Solving Linear Equations, with 3 unknowns












0














$$6x-8y=24$$



$$frac{-2}{3}x+ frac{8}{9}y = m$$



I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.










share|cite|improve this question


















  • 4




    Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
    – Christoph
    Dec 5 '18 at 12:07










  • Do u know Gaussian elimination?
    – Cloud JR
    Dec 5 '18 at 12:07










  • @christoph well asked lol
    – Cloud JR
    Dec 5 '18 at 12:08
















0














$$6x-8y=24$$



$$frac{-2}{3}x+ frac{8}{9}y = m$$



I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.










share|cite|improve this question


















  • 4




    Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
    – Christoph
    Dec 5 '18 at 12:07










  • Do u know Gaussian elimination?
    – Cloud JR
    Dec 5 '18 at 12:07










  • @christoph well asked lol
    – Cloud JR
    Dec 5 '18 at 12:08














0












0








0







$$6x-8y=24$$



$$frac{-2}{3}x+ frac{8}{9}y = m$$



I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.










share|cite|improve this question













$$6x-8y=24$$



$$frac{-2}{3}x+ frac{8}{9}y = m$$



I want to solve for all three of the variables. I did plot this into my calculator and the answer is $$frac{-8}{3}$$ I want to know the process in solving it.







linear-algebra fractions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 12:03









Mattking32Mattking32

11




11








  • 4




    Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
    – Christoph
    Dec 5 '18 at 12:07










  • Do u know Gaussian elimination?
    – Cloud JR
    Dec 5 '18 at 12:07










  • @christoph well asked lol
    – Cloud JR
    Dec 5 '18 at 12:08














  • 4




    Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
    – Christoph
    Dec 5 '18 at 12:07










  • Do u know Gaussian elimination?
    – Cloud JR
    Dec 5 '18 at 12:07










  • @christoph well asked lol
    – Cloud JR
    Dec 5 '18 at 12:08








4




4




Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
– Christoph
Dec 5 '18 at 12:07




Two linear equations will not determine $3$ unknowns uniquely. Furthermore, how is "$-frac{8}{3}$" an answer when asking for the values of $3$ unknowns?
– Christoph
Dec 5 '18 at 12:07












Do u know Gaussian elimination?
– Cloud JR
Dec 5 '18 at 12:07




Do u know Gaussian elimination?
– Cloud JR
Dec 5 '18 at 12:07












@christoph well asked lol
– Cloud JR
Dec 5 '18 at 12:08




@christoph well asked lol
– Cloud JR
Dec 5 '18 at 12:08










2 Answers
2






active

oldest

votes


















1














Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).



However, what you can do is solve for $m$.



Multiplying the second equation through by $-9$ we get:



$$ 6x - 8y = -9m$$



Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$



However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).






share|cite|improve this answer





























    1














    From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$



    It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.






    share|cite|improve this answer





















      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%2f3026984%2fsolving-linear-equations-with-3-unknowns%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














      Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).



      However, what you can do is solve for $m$.



      Multiplying the second equation through by $-9$ we get:



      $$ 6x - 8y = -9m$$



      Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$



      However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).






      share|cite|improve this answer


























        1














        Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).



        However, what you can do is solve for $m$.



        Multiplying the second equation through by $-9$ we get:



        $$ 6x - 8y = -9m$$



        Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$



        However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).






        share|cite|improve this answer
























          1












          1








          1






          Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).



          However, what you can do is solve for $m$.



          Multiplying the second equation through by $-9$ we get:



          $$ 6x - 8y = -9m$$



          Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$



          However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).






          share|cite|improve this answer












          Since you has three unknowns, you need three equations to be able to solve for all three variables (your system of linear equations is underdetermined).



          However, what you can do is solve for $m$.



          Multiplying the second equation through by $-9$ we get:



          $$ 6x - 8y = -9m$$



          Aha! This looks a lot like our first equation: we already know $6x - 8y = 24$. Putting these together gives $-9m = 24$ and hence $$m = -frac{8}{3}$$



          However, without another equation we cannot determine $x$ and $y$ (for example, both $x = 0, y=-3$ and $x = 4, y=0$ satisfy your system of equations. In fact, there will be infinitely many values of $x$ and $y$ satisfying the two equations).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 5 '18 at 12:09









          ODFODF

          1,381510




          1,381510























              1














              From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$



              It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.






              share|cite|improve this answer


























                1














                From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$



                It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.






                share|cite|improve this answer
























                  1












                  1








                  1






                  From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$



                  It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.






                  share|cite|improve this answer












                  From $frac{-2}{3}x+ frac{8}{9}y = m$ we get $-6x+8y=9m$. If we add this to the first equation we get $24+9m=0$, hence $m=frac{-8}{3}.$



                  It remains only one equation: $ 6x-8y=24$. There are infinitely many pairs $(x,y)$ which are solutions of this equation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 12:09









                  FredFred

                  44.4k1845




                  44.4k1845






























                      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%2f3026984%2fsolving-linear-equations-with-3-unknowns%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

                      Berounka

                      Sphinx de Gizeh

                      Different font size/position of beamer's navigation symbols template's content depending on regular/plain...