\left\{u=\frac{4}{5}\,v=\frac{3}{5}\,w=-\frac{2}{5}\right\}$

Check solutions.

$\mathrm{eval}\left(\mathrm{equations}\,\mathrm{solutions}\right)$

$\left\{0=0\,1=1\,3=3\right\}$

Select one solution.

$\mathrm{eval}\left(u\,\mathrm{solutions}\right)$

$\frac{4}{5}$

Assign all solutions.

$\mathrm{assign}\left(\mathrm{solutions}\right)$

$u$

$\frac{4}{5}$

$\left\{\mathrm{solve}\left(\sin \left(x\right)=\cos \left(x\right)-1\,x\,\mathrm{allsolutions}\right)\right\}$

$\left\{2\mathrm{\pi }\mathrm{\_Z2~}\,-\frac{1}{2}\mathrm{\pi }+2\mathrm{\pi }\mathrm{\_Z1~}\right\}$

$r≔\mathrm{solve}\left(\cos \left(x\right)\,x\,\mathrm{allsolutions}\right)$

$r≔\frac{1}{2}\mathrm{\pi }+\mathrm{\pi }\mathrm{\_Z3~}$

Substitute 3 for $\mathrm{\_Z3~}$ in $r$.

$\mathrm{eval}\left(r\,\mathrm{op}\left(\mathrm{indets}\left(r\right)\right)=3\right)$

$\frac{7\mathrm{\pi }}{2}$

$\left\{\mathrm{solve}\left(\left[\sin \left(x\right)\cos \left(x\right)=0\&comma;0<x\&comma;x<10\right]\&comma;x\&comma;\mathrm{allsolutions}\right)\right\}$

$\left\{\left\{x=\mathrm{\pi }\mathrm{\_Z5~}\right\}\&comma;\left\{x=\frac{1}{2}\mathrm{\pi }+\mathrm{\pi }\mathrm{\_Z4~}\right\}\right\}$

$\mathrm{about}\left(\mathrm{\_Z4}\right)$

Originally _Z4, renamed _Z4~:
  is assumed to be: AndProp(integer,RealRange(0,2))

$\mathrm{about}\left(\mathrm{\_Z5}\right)$

Originally _Z5, renamed _Z5~:
  is assumed to be: AndProp(integer,RealRange(1,3))

$\left\{\mathrm{solve}\left(\left[\sin \left(x\right)\cos \left(x\right)=0\&comma;0<x\&comma;x<10\right]\&comma;x\&comma;\mathrm{allsolutions}\&comma;\mathrm{explicit}\right)\right\}$

$\left\{\left\{x=\mathrm{\pi }\right\}\&comma;\left\{x=2\mathrm{\pi }\right\}\&comma;\left\{x=3\mathrm{\pi }\right\}\&comma;\left\{x=\frac{\mathrm{\pi }}{2}\right\}\&comma;\left\{x=\frac{3\mathrm{\pi }}{2}\right\}\&comma;\left\{x=\frac{5\mathrm{\pi }}{2}\right\}\right\}$

equations can be given entry-wise in terms of any rtable

$\mathrm{solve}\left(⟨⟨x+1|y+1⟩\&comma;⟨z+2|w+2⟩⟩=⟨⟨1|1⟩\&comma;⟨2|2⟩⟩\right)$

a constant is interpreted as a constant rtable of the right dimensions

$\mathrm{solve}\left(⟨⟨x+1|y+1⟩\&comma;⟨z+2|w+2⟩⟩=1\right)$

all other dimension mismatches will produce an error

$\mathrm{solve}\left(⟨⟨x+1|y+1⟩\&comma;⟨z+2|w+2⟩⟩=⟨1\&comma;2⟩\right)$

Error, (in SolveTools:-rtableToEquations) invalid input: left- and right-hand sides have incompatible dimensions

While it is possible to solve linear equations as $\mathrm{solve}\left(M·v=b\right)$, it advisable to use LinearAlgebra:-LinearSolve instead.

$\mathrm{restart}$

$\mathrm{solve}\left(\left[b<ax\right]\&comma;\left\{x\right\}\right)$

$\left\{\begin{array}{cc}\left[\left\{x=x\right\}\right]& b<0\wedge a=0\\ \left[\left\{\frac{b}{a}<x\right\}\right]& 0<a\\ \left[\left\{x<\frac{b}{a}\right\}\right]& a<0\\ \left[\right]& \mathrm{otherwise}\end{array}\right.$

$\mathrm{solve}\left(2<\mathrm{abs}\left(x+a\right)+\mathrm{abs}\left(x+b\right)\&comma;x\right)$

$\left\{\begin{array}{cc}\left[\left\{-\frac{a}{2}-\frac{b}{2}+1<x\right\}\&comma;\left\{x<-\frac{a}{2}-\frac{b}{2}-1\right\}\right]& -2+b\le a\wedge a\le 2+b\\ \left[\left\{-b\le x\&comma;x<-a\right\}\&comma;\left\{-a\le x\right\}\&comma;\left\{x<-b\right\}\right]& a<-2+b\\ \left[\left\{-a\le x\&comma;x<-b\right\}\&comma;\left\{-b\le x\right\}\&comma;\left\{x<-a\right\}\right]& 2+b<a\\ \left[\right]& \mathrm{otherwise}\end{array}\right.$

$\mathrm{solve}\left(\left[\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\&comma;a\left(x-1\right)<2+b\left(x-1\right)\right]\&comma;\left[x\right]\right)$

$\left\{\begin{array}{cc}\left[\left[x=3\right]\&comma;\left[x=2\right]\&comma;\left[x=1\right]\right]& a<1+b\\ \left[\left[x=2\right]\&comma;\left[x=1\right]\right]& 1+b\le a\wedge a<2+b\\ \left[\left[x=1\right]\right]& \mathrm{otherwise}\end{array}\right.$

$\mathrm{solve}\left(\left[\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\&comma;a\left(x-1\right)<2+b\left(x-1\right)\right]\&comma;\left[x\right]\&comma;\mathrm{conditionalsolutions}=\mathrm{false}\right)$

Warning, one or more conditional solutions have been discarded

$\left[\left[x=1\right]\right]$