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Can you help me solve this algebra problem?


8th Grade Algebra Problem I Can't SolveA problem with exponent laws…How can I solve for this variable?Age Problem (Equation Writing Help)Help on Algebra ProblemCan someone show me how to solve these matrix?Precalculus algebra problemIf $y=frac {a+bz}{c+dz}$, $z=frac{a+bx}{c+dx}$, $x=frac{a+by}{c+dy}$, then $ad + bc + b^2 + c^2 = 0$Algebra precalculus problemPrecalculus algebra exercise













6












$begingroup$


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.
    3 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 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
    1 hour ago


















6












$begingroup$


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.
    3 hours ago










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 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
    1 hour ago
















6












6








6


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 35 mins ago









Michael Rozenberg

104k1892197




104k1892197










asked 3 hours ago









questions about mathquestions about math

644




644












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










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 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
    1 hour ago




















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










  • $begingroup$
    @VictorS. I need to prove that each expression is equal to x + y + z - a
    $endgroup$
    – questions about math
    3 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
    1 hour ago


















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




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












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






$begingroup$
@VictorS. I need to prove that each expression is equal to x + y + z - a
$endgroup$
– questions about math
3 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
1 hour 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
1 hour ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Denote $k=xyz$. Let us consider the equation
$$
t-frac{ak}{t^3}=btag{*}
$$ where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
$$
t^4-b t^3-ak=0.
$$ By Vieta's formula, the fourth root $w$ satisfies
$$
x+y+z+w=b text{ and } xyzw=-ak.
$$ Since $k=xyzne 0$ it follows
$$
w=-a=b-x-y-z,
$$ hence we get
$$
b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
$$as wanted.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ 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









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Denote $k=xyz$. Let us consider the equation
      $$
      t-frac{ak}{t^3}=btag{*}
      $$       where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
      $$
      t^4-b t^3-ak=0.
      $$       By Vieta's formula, the fourth root $w$ satisfies
      $$
      x+y+z+w=b text{ and } xyzw=-ak.
      $$       Since $k=xyzne 0$ it follows
      $$
      w=-a=b-x-y-z,
      $$       hence we get
      $$
      b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
      $$      as wanted.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Denote $k=xyz$. Let us consider the equation
        $$
        t-frac{ak}{t^3}=btag{*}
        $$         where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
        $$
        t^4-b t^3-ak=0.
        $$         By Vieta's formula, the fourth root $w$ satisfies
        $$
        x+y+z+w=b text{ and } xyzw=-ak.
        $$         Since $k=xyzne 0$ it follows
        $$
        w=-a=b-x-y-z,
        $$         hence we get
        $$
        b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
        $$        as wanted.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Denote $k=xyz$. Let us consider the equation
          $$
          t-frac{ak}{t^3}=btag{*}
          $$           where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
          $$
          t^4-b t^3-ak=0.
          $$           By Vieta's formula, the fourth root $w$ satisfies
          $$
          x+y+z+w=b text{ and } xyzw=-ak.
          $$           Since $k=xyzne 0$ it follows
          $$
          w=-a=b-x-y-z,
          $$           hence we get
          $$
          b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
          $$          as wanted.






          share|cite|improve this answer











          $endgroup$



          Denote $k=xyz$. Let us consider the equation
          $$
          t-frac{ak}{t^3}=btag{*}
          $$           where $b =x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2},$ and $x,y,z$ are distinct, non-zero roots of $(*)$. Note that $(*)$ can be written as a 4-th degree polynomial equation
          $$
          t^4-b t^3-ak=0.
          $$           By Vieta's formula, the fourth root $w$ satisfies
          $$
          x+y+z+w=b text{ and } xyzw=-ak.
          $$           Since $k=xyzne 0$ it follows
          $$
          w=-a=b-x-y-z,
          $$           hence we get
          $$
          b=x+y+z-a=x-frac{ayz}{x^2}=y-frac{azx}{y^2}=z-frac{axy}{z^2}
          $$          as wanted.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 15 mins ago

























          answered 49 mins ago









          SongSong

          14.5k1635




          14.5k1635























              1












              $begingroup$

              $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ 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









              $endgroup$


















                1












                $begingroup$

                $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ 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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$x-frac{ayz}{x^2}=y-frac{axz}{y^2}$$ 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 45 mins ago









                  Michael RozenbergMichael Rozenberg

                  104k1892197




                  104k1892197






























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