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Solving a linear system of reciprocals.


Solving system of multivariable 2nd-degree polynomialsProblem with system of equationsSolving a system of equations with variables in denominator.Solving system of non-linear equations.Solving a combined system of linear and bilinear equationsSolving system of linear equations ( 4 variables, 3 equations)Solving a system of 4 non-linear equationsSolving a linear system of congruencesEasier solution for system of equations.Solve the following system of equations - (4).













2












$begingroup$


Solve for $begin{cases}frac{1}{x} +frac{1}{y}+frac{1}{z}=0\frac{4}{x} +frac{3}{y}+frac{2}{z}=5\frac{3}{x} +frac{2}{y}+frac{4}{z}=-4end{cases}$



I turn the equations into $begin{cases}yz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzend{cases}$



Not sure if I am doing fine






systems-of-equations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Solve for $begin{cases}frac{1}{x} +frac{1}{y}+frac{1}{z}=0\frac{4}{x} +frac{3}{y}+frac{2}{z}=5\frac{3}{x} +frac{2}{y}+frac{4}{z}=-4end{cases}$



    I turn the equations into $begin{cases}yz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzend{cases}$



    Not sure if I am doing fine






    systems-of-equations







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Solve for $begin{cases}frac{1}{x} +frac{1}{y}+frac{1}{z}=0\frac{4}{x} +frac{3}{y}+frac{2}{z}=5\frac{3}{x} +frac{2}{y}+frac{4}{z}=-4end{cases}$



      I turn the equations into $begin{cases}yz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzend{cases}$



      Not sure if I am doing fine






      systems-of-equations







      share|cite|improve this question











      $endgroup$




      Solve for $begin{cases}frac{1}{x} +frac{1}{y}+frac{1}{z}=0\frac{4}{x} +frac{3}{y}+frac{2}{z}=5\frac{3}{x} +frac{2}{y}+frac{4}{z}=-4end{cases}$



      I turn the equations into $begin{cases}yz+xz+xy=0\4yz+3xz+2xy=5xyz\3yz+2xz+4xy=-4xyzend{cases}$



      Not sure if I am doing fine






      systems-of-equations




      systems-of-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 31 mins ago









      Cameron Buie

      87.6k773162




      87.6k773162










      asked 34 mins ago









      DavidDavid

      764




      764






















          4 Answers
          4






          active

          oldest

          votes


















          1












          $begingroup$

          If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begin{cases}t +u+v=0\4t +3u+2v=5\3t +2u+4v=-4end{cases}$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.






          share|cite|improve this answer









          $endgroup$





















            6












            $begingroup$

            It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              how about this, let:
              $$X=frac1x,,Y=frac1y,,Z=frac1z$$
              and it is much easier to solve the following:
              $$begin{pmatrix}
              1&1&1\
              4&3&2\
              3&2&4
              end{pmatrix}
              begin{pmatrix}
              X\
              Y\
              Z
              end{pmatrix}=
              begin{pmatrix}
              0\5\-4
              end{pmatrix}
              $$






              share|cite|improve this answer









              $endgroup$





















                1












                $begingroup$

                Observe that the system's matrix of the reciprocal of the unknowns is:



                $$A=begin{pmatrix}1&1&1&0\
                4&3&2&5\
                3&2&4&-4end{pmatrix}stackrel{R_2-4R_1,,R_3-3R_1}longrightarrowbegin{pmatrix}1&1&1&0\
                0&-1&-2&5\
                0&-1&-1&-4end{pmatrix}stackrel{R_3-R_2}longrightarrow$$



                $$begin{pmatrix}1&1&1&0\
                0&-1&-2&5\
                0&0&1&-9end{pmatrix}$$



                Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                  $endgroup$
                  – Bernard
                  21 mins ago














                Your Answer








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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                1












                $begingroup$

                If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begin{cases}t +u+v=0\4t +3u+2v=5\3t +2u+4v=-4end{cases}$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begin{cases}t +u+v=0\4t +3u+2v=5\3t +2u+4v=-4end{cases}$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begin{cases}t +u+v=0\4t +3u+2v=5\3t +2u+4v=-4end{cases}$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.






                    share|cite|improve this answer









                    $endgroup$



                    If the reciprocals are freaking you out, just let $t=frac1x,u=frac1y,v=frac1z,$ so you have the system $$begin{cases}t +u+v=0\4t +3u+2v=5\3t +2u+4v=-4end{cases}$$ Once you've solved this, as long as none of $t,u,v$ is $0,$ you can simply let $x=frac1t,y=frac1u,z=frac1v.$ If one or more of $t,u,v$ is $0,$ then the system has no solution.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 28 mins ago









                    Cameron BuieCameron Buie

                    87.6k773162




                    87.6k773162























                        6












                        $begingroup$

                        It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.






                        share|cite|improve this answer









                        $endgroup$


















                          6












                          $begingroup$

                          It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.






                          share|cite|improve this answer









                          $endgroup$
















                            6












                            6








                            6





                            $begingroup$

                            It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.






                            share|cite|improve this answer









                            $endgroup$



                            It's much easier to solve the linear system for the reciprocals then take the reciprocal to get the result.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 33 mins ago









                            Matt SamuelMatt Samuel

                            39.6k63870




                            39.6k63870























                                1












                                $begingroup$

                                how about this, let:
                                $$X=frac1x,,Y=frac1y,,Z=frac1z$$
                                and it is much easier to solve the following:
                                $$begin{pmatrix}
                                1&1&1\
                                4&3&2\
                                3&2&4
                                end{pmatrix}
                                begin{pmatrix}
                                X\
                                Y\
                                Z
                                end{pmatrix}=
                                begin{pmatrix}
                                0\5\-4
                                end{pmatrix}
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  how about this, let:
                                  $$X=frac1x,,Y=frac1y,,Z=frac1z$$
                                  and it is much easier to solve the following:
                                  $$begin{pmatrix}
                                  1&1&1\
                                  4&3&2\
                                  3&2&4
                                  end{pmatrix}
                                  begin{pmatrix}
                                  X\
                                  Y\
                                  Z
                                  end{pmatrix}=
                                  begin{pmatrix}
                                  0\5\-4
                                  end{pmatrix}
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    how about this, let:
                                    $$X=frac1x,,Y=frac1y,,Z=frac1z$$
                                    and it is much easier to solve the following:
                                    $$begin{pmatrix}
                                    1&1&1\
                                    4&3&2\
                                    3&2&4
                                    end{pmatrix}
                                    begin{pmatrix}
                                    X\
                                    Y\
                                    Z
                                    end{pmatrix}=
                                    begin{pmatrix}
                                    0\5\-4
                                    end{pmatrix}
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    how about this, let:
                                    $$X=frac1x,,Y=frac1y,,Z=frac1z$$
                                    and it is much easier to solve the following:
                                    $$begin{pmatrix}
                                    1&1&1\
                                    4&3&2\
                                    3&2&4
                                    end{pmatrix}
                                    begin{pmatrix}
                                    X\
                                    Y\
                                    Z
                                    end{pmatrix}=
                                    begin{pmatrix}
                                    0\5\-4
                                    end{pmatrix}
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 27 mins ago









                                    Henry LeeHenry Lee

                                    2,221319




                                    2,221319























                                        1












                                        $begingroup$

                                        Observe that the system's matrix of the reciprocal of the unknowns is:



                                        $$A=begin{pmatrix}1&1&1&0\
                                        4&3&2&5\
                                        3&2&4&-4end{pmatrix}stackrel{R_2-4R_1,,R_3-3R_1}longrightarrowbegin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&-1&-1&-4end{pmatrix}stackrel{R_3-R_2}longrightarrow$$



                                        $$begin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&0&1&-9end{pmatrix}$$



                                        Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$






                                        share|cite|improve this answer









                                        $endgroup$













                                        • $begingroup$
                                          There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                                          $endgroup$
                                          – Bernard
                                          21 mins ago


















                                        1












                                        $begingroup$

                                        Observe that the system's matrix of the reciprocal of the unknowns is:



                                        $$A=begin{pmatrix}1&1&1&0\
                                        4&3&2&5\
                                        3&2&4&-4end{pmatrix}stackrel{R_2-4R_1,,R_3-3R_1}longrightarrowbegin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&-1&-1&-4end{pmatrix}stackrel{R_3-R_2}longrightarrow$$



                                        $$begin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&0&1&-9end{pmatrix}$$



                                        Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$






                                        share|cite|improve this answer









                                        $endgroup$













                                        • $begingroup$
                                          There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                                          $endgroup$
                                          – Bernard
                                          21 mins ago
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        Observe that the system's matrix of the reciprocal of the unknowns is:



                                        $$A=begin{pmatrix}1&1&1&0\
                                        4&3&2&5\
                                        3&2&4&-4end{pmatrix}stackrel{R_2-4R_1,,R_3-3R_1}longrightarrowbegin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&-1&-1&-4end{pmatrix}stackrel{R_3-R_2}longrightarrow$$



                                        $$begin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&0&1&-9end{pmatrix}$$



                                        Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$






                                        share|cite|improve this answer









                                        $endgroup$



                                        Observe that the system's matrix of the reciprocal of the unknowns is:



                                        $$A=begin{pmatrix}1&1&1&0\
                                        4&3&2&5\
                                        3&2&4&-4end{pmatrix}stackrel{R_2-4R_1,,R_3-3R_1}longrightarrowbegin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&-1&-1&-4end{pmatrix}stackrel{R_3-R_2}longrightarrow$$



                                        $$begin{pmatrix}1&1&1&0\
                                        0&-1&-2&5\
                                        0&0&1&-9end{pmatrix}$$



                                        Try to finish the exercise now, taking into account that the third column represents $;cfrac 1z;$ , the second $;cfrac1y;$ and the first one $;cfrac1x;$







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 26 mins ago









                                        DonAntonioDonAntonio

                                        180k1495234




                                        180k1495234












                                        • $begingroup$
                                          There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                                          $endgroup$
                                          – Bernard
                                          21 mins ago




















                                        • $begingroup$
                                          There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                                          $endgroup$
                                          – Bernard
                                          21 mins ago


















                                        $begingroup$
                                        There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                                        $endgroup$
                                        – Bernard
                                        21 mins ago






                                        $begingroup$
                                        There's an error at the 1st step: the third row should be $;0;-1; color{red}{+}1;-4$.
                                        $endgroup$
                                        – Bernard
                                        21 mins ago




















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                                        Do I have Disadvantage attacking with an off-hand weapon? My body leaves; my core can stay What information about me do stores get via my credit card? Using dividends to reduce short term capital gains? different output for groups and groups USERNAME after adding a username to a group Why can't devices on different VLANs, but on the same subnet, communicate? Accepted by European university, rejected by all American ones I applied to? Possible reasons? Didn't get enough time to take a Coding Test - what to do now? Is 'stolen' appropriate word? Can I visit the Trinity College (Cambridge) library and see some of their r...
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