simple proof algebra question on' or'












0












$begingroup$


If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.



Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks






proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    No, you need to prove the complete "or" condition.
    $endgroup$
    – Yves Daoust
    Dec 4 '18 at 17:31










  • $begingroup$
    It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
    $endgroup$
    – user3482749
    Dec 4 '18 at 17:32
















0












$begingroup$


If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.



Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks






proof-writing







share|cite|improve this question











$endgroup$












  • $begingroup$
    No, you need to prove the complete "or" condition.
    $endgroup$
    – Yves Daoust
    Dec 4 '18 at 17:31










  • $begingroup$
    It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
    $endgroup$
    – user3482749
    Dec 4 '18 at 17:32














0












0








0





$begingroup$


If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.



Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks






proof-writing







share|cite|improve this question











$endgroup$




If $x^2+5y=y^2+5x$ then $x=y$ or $x+y=5$, where $x$ and $y$ are real numbers . Prove this statement.



Can someone help me with this problem or how to approach it? I can get x=y Does this mean i have proved the statement because it is an 'or'? Thanks






proof-writing




proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 17:31









user3482749

4,271919




4,271919










asked Dec 4 '18 at 17:28









HarryHarry

253




253












  • $begingroup$
    No, you need to prove the complete "or" condition.
    $endgroup$
    – Yves Daoust
    Dec 4 '18 at 17:31










  • $begingroup$
    It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
    $endgroup$
    – user3482749
    Dec 4 '18 at 17:32


















  • $begingroup$
    No, you need to prove the complete "or" condition.
    $endgroup$
    – Yves Daoust
    Dec 4 '18 at 17:31










  • $begingroup$
    It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
    $endgroup$
    – user3482749
    Dec 4 '18 at 17:32
















$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31




$begingroup$
No, you need to prove the complete "or" condition.
$endgroup$
– Yves Daoust
Dec 4 '18 at 17:31












$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32




$begingroup$
It would, but you won't be able to, because it isn't true. For example, $x = 0, y = 5$ solves that equation, so whatever you've done to get $x = y$ is wrong.
$endgroup$
– user3482749
Dec 4 '18 at 17:32










2 Answers
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$begingroup$

$$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
\iff(x+y-5)(x-y)=0.$$



Now a product is zero when either factor is zero, that means



$$x+y-5=0text{ or }x-y=0.$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Hint



    $$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      2












      $begingroup$

      $$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
      \iff(x+y-5)(x-y)=0.$$



      Now a product is zero when either factor is zero, that means



      $$x+y-5=0text{ or }x-y=0.$$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        $$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
        \iff(x+y-5)(x-y)=0.$$



        Now a product is zero when either factor is zero, that means



        $$x+y-5=0text{ or }x-y=0.$$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          $$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
          \iff(x+y-5)(x-y)=0.$$



          Now a product is zero when either factor is zero, that means



          $$x+y-5=0text{ or }x-y=0.$$






          share|cite|improve this answer











          $endgroup$



          $$x^2+5y=y^2+5xiff x^2-y^2=5x-5yiff(x+y)(x-y)=5(x-y)
          \iff(x+y-5)(x-y)=0.$$



          Now a product is zero when either factor is zero, that means



          $$x+y-5=0text{ or }x-y=0.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 4 '18 at 18:07

























          answered Dec 4 '18 at 17:33









          Yves DaoustYves Daoust

          126k672226




          126k672226























              0












              $begingroup$

              Hint



              $$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Hint



                $$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Hint



                  $$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$






                  share|cite|improve this answer









                  $endgroup$



                  Hint



                  $$x^2+5y=y^2+5xiff y^2-5y+5x-x^2=0iff y=dfrac{5pm |2x-5|}{2}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 17:32









                  mflmfl

                  26.3k12141




                  26.3k12141






























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