Don't understand an algebraic answer [closed]












1












$begingroup$


Question:



$a+frac{10b}{a^2+b^2} = 5$



$b+frac{10a}{a^2+b^2} = 4$



Prove $ab$ is not equal to 0.



Thanks for your help.






algebra-precalculus systems-of-equations







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$endgroup$



closed as off-topic by Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz Jan 19 at 7:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 3




    $begingroup$
    Hint: if $ab=0$ Then either a or b is zero. Inserting for example $a=0$ in both equations leads to an impossibile system of equations for $b$
    $endgroup$
    – b00n heT
    Jan 18 at 7:22












  • $begingroup$
    Do we know that the set here is an integral domain? Maybe it's irrelevant, but we can use $ab=0 Rightarrow a= 0$ or $b=0 $ only when we working in integral domains, Right??
    $endgroup$
    – Mayank Mishra
    Jan 18 at 7:58












  • $begingroup$
    @Rick Given the level of this question, I think it is safe to assume that we are working with the real numbers.
    $endgroup$
    – Math1000
    Jan 18 at 17:35
















1












$begingroup$


Question:



$a+frac{10b}{a^2+b^2} = 5$



$b+frac{10a}{a^2+b^2} = 4$



Prove $ab$ is not equal to 0.



Thanks for your help.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$



closed as off-topic by Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz Jan 19 at 7:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 3




    $begingroup$
    Hint: if $ab=0$ Then either a or b is zero. Inserting for example $a=0$ in both equations leads to an impossibile system of equations for $b$
    $endgroup$
    – b00n heT
    Jan 18 at 7:22












  • $begingroup$
    Do we know that the set here is an integral domain? Maybe it's irrelevant, but we can use $ab=0 Rightarrow a= 0$ or $b=0 $ only when we working in integral domains, Right??
    $endgroup$
    – Mayank Mishra
    Jan 18 at 7:58












  • $begingroup$
    @Rick Given the level of this question, I think it is safe to assume that we are working with the real numbers.
    $endgroup$
    – Math1000
    Jan 18 at 17:35














1












1








1





$begingroup$


Question:



$a+frac{10b}{a^2+b^2} = 5$



$b+frac{10a}{a^2+b^2} = 4$



Prove $ab$ is not equal to 0.



Thanks for your help.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$




Question:



$a+frac{10b}{a^2+b^2} = 5$



$b+frac{10a}{a^2+b^2} = 4$



Prove $ab$ is not equal to 0.



Thanks for your help.






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








edited Jan 18 at 7:37









Michael Rozenberg

111k1897201




111k1897201










asked Jan 18 at 7:17









Chin Hong TanChin Hong Tan

305




305




closed as off-topic by Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz Jan 19 at 7:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz Jan 19 at 7:58


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eric Wofsey, Robert Z, José Carlos Santos, Brandon Carter, A. Pongrácz

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 3




    $begingroup$
    Hint: if $ab=0$ Then either a or b is zero. Inserting for example $a=0$ in both equations leads to an impossibile system of equations for $b$
    $endgroup$
    – b00n heT
    Jan 18 at 7:22












  • $begingroup$
    Do we know that the set here is an integral domain? Maybe it's irrelevant, but we can use $ab=0 Rightarrow a= 0$ or $b=0 $ only when we working in integral domains, Right??
    $endgroup$
    – Mayank Mishra
    Jan 18 at 7:58












  • $begingroup$
    @Rick Given the level of this question, I think it is safe to assume that we are working with the real numbers.
    $endgroup$
    – Math1000
    Jan 18 at 17:35














  • 3




    $begingroup$
    Hint: if $ab=0$ Then either a or b is zero. Inserting for example $a=0$ in both equations leads to an impossibile system of equations for $b$
    $endgroup$
    – b00n heT
    Jan 18 at 7:22












  • $begingroup$
    Do we know that the set here is an integral domain? Maybe it's irrelevant, but we can use $ab=0 Rightarrow a= 0$ or $b=0 $ only when we working in integral domains, Right??
    $endgroup$
    – Mayank Mishra
    Jan 18 at 7:58












  • $begingroup$
    @Rick Given the level of this question, I think it is safe to assume that we are working with the real numbers.
    $endgroup$
    – Math1000
    Jan 18 at 17:35








3




3




$begingroup$
Hint: if $ab=0$ Then either a or b is zero. Inserting for example $a=0$ in both equations leads to an impossibile system of equations for $b$
$endgroup$
– b00n heT
Jan 18 at 7:22






$begingroup$
Hint: if $ab=0$ Then either a or b is zero. Inserting for example $a=0$ in both equations leads to an impossibile system of equations for $b$
$endgroup$
– b00n heT
Jan 18 at 7:22














$begingroup$
Do we know that the set here is an integral domain? Maybe it's irrelevant, but we can use $ab=0 Rightarrow a= 0$ or $b=0 $ only when we working in integral domains, Right??
$endgroup$
– Mayank Mishra
Jan 18 at 7:58






$begingroup$
Do we know that the set here is an integral domain? Maybe it's irrelevant, but we can use $ab=0 Rightarrow a= 0$ or $b=0 $ only when we working in integral domains, Right??
$endgroup$
– Mayank Mishra
Jan 18 at 7:58














$begingroup$
@Rick Given the level of this question, I think it is safe to assume that we are working with the real numbers.
$endgroup$
– Math1000
Jan 18 at 17:35




$begingroup$
@Rick Given the level of this question, I think it is safe to assume that we are working with the real numbers.
$endgroup$
– Math1000
Jan 18 at 17:35










2 Answers
2






active

oldest

votes


















2












$begingroup$

Let $a=0$.



Thus, from the first equation we obtain $b=2$ and from the second we obtain $b=4$, which is a contradiction.



By the same way we can get a contradiction for $b=0$.



Thus, $abneq0.$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Multiplying both equations, rearranging and factoring gives
    $$ableft(1+frac {100}{(a^2+b^2)^2}right)=10$$
    hence $abneq 0$.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Let $a=0$.



      Thus, from the first equation we obtain $b=2$ and from the second we obtain $b=4$, which is a contradiction.



      By the same way we can get a contradiction for $b=0$.



      Thus, $abneq0.$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Let $a=0$.



        Thus, from the first equation we obtain $b=2$ and from the second we obtain $b=4$, which is a contradiction.



        By the same way we can get a contradiction for $b=0$.



        Thus, $abneq0.$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Let $a=0$.



          Thus, from the first equation we obtain $b=2$ and from the second we obtain $b=4$, which is a contradiction.



          By the same way we can get a contradiction for $b=0$.



          Thus, $abneq0.$






          share|cite|improve this answer









          $endgroup$



          Let $a=0$.



          Thus, from the first equation we obtain $b=2$ and from the second we obtain $b=4$, which is a contradiction.



          By the same way we can get a contradiction for $b=0$.



          Thus, $abneq0.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 7:27









          Michael RozenbergMichael Rozenberg

          111k1897201




          111k1897201























              1












              $begingroup$

              Multiplying both equations, rearranging and factoring gives
              $$ableft(1+frac {100}{(a^2+b^2)^2}right)=10$$
              hence $abneq 0$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Multiplying both equations, rearranging and factoring gives
                $$ableft(1+frac {100}{(a^2+b^2)^2}right)=10$$
                hence $abneq 0$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Multiplying both equations, rearranging and factoring gives
                  $$ableft(1+frac {100}{(a^2+b^2)^2}right)=10$$
                  hence $abneq 0$.






                  share|cite|improve this answer









                  $endgroup$



                  Multiplying both equations, rearranging and factoring gives
                  $$ableft(1+frac {100}{(a^2+b^2)^2}right)=10$$
                  hence $abneq 0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 18:06









                  hypergeometrichypergeometric

                  17.9k1762




                  17.9k1762















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