Solving Linear Equations, with 3 unknowns











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$$6x-8y=24$$



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



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.










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  • 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 at 12:07










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










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















up vote
0
down vote

favorite












$$6x-8y=24$$



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



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 at 12:07










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










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













up vote
0
down vote

favorite









up vote
0
down vote

favorite











$$6x-8y=24$$



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



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






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asked Dec 5 at 12:03









Mattking32

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 at 12:07










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










  • @christoph well asked lol
    – Cloud JR
    Dec 5 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 at 12:07










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










  • @christoph well asked lol
    – Cloud JR
    Dec 5 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 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 at 12:07












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




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












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




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










2 Answers
2






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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




























    up vote
    1
    down vote













    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





















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      2 Answers
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      2 Answers
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      up vote
      1
      down vote













      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

























        up vote
        1
        down vote













        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























          up vote
          1
          down vote










          up vote
          1
          down vote









          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 at 12:09









          ODF

          78438




          78438






















              up vote
              1
              down vote













              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

























                up vote
                1
                down vote













                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























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  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 at 12:09









                  Fred

                  43.8k1644




                  43.8k1644






























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