Very Hard System of Equations












-1












$begingroup$


Solve the system of equations:



begin{cases}
sqrt{xy}(x + 3y)(3x + y) = 14 \
(x + y)(x^2 + y^2 + 14xy) = 36.
end{cases}



Suppose $x + y = m$ and $xy = n$. So I get



begin{cases}
(3m^2+4n) sqrt n = 14 \
m^3+12mn=36
end{cases}



So can express $m$ in terms of $n$ but I want only real solution.
Is there a faster way?






algebra-precalculus systems-of-equations substitution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It will help you greatly to note that these equations are transposable, that is, $(x,y)=(X,Y)$ is a solution, so too is $(x,y)=(Y,X)$.
    $endgroup$
    – Rhys Hughes
    Jan 18 at 3:22
















-1












$begingroup$


Solve the system of equations:



begin{cases}
sqrt{xy}(x + 3y)(3x + y) = 14 \
(x + y)(x^2 + y^2 + 14xy) = 36.
end{cases}



Suppose $x + y = m$ and $xy = n$. So I get



begin{cases}
(3m^2+4n) sqrt n = 14 \
m^3+12mn=36
end{cases}



So can express $m$ in terms of $n$ but I want only real solution.
Is there a faster way?






algebra-precalculus systems-of-equations substitution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It will help you greatly to note that these equations are transposable, that is, $(x,y)=(X,Y)$ is a solution, so too is $(x,y)=(Y,X)$.
    $endgroup$
    – Rhys Hughes
    Jan 18 at 3:22














-1












-1








-1





$begingroup$


Solve the system of equations:



begin{cases}
sqrt{xy}(x + 3y)(3x + y) = 14 \
(x + y)(x^2 + y^2 + 14xy) = 36.
end{cases}



Suppose $x + y = m$ and $xy = n$. So I get



begin{cases}
(3m^2+4n) sqrt n = 14 \
m^3+12mn=36
end{cases}



So can express $m$ in terms of $n$ but I want only real solution.
Is there a faster way?






algebra-precalculus systems-of-equations substitution







share|cite|improve this question











$endgroup$




Solve the system of equations:



begin{cases}
sqrt{xy}(x + 3y)(3x + y) = 14 \
(x + y)(x^2 + y^2 + 14xy) = 36.
end{cases}



Suppose $x + y = m$ and $xy = n$. So I get



begin{cases}
(3m^2+4n) sqrt n = 14 \
m^3+12mn=36
end{cases}



So can express $m$ in terms of $n$ but I want only real solution.
Is there a faster way?






algebra-precalculus systems-of-equations substitution




algebra-precalculus systems-of-equations substitution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 12:41









Michael Rozenberg

103k1891195




103k1891195










asked Jan 18 at 2:44









HeartHeart

28918




28918








  • 1




    $begingroup$
    It will help you greatly to note that these equations are transposable, that is, $(x,y)=(X,Y)$ is a solution, so too is $(x,y)=(Y,X)$.
    $endgroup$
    – Rhys Hughes
    Jan 18 at 3:22














  • 1




    $begingroup$
    It will help you greatly to note that these equations are transposable, that is, $(x,y)=(X,Y)$ is a solution, so too is $(x,y)=(Y,X)$.
    $endgroup$
    – Rhys Hughes
    Jan 18 at 3:22








1




1




$begingroup$
It will help you greatly to note that these equations are transposable, that is, $(x,y)=(X,Y)$ is a solution, so too is $(x,y)=(Y,X)$.
$endgroup$
– Rhys Hughes
Jan 18 at 3:22




$begingroup$
It will help you greatly to note that these equations are transposable, that is, $(x,y)=(X,Y)$ is a solution, so too is $(x,y)=(Y,X)$.
$endgroup$
– Rhys Hughes
Jan 18 at 3:22










2 Answers
2






active

oldest

votes


















0












$begingroup$

Let $x+y=2ksqrt{xy}.$
Thus, by AM-GM $kgeq1$ and from the first and the second equations we obtain:
$$18sqrt{xy}(3x^2+3y^2+10xy)=7(x+y)(x^2+y^2+14xy)$$ or
$$18(12k^2+4)=7cdot2k(4k^2+12)$$ or
$$7k^3-27k^2+21k-9=0$$ or
$$7k^3-21k^2-6k^2+18k+3k-9=0$$ or
$$(k-3)(7k^2-6k+3)=0,$$ which gives $k=3$ and $$x+y=6sqrt{xy}.$$
Can you end it now?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes thank you for good solution :)
    $endgroup$
    – Heart
    Jan 18 at 14:34










  • $begingroup$
    @Heart You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 18 at 17:02



















2












$begingroup$

Solving the second equation for $n$ we get $$n=-1/12,{frac {{m}^{3}-36}{m}}$$ for $$mneq 0$$ plugging this in your first equation we get
$$1/6, left( 3,{m}^{2}-1/3,{frac {{m}^{3}-36}{m}} right) sqrt {-3
,{frac {{m}^{3}-36}{m}}}=14
$$
so now we square and factorize this equation we obtain:
$$-16, left( m-3 right) left( {m}^{2}+3 right) left( {m}^{2}+3,
m+3 right) left( {m}^{2}-3,m+3 right) left( {m}^{2}+3,m+9
right)
=0$$
I hope you will find all solutions.






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









    0












    $begingroup$

    Let $x+y=2ksqrt{xy}.$
    Thus, by AM-GM $kgeq1$ and from the first and the second equations we obtain:
    $$18sqrt{xy}(3x^2+3y^2+10xy)=7(x+y)(x^2+y^2+14xy)$$ or
    $$18(12k^2+4)=7cdot2k(4k^2+12)$$ or
    $$7k^3-27k^2+21k-9=0$$ or
    $$7k^3-21k^2-6k^2+18k+3k-9=0$$ or
    $$(k-3)(7k^2-6k+3)=0,$$ which gives $k=3$ and $$x+y=6sqrt{xy}.$$
    Can you end it now?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes thank you for good solution :)
      $endgroup$
      – Heart
      Jan 18 at 14:34










    • $begingroup$
      @Heart You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 18 at 17:02
















    0












    $begingroup$

    Let $x+y=2ksqrt{xy}.$
    Thus, by AM-GM $kgeq1$ and from the first and the second equations we obtain:
    $$18sqrt{xy}(3x^2+3y^2+10xy)=7(x+y)(x^2+y^2+14xy)$$ or
    $$18(12k^2+4)=7cdot2k(4k^2+12)$$ or
    $$7k^3-27k^2+21k-9=0$$ or
    $$7k^3-21k^2-6k^2+18k+3k-9=0$$ or
    $$(k-3)(7k^2-6k+3)=0,$$ which gives $k=3$ and $$x+y=6sqrt{xy}.$$
    Can you end it now?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes thank you for good solution :)
      $endgroup$
      – Heart
      Jan 18 at 14:34










    • $begingroup$
      @Heart You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 18 at 17:02














    0












    0








    0





    $begingroup$

    Let $x+y=2ksqrt{xy}.$
    Thus, by AM-GM $kgeq1$ and from the first and the second equations we obtain:
    $$18sqrt{xy}(3x^2+3y^2+10xy)=7(x+y)(x^2+y^2+14xy)$$ or
    $$18(12k^2+4)=7cdot2k(4k^2+12)$$ or
    $$7k^3-27k^2+21k-9=0$$ or
    $$7k^3-21k^2-6k^2+18k+3k-9=0$$ or
    $$(k-3)(7k^2-6k+3)=0,$$ which gives $k=3$ and $$x+y=6sqrt{xy}.$$
    Can you end it now?






    share|cite|improve this answer









    $endgroup$



    Let $x+y=2ksqrt{xy}.$
    Thus, by AM-GM $kgeq1$ and from the first and the second equations we obtain:
    $$18sqrt{xy}(3x^2+3y^2+10xy)=7(x+y)(x^2+y^2+14xy)$$ or
    $$18(12k^2+4)=7cdot2k(4k^2+12)$$ or
    $$7k^3-27k^2+21k-9=0$$ or
    $$7k^3-21k^2-6k^2+18k+3k-9=0$$ or
    $$(k-3)(7k^2-6k+3)=0,$$ which gives $k=3$ and $$x+y=6sqrt{xy}.$$
    Can you end it now?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 18 at 7:06









    Michael RozenbergMichael Rozenberg

    103k1891195




    103k1891195












    • $begingroup$
      Yes thank you for good solution :)
      $endgroup$
      – Heart
      Jan 18 at 14:34










    • $begingroup$
      @Heart You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 18 at 17:02


















    • $begingroup$
      Yes thank you for good solution :)
      $endgroup$
      – Heart
      Jan 18 at 14:34










    • $begingroup$
      @Heart You are welcome!
      $endgroup$
      – Michael Rozenberg
      Jan 18 at 17:02
















    $begingroup$
    Yes thank you for good solution :)
    $endgroup$
    – Heart
    Jan 18 at 14:34




    $begingroup$
    Yes thank you for good solution :)
    $endgroup$
    – Heart
    Jan 18 at 14:34












    $begingroup$
    @Heart You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 18 at 17:02




    $begingroup$
    @Heart You are welcome!
    $endgroup$
    – Michael Rozenberg
    Jan 18 at 17:02











    2












    $begingroup$

    Solving the second equation for $n$ we get $$n=-1/12,{frac {{m}^{3}-36}{m}}$$ for $$mneq 0$$ plugging this in your first equation we get
    $$1/6, left( 3,{m}^{2}-1/3,{frac {{m}^{3}-36}{m}} right) sqrt {-3
    ,{frac {{m}^{3}-36}{m}}}=14
    $$
    so now we square and factorize this equation we obtain:
    $$-16, left( m-3 right) left( {m}^{2}+3 right) left( {m}^{2}+3,
    m+3 right) left( {m}^{2}-3,m+3 right) left( {m}^{2}+3,m+9
    right)
    =0$$
    I hope you will find all solutions.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Solving the second equation for $n$ we get $$n=-1/12,{frac {{m}^{3}-36}{m}}$$ for $$mneq 0$$ plugging this in your first equation we get
      $$1/6, left( 3,{m}^{2}-1/3,{frac {{m}^{3}-36}{m}} right) sqrt {-3
      ,{frac {{m}^{3}-36}{m}}}=14
      $$
      so now we square and factorize this equation we obtain:
      $$-16, left( m-3 right) left( {m}^{2}+3 right) left( {m}^{2}+3,
      m+3 right) left( {m}^{2}-3,m+3 right) left( {m}^{2}+3,m+9
      right)
      =0$$
      I hope you will find all solutions.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Solving the second equation for $n$ we get $$n=-1/12,{frac {{m}^{3}-36}{m}}$$ for $$mneq 0$$ plugging this in your first equation we get
        $$1/6, left( 3,{m}^{2}-1/3,{frac {{m}^{3}-36}{m}} right) sqrt {-3
        ,{frac {{m}^{3}-36}{m}}}=14
        $$
        so now we square and factorize this equation we obtain:
        $$-16, left( m-3 right) left( {m}^{2}+3 right) left( {m}^{2}+3,
        m+3 right) left( {m}^{2}-3,m+3 right) left( {m}^{2}+3,m+9
        right)
        =0$$
        I hope you will find all solutions.






        share|cite|improve this answer









        $endgroup$



        Solving the second equation for $n$ we get $$n=-1/12,{frac {{m}^{3}-36}{m}}$$ for $$mneq 0$$ plugging this in your first equation we get
        $$1/6, left( 3,{m}^{2}-1/3,{frac {{m}^{3}-36}{m}} right) sqrt {-3
        ,{frac {{m}^{3}-36}{m}}}=14
        $$
        so now we square and factorize this equation we obtain:
        $$-16, left( m-3 right) left( {m}^{2}+3 right) left( {m}^{2}+3,
        m+3 right) left( {m}^{2}-3,m+3 right) left( {m}^{2}+3,m+9
        right)
        =0$$
        I hope you will find all solutions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 3:23









        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

        75.2k42865




        75.2k42865






























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