# Return to Answer

 3 added 485 characters in body edited Nov 6 '12 at 18:26 Artes 46.3k99 gold badges128128 silver badges212212 bronze badges The second itemfirst items of More Information in the documentation of Solve says : Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].  The system expr in Solve[expr,vars] can be any logical combination of: lhs == rhs equations lhs != rhs inequations lhs > rhs or lhs >= rhs inequalities expr ∈ dom domain specifications ForAll[x,cond,expr] universal quantifiers Exists[x,cond,expr] existential quantifiers  Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars]. Every expri can be an equation or an inequality in the variables vars, inequality as well as an expression testtests like e.g. Positive[x]Positive or NonNegative[x]Negative etc., thus we can do simply e.g. Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x], but to get only the list of solutions (without Rules ) there are at least two ways: using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 : x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]  {1, 2, 3}  using Part (shorthand [[]]) with e.g. x > 0 or with the expression test like Positive, NonNegative etc.: Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]   {1, 2, 3}  using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 : x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]  {1, 2, 3}  using Part (shorthand [[]]) with e.g. x > 0 or with an expression test like Positive, NonNegative etc.: Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]   {1, 2, 3}  WeThe above ways can be mixed, e.g. : x /. Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][]. We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use it like e.g. The second item of More Information in the documentation of Solve says : Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].  Every expri can be an equation or an inequality in the variables vars, as well as an expression test like e.g. Positive[x] or NonNegative[x] etc., thus we can do simply Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x], but to get only the list of solutions (without Rules ) there are at least two ways: using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 : x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]  {1, 2, 3}  using Part (shorthand [[]]) with e.g. x > 0 or with the expression test like Positive, NonNegative etc.: Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]   {1, 2, 3}  We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use e.g. The first items of More Information in the documentation of Solve says : The system expr in Solve[expr,vars] can be any logical combination of: lhs == rhs equations lhs != rhs inequations lhs > rhs or lhs >= rhs inequalities expr ∈ dom domain specifications ForAll[x,cond,expr] universal quantifiers Exists[x,cond,expr] existential quantifiers  Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars]. Every expri can be an equation, inequality as well as an expression tests like e.g. Positive or Negative etc., thus we can do simply e.g. Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x], but to get only the list of solutions (without Rules ) there are at least two ways: using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 : x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]  {1, 2, 3}  using Part (shorthand [[]]) with e.g. x > 0 or with an expression test like Positive, NonNegative etc.: Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]   {1, 2, 3}  The above ways can be mixed, e.g. : x /. Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][]. We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use it like e.g. 2 extended discussion edited Nov 6 '12 at 17:11 Artes 46.3k99 gold badges128128 silver badges212212 bronze badges The second item of More Information in the documentation of Solve says : Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].  Every expri can be equations an equation or inequalitiesan inequality in the variables of vars, as well as an expression test like e.g. Positive[x] or NonNegative[x] etc., thus we can do simply Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x], but to get only the list of solutions (without Rules ) there are at least two ways: Solve[ x^2 - 1 == 0 && x > 0, x ]  {{x -> 1}}  using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 : x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]  {1, 2, 3}  using Part (shorthand [[]]) with e.g. x > 0 or with the expression test like Positive, NonNegative etc.: Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]   {1, 2, 3}  We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use e.g. Reduce[x^2 - 1 == 0 && x > 0, x]. Reduce[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][[All, 2]]  The second item of More Information in the documentation of Solve says : Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].  expri can be equations or inequalities in the variables of vars, thus we can do simply : Solve[ x^2 - 1 == 0 && x > 0, x ]  {{x -> 1}}  We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use e.g. Reduce[x^2 - 1 == 0 && x > 0, x]. The second item of More Information in the documentation of Solve says : Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].  Every expri can be an equation or an inequality in the variables vars, as well as an expression test like e.g. Positive[x] or NonNegative[x] etc., thus we can do simply Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x], but to get only the list of solutions (without Rules ) there are at least two ways: using ReplaceAll (shorthand /.) (mentioned by Markus Roellig) with the condition x > 0 : x/.Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x]  {1, 2, 3}  using Part (shorthand [[]]) with e.g. x > 0 or with the expression test like Positive, NonNegative etc.: Solve[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && Positive[x], x][[All, 1, 2]]   {1, 2, 3}  We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use e.g. Reduce[-36 + 49 x^2 - 14 x^4 + x^6 == 0 && x > 0, x][[All, 2]]  1 answered Nov 6 '12 at 14:19 Artes 46.3k99 gold badges128128 silver badges212212 bronze badges The second item of More Information in the documentation of Solve says : Solve[{ expr1, expr2,...},vars] is equivalent to Solve[ expr1 && expr2 &&...,vars].  expri can be equations or inequalities in the variables of vars, thus we can do simply : Solve[ x^2 - 1 == 0 && x > 0, x ]  {{x -> 1}}  We needn't point out the domain Reals since the condition x > 0 implies that x is a positive and real number. The same concerns Reduce, i.e. use e.g. Reduce[x^2 - 1 == 0 && x > 0, x].