Can you help me solve this algebra problem?












7












$begingroup$


Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    8 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    8 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    6 hours ago


















7












$begingroup$


Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    8 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    8 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    6 hours ago
















7












7








7


1



$begingroup$


Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






algebra-precalculus systems-of-equations







share|cite|improve this question











$endgroup$




Hi I need to solve this problem and I don’t know how. I’d appreciate your help.




If $x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}$ and $xneq yneq z$, then $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$




I think I need to



$x^3 - ayz = x^2k$



$y^3 - azx = y^2k,$



$z^3 - axy = z^2k$



then to multiply both sides of each equality by some quantity, add them all together and factor but I don’t know how to find that quantity.






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 5 hours ago









Michael Rozenberg

105k1892197




105k1892197










asked 8 hours ago









questions about mathquestions about math

714




714












  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    8 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    8 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    6 hours ago




















  • $begingroup$
    Are you trying to prove the last equality? It's not clear.
    $endgroup$
    – Victor S.
    8 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    8 hours ago












  • $begingroup$
    This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
    $endgroup$
    – Dr. Mathva
    6 hours ago


















$begingroup$
Are you trying to prove the last equality? It's not clear.
$endgroup$
– Victor S.
8 hours ago




$begingroup$
Are you trying to prove the last equality? It's not clear.
$endgroup$
– Victor S.
8 hours ago












$begingroup$
@VictorS. I need to prove that each expression is equal to x + y + z - a
$endgroup$
– questions about math
8 hours ago






$begingroup$
@VictorS. I need to prove that each expression is equal to x + y + z - a
$endgroup$
– questions about math
8 hours ago














$begingroup$
This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
$endgroup$
– Dr. Mathva
6 hours ago






$begingroup$
This isn't an answer. However, it could help... $$$$Observe that $$(x+y+z-a)-bigg(x-frac{ayz}{x^2}bigg)=frac{ayz+yx^2+zx^2-ax^2}{x^2}=frac{a(yz-x^2)+x^2(y+z)}{x^2}$$ This equals zero iff $$a(x^2-yz)=x^2(y+z)iff a-frac{ayz}{x^2}=y+z$$ $$iff a·bigg(frac{x^2-yz}{x^2}bigg)=y+ziff a=frac{x^2y+x^2z}{x^2-yz}$$
$endgroup$
– Dr. Mathva
6 hours ago












2 Answers
2






active

oldest

votes


















2












$begingroup$

$$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
$$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
$$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
$$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
$$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
$$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
$$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
$$x=y=z=0,$$ which is impossible.



Id est, the given is wrong, which says that
$$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
    $endgroup$
    – Marwan Mizuri
    3 hours ago






  • 1




    $begingroup$
    @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
    $endgroup$
    – Michael Rozenberg
    3 hours ago



















6












$begingroup$

Denote $k=xyz$ and $b$ the common value of $x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}$. We can see that the equation
$$
t-frac{ak}{t^3}=btag{*}
$$ is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
$$
t^4-b t^3-ak=0.
$$ By Vieta's formula, the other root $w$ satisfies
$$
x+y+z+w=b,quad xyzw=-ak.
$$ Since $k=xyzne 0$ it follows that
$$
w=-a=b-x-y-z,
$$ hence we get
$$
x-frac{ayz}{x^2}= y-frac{azx}{y^2}=z-frac{axy}{z^2}=b=x+y+z-a.
$$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
    $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
    $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
    Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
    $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
    Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
    $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
    $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
    $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
    $$x=y=z=0,$$ which is impossible.



    Id est, the given is wrong, which says that
    $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
      $endgroup$
      – Marwan Mizuri
      3 hours ago






    • 1




      $begingroup$
      @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
      $endgroup$
      – Michael Rozenberg
      3 hours ago
















    2












    $begingroup$

    $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
    $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
    $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
    Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
    $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
    Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
    $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
    $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
    $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
    $$x=y=z=0,$$ which is impossible.



    Id est, the given is wrong, which says that
    $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
      $endgroup$
      – Marwan Mizuri
      3 hours ago






    • 1




      $begingroup$
      @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
      $endgroup$
      – Michael Rozenberg
      3 hours ago














    2












    2








    2





    $begingroup$

    $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
    $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
    $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
    Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
    $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
    Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
    $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
    $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
    $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
    $$x=y=z=0,$$ which is impossible.



    Id est, the given is wrong, which says that
    $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.






    share|cite|improve this answer









    $endgroup$



    $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ gives
    $$x-y+azleft(frac{x}{y^2}-frac{y}{x^2}right)=0$$ or
    $$1+frac{az(x^2+xy+y^2)}{x^2y^2}=0.$$
    Similarly, $$frac{ax(y^2+yz+z^2)}{y^2z^2}=-1$$ and
    $$frac{ay(x^2+xz+z^2)}{x^2z^2}=-1.$$
    Thus, $$x^3(y^2+yz+z^2)=y^3(x^2+xz+z^2)$$ or
    $$(x-y)(x^2y^2+xyz(x+y)+z^2(x^2+xy+y^2))=0$$ or
    $$sum_{cyc}(x^2y^2+x^2yz)=0$$ or
    $$sum_{cyc}z^2(x+y)^2=0,$$ which gives $$x+y=x+z=y+z=0$$ or
    $$x=y=z=0,$$ which is impossible.



    Id est, the given is wrong, which says that
    $$x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2}Rightarrow x - frac{ayz}{x^2} = y - frac{azx}{y^2} = z - frac{axy}{z^2} = x + y + z - a$$ is true.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 5 hours ago









    Michael RozenbergMichael Rozenberg

    105k1892197




    105k1892197












    • $begingroup$
      So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
      $endgroup$
      – Marwan Mizuri
      3 hours ago






    • 1




      $begingroup$
      @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
      $endgroup$
      – Michael Rozenberg
      3 hours ago


















    • $begingroup$
      So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
      $endgroup$
      – Marwan Mizuri
      3 hours ago






    • 1




      $begingroup$
      @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
      $endgroup$
      – Michael Rozenberg
      3 hours ago
















    $begingroup$
    So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
    $endgroup$
    – Marwan Mizuri
    3 hours ago




    $begingroup$
    So, taking into account Song's answer, does this mean that the premise of the question is wrong, but if we were to ignore that then the conclusion is correct?
    $endgroup$
    – Marwan Mizuri
    3 hours ago




    1




    1




    $begingroup$
    @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
    $endgroup$
    – Michael Rozenberg
    3 hours ago




    $begingroup$
    @Marwan Mizuri The Song's answer is beautiful with assuming that these $x$, $y$ and $z$ there are exist. But it's not so.
    $endgroup$
    – Michael Rozenberg
    3 hours ago











    6












    $begingroup$

    Denote $k=xyz$ and $b$ the common value of $x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}$. We can see that the equation
    $$
    t-frac{ak}{t^3}=btag{*}
    $$     is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
    $$
    t^4-b t^3-ak=0.
    $$     By Vieta's formula, the other root $w$ satisfies
    $$
    x+y+z+w=b,quad xyzw=-ak.
    $$     Since $k=xyzne 0$ it follows that
    $$
    w=-a=b-x-y-z,
    $$     hence we get
    $$
    x-frac{ayz}{x^2}= y-frac{azx}{y^2}=z-frac{axy}{z^2}=b=x+y+z-a.
    $$






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      Denote $k=xyz$ and $b$ the common value of $x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}$. We can see that the equation
      $$
      t-frac{ak}{t^3}=btag{*}
      $$       is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
      $$
      t^4-b t^3-ak=0.
      $$       By Vieta's formula, the other root $w$ satisfies
      $$
      x+y+z+w=b,quad xyzw=-ak.
      $$       Since $k=xyzne 0$ it follows that
      $$
      w=-a=b-x-y-z,
      $$       hence we get
      $$
      x-frac{ayz}{x^2}= y-frac{azx}{y^2}=z-frac{axy}{z^2}=b=x+y+z-a.
      $$






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Denote $k=xyz$ and $b$ the common value of $x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}$. We can see that the equation
        $$
        t-frac{ak}{t^3}=btag{*}
        $$         is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
        $$
        t^4-b t^3-ak=0.
        $$         By Vieta's formula, the other root $w$ satisfies
        $$
        x+y+z+w=b,quad xyzw=-ak.
        $$         Since $k=xyzne 0$ it follows that
        $$
        w=-a=b-x-y-z,
        $$         hence we get
        $$
        x-frac{ayz}{x^2}= y-frac{azx}{y^2}=z-frac{axy}{z^2}=b=x+y+z-a.
        $$






        share|cite|improve this answer











        $endgroup$



        Denote $k=xyz$ and $b$ the common value of $x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}$. We can see that the equation
        $$
        t-frac{ak}{t^3}=btag{*}
        $$         is satisfied by $t=x,y,z$. Hence $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a polynomial equation of degree 4
        $$
        t^4-b t^3-ak=0.
        $$         By Vieta's formula, the other root $w$ satisfies
        $$
        x+y+z+w=b,quad xyzw=-ak.
        $$         Since $k=xyzne 0$ it follows that
        $$
        w=-a=b-x-y-z,
        $$         hence we get
        $$
        x-frac{ayz}{x^2}= y-frac{azx}{y^2}=z-frac{axy}{z^2}=b=x+y+z-a.
        $$







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        edited 4 hours ago

























        answered 5 hours ago









        SongSong

        14.5k1635




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