Rational solution to a system of equations












2














Some context.




  • By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.


  • For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:



$$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$



The question.




Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where



$$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$



Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?




I believe the answer to be negative.





What I tried.



For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.



Lemma. The answer to the question is negative, if, and only if,



$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$



I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.



Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.



We have



$$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$



If we assume (to try to get somewhere) that



$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$



then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that



$$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$



which can be rewrite



$$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$



So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.



Let's compute $det(M)$ then:



$$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$



with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since



$$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$



and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have



$$A=B=C=D=0.$$



The computations give



$$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



The question can now be reformulated as follow:




Does the system $(mathscr S)$ has rational solutions?




If we try to solve the system, we can end up with this expression:



$frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$



The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:



enter image description here





Final remarks.



This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.










share|cite|improve this question



























    2














    Some context.




    • By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.


    • For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:



    $$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$



    The question.




    Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where



    $$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$



    Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?




    I believe the answer to be negative.





    What I tried.



    For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.



    Lemma. The answer to the question is negative, if, and only if,



    $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$



    I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.



    Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.



    We have



    $$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$



    If we assume (to try to get somewhere) that



    $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$



    then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that



    $$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$



    which can be rewrite



    $$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$



    So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.



    Let's compute $det(M)$ then:



    $$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$



    with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since



    $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$



    and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have



    $$A=B=C=D=0.$$



    The computations give



    $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



    The question can now be reformulated as follow:




    Does the system $(mathscr S)$ has rational solutions?




    If we try to solve the system, we can end up with this expression:



    $frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$



    The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:



    enter image description here





    Final remarks.



    This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.










    share|cite|improve this question

























      2












      2








      2







      Some context.




      • By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.


      • For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:



      $$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$



      The question.




      Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where



      $$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$



      Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?




      I believe the answer to be negative.





      What I tried.



      For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.



      Lemma. The answer to the question is negative, if, and only if,



      $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$



      I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.



      Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.



      We have



      $$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$



      If we assume (to try to get somewhere) that



      $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$



      then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that



      $$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$



      which can be rewrite



      $$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$



      So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.



      Let's compute $det(M)$ then:



      $$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$



      with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since



      $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$



      and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have



      $$A=B=C=D=0.$$



      The computations give



      $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



      The question can now be reformulated as follow:




      Does the system $(mathscr S)$ has rational solutions?




      If we try to solve the system, we can end up with this expression:



      $frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$



      The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:



      enter image description here





      Final remarks.



      This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.










      share|cite|improve this question













      Some context.




      • By rational subspace, I mean a subspace of $mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $mathbb Q^5$.


      • For instance, the vector $v=(0,pi,3pi,pi-pi^2)$ is in a rational subspace of dimension $2$ since:



      $$v=pibegin{pmatrix} 0 \ 1 \ 3 \ 1end{pmatrix}+pi^2begin{pmatrix} 0 \ 0 \ 0 \ -1end{pmatrix}.$$



      The question.




      Let $A=mathrm{Span}(Y_1,Y_2,Y_3)$ where



      $$Y_1=begin{pmatrix} 1 \ 0 \ 0 \ sqrt 6 \ sqrt {15} end{pmatrix},quad Y_2=begin{pmatrix} 0 \ sqrt{6} \ 0 \ -sqrt {15} \ sqrt {10} end{pmatrix}quadtext { and }quad Y_3=begin{pmatrix} 0 \ 0 \ sqrt{15} \ sqrt {10} \ sqrt {6} end{pmatrix}.$$



      Does there exist a rational subspace $B$ of $mathbb R^5$, of dimension $2$, such that $Acap Bne{0}$?




      I believe the answer to be negative.





      What I tried.



      For $Yin A$, let's denote by $Y_1,ldots,Y_5inmathbb R$ its coordinates. We can prove the following lemma.



      Lemma. The answer to the question is negative, if, and only if,



      $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(Y_1,ldots,Y_5)ge 3.$$



      I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $mathbb R^5$ of dimension $2$, such that $Acap Bne{0}$.



      Let $Y:=alpha Y_1+beta Y_2+gamma Y_3$ be a vector in $Asetminus{0}$.



      We have



      $$alpha Y_1+beta Y_2+gamma Y_3=begin{pmatrix} alpha \ betasqrt 6 \ gamma sqrt {15} \ alphasqrt 6-betasqrt {15} +gammasqrt{10} \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}end{pmatrix}.$$



      If we assume (to try to get somewhere) that



      $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(alpha,sqrt{6}beta)=2,$$



      then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,ldots,x_6inmathbb Q$ such that



      $$begin{cases}gamma sqrt {15}=x_1alpha+x_2betasqrt 6 \ alphasqrt 6-betasqrt {15} +gammasqrt{10}= x_3alpha+x_4betasqrt 6 \ alphasqrt{15}+betasqrt{10}+gammasqrt{6}=x_5alpha+x_6betasqrt 6end{cases}$$



      which can be rewrite



      $$MX=0quadtext{ with } quad M=begin{pmatrix} -x_1 & -x_2sqrt 6 & sqrt{15} \ sqrt{6}-x_3 & -sqrt{15}-x_4sqrt 6 & sqrt{10} \ sqrt{15}-x_5 & sqrt{10}- x_6sqrt 6 & sqrt 6 end{pmatrix}quad text{ and }quad X=begin{pmatrix} alpha \ beta\ gammaend{pmatrix}.$$



      So if $det(M)ne 0$, we have won, since $(alpha,beta,gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.



      Let's compute $det(M)$ then:



      $$det(M)=Asqrt{15}+Bsqrt{10}+Csqrt 6+D,$$



      with $A,B,C,Dinmathbb Q$ depending on the $x_i$. Since



      $$dim_{mathbb Q}(mathrm{Span}_{mathbb Q}(sqrt 6,sqrt{10},sqrt{15}))=3,$$



      and $A,B,C,Dinmathbb Q$, if we assume $det(M)=0$, we must have



      $$A=B=C=D=0.$$



      The computations give



      $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



      The question can now be reformulated as follow:




      Does the system $(mathscr S)$ has rational solutions?




      If we try to solve the system, we can end up with this expression:



      $frac{20 , x_{1} x_{3}^{2} x_{5} + 12 , x_{1}^{3} - 30 , x_{1}^{2} x_{3} + 20 , x_{1}^{2} x_{5} + 30 , x_{3}^{2} x_{5} - 30 , x_{1} x_{5}^{2} - 75 , x_{3} x_{5}^{2} + 186 , x_{1}^{2} - 225 , x_{3}^{2} + 120 , x_{1} x_{5} - 90 , x_{5}^{2} + 450 , x_{1} + 1125 , x_{3} + 180 , x_{5}}{2 , x_{1} x_{5} + 5 , x_{3} x_{5} + 6 , x_{5}}=0.$



      The question is then to understand if this surface in $mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:



      enter image description here





      Final remarks.



      This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.







      linear-algebra number-theory vector-spaces diophantine-equations rational-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 1 '18 at 14:31









      E. Joseph

      11.6k82856




      11.6k82856






















          2 Answers
          2






          active

          oldest

          votes


















          1














          If your question boils down to the system,



          $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



          then, YES, this system has infinitely many rational solutions. If you let,



          $$x_1 = x_4 x_5-x_3 x_6$$
          $$x_2 = (5/6)(x_3-3x_4)$$
          $$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$



          This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,



          $$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$



          You simply solve the linear equation,



          $$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$



          for $x_6$, thus,



          $$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$



          with free parameter $x_3$.






          share|cite|improve this answer





























            1














            Above simultaneous equations shown below has numerical solutions:



            $(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$



            $x_1$=$(-81255/10666)$



            $x_2$=(32315/504)



            $x_3$=$(2)$



            $x_4$=$(-6295/252)$



            $x_5$=$(-20790/5333)$



            $x_6$=$(105/2)$



            The solution provided by Tito Piezas in this context



            actually requires solving a cubic equation instead of



            a quadratic equation as mentioned by him.



            (Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.






            share|cite|improve this answer























            • Thanks a lot, I really appreciate the explicit solution.
              – E. Joseph
              Dec 2 '18 at 19:22










            • You are welcome
              – Sam
              Dec 2 '18 at 20:40










            • In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
              – Tito Piezas III
              2 days ago











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021415%2frational-solution-to-a-system-of-equations%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            If your question boils down to the system,



            $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



            then, YES, this system has infinitely many rational solutions. If you let,



            $$x_1 = x_4 x_5-x_3 x_6$$
            $$x_2 = (5/6)(x_3-3x_4)$$
            $$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$



            This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,



            $$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$



            You simply solve the linear equation,



            $$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$



            for $x_6$, thus,



            $$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$



            with free parameter $x_3$.






            share|cite|improve this answer


























              1














              If your question boils down to the system,



              $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



              then, YES, this system has infinitely many rational solutions. If you let,



              $$x_1 = x_4 x_5-x_3 x_6$$
              $$x_2 = (5/6)(x_3-3x_4)$$
              $$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$



              This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,



              $$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$



              You simply solve the linear equation,



              $$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$



              for $x_6$, thus,



              $$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$



              with free parameter $x_3$.






              share|cite|improve this answer
























                1












                1








                1






                If your question boils down to the system,



                $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



                then, YES, this system has infinitely many rational solutions. If you let,



                $$x_1 = x_4 x_5-x_3 x_6$$
                $$x_2 = (5/6)(x_3-3x_4)$$
                $$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$



                This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,



                $$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$



                You simply solve the linear equation,



                $$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$



                for $x_6$, thus,



                $$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$



                with free parameter $x_3$.






                share|cite|improve this answer












                If your question boils down to the system,



                $$(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$$



                then, YES, this system has infinitely many rational solutions. If you let,



                $$x_1 = x_4 x_5-x_3 x_6$$
                $$x_2 = (5/6)(x_3-3x_4)$$
                $$x_5 =frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$



                This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,



                $$text{Poly}_1 x_4^2+text{Poly}_2 x_4+text{Poly}_3=0$$



                You simply solve the linear equation,



                $$text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$



                for $x_6$, thus,



                $$x_4 =-frac{text{Poly}_3}{text{Poly}_2}$$



                with free parameter $x_3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 15:28









                Tito Piezas III

                26.8k365169




                26.8k365169























                    1














                    Above simultaneous equations shown below has numerical solutions:



                    $(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$



                    $x_1$=$(-81255/10666)$



                    $x_2$=(32315/504)



                    $x_3$=$(2)$



                    $x_4$=$(-6295/252)$



                    $x_5$=$(-20790/5333)$



                    $x_6$=$(105/2)$



                    The solution provided by Tito Piezas in this context



                    actually requires solving a cubic equation instead of



                    a quadratic equation as mentioned by him.



                    (Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.






                    share|cite|improve this answer























                    • Thanks a lot, I really appreciate the explicit solution.
                      – E. Joseph
                      Dec 2 '18 at 19:22










                    • You are welcome
                      – Sam
                      Dec 2 '18 at 20:40










                    • In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
                      – Tito Piezas III
                      2 days ago
















                    1














                    Above simultaneous equations shown below has numerical solutions:



                    $(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$



                    $x_1$=$(-81255/10666)$



                    $x_2$=(32315/504)



                    $x_3$=$(2)$



                    $x_4$=$(-6295/252)$



                    $x_5$=$(-20790/5333)$



                    $x_6$=$(105/2)$



                    The solution provided by Tito Piezas in this context



                    actually requires solving a cubic equation instead of



                    a quadratic equation as mentioned by him.



                    (Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.






                    share|cite|improve this answer























                    • Thanks a lot, I really appreciate the explicit solution.
                      – E. Joseph
                      Dec 2 '18 at 19:22










                    • You are welcome
                      – Sam
                      Dec 2 '18 at 20:40










                    • In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
                      – Tito Piezas III
                      2 days ago














                    1












                    1








                    1






                    Above simultaneous equations shown below has numerical solutions:



                    $(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$



                    $x_1$=$(-81255/10666)$



                    $x_2$=(32315/504)



                    $x_3$=$(2)$



                    $x_4$=$(-6295/252)$



                    $x_5$=$(-20790/5333)$



                    $x_6$=$(105/2)$



                    The solution provided by Tito Piezas in this context



                    actually requires solving a cubic equation instead of



                    a quadratic equation as mentioned by him.



                    (Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.






                    share|cite|improve this answer














                    Above simultaneous equations shown below has numerical solutions:



                    $(mathscr S)quad begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \ -3x_4x_5+3x_3x_6+3x_1=0 \ 6x_2-5x_3+15x_4=0 \ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.end{cases}$



                    $x_1$=$(-81255/10666)$



                    $x_2$=(32315/504)



                    $x_3$=$(2)$



                    $x_4$=$(-6295/252)$



                    $x_5$=$(-20790/5333)$



                    $x_6$=$(105/2)$



                    The solution provided by Tito Piezas in this context



                    actually requires solving a cubic equation instead of



                    a quadratic equation as mentioned by him.



                    (Note by Tito Piezas): My solution is a cubic in the variable $x_3$, but only a quadratic in $color{red}{x_4}$. This is easily verified using Mathematica.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago









                    Tito Piezas III

                    26.8k365169




                    26.8k365169










                    answered Dec 2 '18 at 15:40









                    Sam

                    191




                    191












                    • Thanks a lot, I really appreciate the explicit solution.
                      – E. Joseph
                      Dec 2 '18 at 19:22










                    • You are welcome
                      – Sam
                      Dec 2 '18 at 20:40










                    • In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
                      – Tito Piezas III
                      2 days ago


















                    • Thanks a lot, I really appreciate the explicit solution.
                      – E. Joseph
                      Dec 2 '18 at 19:22










                    • You are welcome
                      – Sam
                      Dec 2 '18 at 20:40










                    • In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
                      – Tito Piezas III
                      2 days ago
















                    Thanks a lot, I really appreciate the explicit solution.
                    – E. Joseph
                    Dec 2 '18 at 19:22




                    Thanks a lot, I really appreciate the explicit solution.
                    – E. Joseph
                    Dec 2 '18 at 19:22












                    You are welcome
                    – Sam
                    Dec 2 '18 at 20:40




                    You are welcome
                    – Sam
                    Dec 2 '18 at 20:40












                    In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
                    – Tito Piezas III
                    2 days ago




                    In my answer, I was careful to say it is a quadratic in the variable $color{red}{x_4}$. Your comment applies to the variable $x_3$.
                    – Tito Piezas III
                    2 days ago


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021415%2frational-solution-to-a-system-of-equations%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Wiesbaden

                    Marschland

                    Dieringhausen