Solve the system of equations in the set of real numbers.












5














Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$



I got:



$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$



However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).



How can I solve this problem or how should I approach it?






algebra-precalculus systems-of-equations







share|cite|improve this question


















  • 1




    To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
    – Card_Trick
    Dec 18 '18 at 22:40






  • 1




    @Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
    – Don Thousand
    Dec 18 '18 at 22:41






  • 1




    I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
    – Card_Trick
    Dec 18 '18 at 22:42










  • @Card_Trick Pero is concluding that the $0$ solution is extraneous.
    – The Great Duck
    Dec 19 '18 at 5:09
















5














Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$



I got:



$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$



However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).



How can I solve this problem or how should I approach it?






algebra-precalculus systems-of-equations







share|cite|improve this question


















  • 1




    To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
    – Card_Trick
    Dec 18 '18 at 22:40






  • 1




    @Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
    – Don Thousand
    Dec 18 '18 at 22:41






  • 1




    I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
    – Card_Trick
    Dec 18 '18 at 22:42










  • @Card_Trick Pero is concluding that the $0$ solution is extraneous.
    – The Great Duck
    Dec 19 '18 at 5:09














5












5








5







Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$



I got:



$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$



However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).



How can I solve this problem or how should I approach it?






algebra-precalculus systems-of-equations







share|cite|improve this question













Solve the system of equations in the set of real numbers:
$$begin{cases}
frac1x + frac1{y+z} = frac13 \
frac1y + frac1{x+z} = frac15 \
frac1z + frac1{x+y} = frac17
end{cases}$$



I got:



$$begin{cases}
3(x+y+z)=x(y+z) \
5(x+y+z)=y(x+z) \
7(x+y+z)=z(x+y)
end{cases}$$



However, no matter how I continue from here, I always get $x=y=z=0$, which cannot be true; or I get a new system of equations, but still with 3 variables (which I cannot solve).



How can I solve this problem or how should I approach it?






algebra-precalculus systems-of-equations




algebra-precalculus systems-of-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 22:35









Pero

1307




1307








  • 1




    To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
    – Card_Trick
    Dec 18 '18 at 22:40






  • 1




    @Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
    – Don Thousand
    Dec 18 '18 at 22:41






  • 1




    I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
    – Card_Trick
    Dec 18 '18 at 22:42










  • @Card_Trick Pero is concluding that the $0$ solution is extraneous.
    – The Great Duck
    Dec 19 '18 at 5:09














  • 1




    To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
    – Card_Trick
    Dec 18 '18 at 22:40






  • 1




    @Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
    – Don Thousand
    Dec 18 '18 at 22:41






  • 1




    I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
    – Card_Trick
    Dec 18 '18 at 22:42










  • @Card_Trick Pero is concluding that the $0$ solution is extraneous.
    – The Great Duck
    Dec 19 '18 at 5:09








1




1




To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40




To be honest, I haven't bothered to retrace your computations. But why wouldn't concluding that $x=y=z=0$ from the second set of equations be invalid? It might very well be that your first set of equations does not have any solutions...
– Card_Trick
Dec 18 '18 at 22:40




1




1




@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41




@Card_Trick Not necessarily. The multiplication by $x+y+z$ can add extraneous solutions.
– Don Thousand
Dec 18 '18 at 22:41




1




1




I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42




I wasn't questioning that, I was just wondering why Pero was concluding that his conclusion cannot be.
– Card_Trick
Dec 18 '18 at 22:42












@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09




@Card_Trick Pero is concluding that the $0$ solution is extraneous.
– The Great Duck
Dec 19 '18 at 5:09










2 Answers
2






active

oldest

votes


















7














Let $x+y+z=m$



Adding all the equations, we get,



$xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$



Subtracting each equation one by one from this, we get,



$xy=frac{m}{2}$



$yz=frac{9m}{2}$



$zx=frac{5m}{2}$



Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
$$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
$$Longrightarrow x:y:z=5:9:45$$
Now, let $x=5k, y=9k, z=45k$ and get the result.



Hope it is helpful






share|cite|improve this answer





























    5














    We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!






    share|cite|improve this answer





















    • How do we get $z=9x, z=5y$?
      – Pero
      Dec 18 '18 at 22:58










    • Divide the three equations I've provided from each other.
      – Don Thousand
      Dec 18 '18 at 23:00











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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

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    7














    Let $x+y+z=m$



    Adding all the equations, we get,



    $xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$



    Subtracting each equation one by one from this, we get,



    $xy=frac{m}{2}$



    $yz=frac{9m}{2}$



    $zx=frac{5m}{2}$



    Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
    $$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
    $$Longrightarrow x:y:z=5:9:45$$
    Now, let $x=5k, y=9k, z=45k$ and get the result.



    Hope it is helpful






    share|cite|improve this answer


























      7














      Let $x+y+z=m$



      Adding all the equations, we get,



      $xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$



      Subtracting each equation one by one from this, we get,



      $xy=frac{m}{2}$



      $yz=frac{9m}{2}$



      $zx=frac{5m}{2}$



      Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
      $$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
      $$Longrightarrow x:y:z=5:9:45$$
      Now, let $x=5k, y=9k, z=45k$ and get the result.



      Hope it is helpful






      share|cite|improve this answer
























        7












        7








        7






        Let $x+y+z=m$



        Adding all the equations, we get,



        $xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$



        Subtracting each equation one by one from this, we get,



        $xy=frac{m}{2}$



        $yz=frac{9m}{2}$



        $zx=frac{5m}{2}$



        Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
        $$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
        $$Longrightarrow x:y:z=5:9:45$$
        Now, let $x=5k, y=9k, z=45k$ and get the result.



        Hope it is helpful






        share|cite|improve this answer












        Let $x+y+z=m$



        Adding all the equations, we get,



        $xy+yz+zx=frac{3m+5m+7m}{2}=frac{15m}{2}$



        Subtracting each equation one by one from this, we get,



        $xy=frac{m}{2}$



        $yz=frac{9m}{2}$



        $zx=frac{5m}{2}$



        Dividing by $ xyz$, we get, $$frac{1}{x}:frac{1}{y}:frac{1}{z}=9:5:1$$
        $$Longrightarrow x:y:z=frac{1}{9}:frac{1}{5}:frac{1}{1}$$
        $$Longrightarrow x:y:z=5:9:45$$
        Now, let $x=5k, y=9k, z=45k$ and get the result.



        Hope it is helpful







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 22:59









        Martund

        1,405212




        1,405212























            5














            We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!






            share|cite|improve this answer





















            • How do we get $z=9x, z=5y$?
              – Pero
              Dec 18 '18 at 22:58










            • Divide the three equations I've provided from each other.
              – Don Thousand
              Dec 18 '18 at 23:00
















            5














            We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!






            share|cite|improve this answer





















            • How do we get $z=9x, z=5y$?
              – Pero
              Dec 18 '18 at 22:58










            • Divide the three equations I've provided from each other.
              – Don Thousand
              Dec 18 '18 at 23:00














            5












            5








            5






            We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!






            share|cite|improve this answer












            We know that via your equations, $$frac{15}2(x+y+z)=xy+yz+xz$$Hence, $$xy=frac12(x+y+z)$$$$yz=frac92(x+y+z)$$$$xz=frac52(x+y+z)$$So, assuming $x+y+zneq0$, $z=9x$, $z=5y$. Try using this to move forward!







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 18 '18 at 22:44









            Don Thousand

            4,277734




            4,277734












            • How do we get $z=9x, z=5y$?
              – Pero
              Dec 18 '18 at 22:58










            • Divide the three equations I've provided from each other.
              – Don Thousand
              Dec 18 '18 at 23:00


















            • How do we get $z=9x, z=5y$?
              – Pero
              Dec 18 '18 at 22:58










            • Divide the three equations I've provided from each other.
              – Don Thousand
              Dec 18 '18 at 23:00
















            How do we get $z=9x, z=5y$?
            – Pero
            Dec 18 '18 at 22:58




            How do we get $z=9x, z=5y$?
            – Pero
            Dec 18 '18 at 22:58












            Divide the three equations I've provided from each other.
            – Don Thousand
            Dec 18 '18 at 23:00




            Divide the three equations I've provided from each other.
            – Don Thousand
            Dec 18 '18 at 23:00


















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