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Thread: system of equations - (difficult problem)

  1. #1
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    system of equations - (difficult problem)

    How does one solve the system of equations:

    x = 3y - 4y^3
    y = 3x - 4x^3

    The problem statement is pretty simple but by using conventional method of replacing variables I get a 9th degree equation in y or x - depending which variable you replace - (impossible to solve normally).
    The best I can do is bring that 9th degree equation down to a 6th degree equation (which is still impossible to solve).
    Anybody have any pointers for me to proceed in the right direction?
    Thanks a lot.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    An idea :

    $\displaystyle \left\{
    \begin{array}{lc}
    x=3y-4y^3&(1)\\
    y=3x-4x^3&(2)\\
    \end{array}\right. \implies \left\{
    \begin{array}{lc}
    x+y=3(x+y)-4(y^3+x^3)&(1)+(2)\\
    x-y=3(x-y)-4(y^3-x^3)&(1)-(2)\\
    \end{array}\right.$

    Now, you may use that $\displaystyle x^3+y^3=(x+y)(\ldots)$ and that $\displaystyle x^3-y^3=(x-y)(\ldots)$. I let you find what the dots have to be replaced by.

    Hope that helps
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  3. #3
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    Quote Originally Posted by tombrownington View Post
    How does one solve the system of equations:

    x = 3y - 4y^3
    y = 3x - 4x^3

    The problem statement is pretty simple but by using conventional method of replacing variables I get a 9th degree equation in y or x - depending which variable you replace - (impossible to solve normally).
    The best I can do is bring that 9th degree equation down to a 6th degree equation (which is still impossible to solve).
    Anybody have any pointers for me to proceed in the right direction?
    Thanks a lot.
    You can also observe that substituting $\displaystyle y = \sin \theta$ yields the equation $\displaystyle \sin \theta = \sin 9\theta$. You can then solve the trig equation.

    But this may not be the complete solution set
    For that we have to prove that $\displaystyle |x|,|y| \leq 1$ first...
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  4. #4
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    An idea :

    $\displaystyle \left\{
    \begin{array}{lc}
    x=3y-4y^3&(1)\\
    y=3x-4x^3&(2)\\
    \end{array}\right. \implies \left\{
    \begin{array}{lc}
    x+y=3(x+y)-4(y^3+x^3)&(1)+(2)\\
    x-y=3(x-y)-4(y^3-x^3)&(1)-(2)\\
    \end{array}\right.$

    Now, you may use that $\displaystyle x^3+y^3=(x+y)(\ldots)$ and that $\displaystyle x^3-y^3=(x-y)(\ldots)$. I let you find what the dots have to be replaced by.

    Hope that helps
    Hi flying squirrel,
    What you say is definitely helpful. Thanks for the insight.
    By putting x = y in one case, and x = -y in another I get 5 solutions for the system of equations. But what do I do with the (x^2-xy+y^2) part of the expression in one case, and the similar expression in the other case (the one you asked me to fill up)? I can't seem to get solutions from these. Any pointers in the right direction?
    Thanks!
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by tombrownington View Post
    I can't seem to get solutions from these. Any pointers in the right direction?
    Thanks!
    We had :

    $\displaystyle \left\{
    \begin{array}{lc}
    x+y=3(x+y)-4(y^3+x^3) & (1)+(2)\\
    x-y=3(x-y)-4(y^3-x^3) & (1)-(2)
    \end{array}\right.
    $

    Simplifying and factoring gives :

    $\displaystyle \left\{
    \begin{array}{l}
    -2(x+y)=-4(y^3+x^3)=-4(x+y)(y^2-xy+x^2) \\
    -2(x-y)=-4(y^3-x^3)=-4(y-x)(y^2+xy+x^2)
    \end{array}\right.
    $

    If $\displaystyle (x+y)(x-y)\neq 0$,

    $\displaystyle
    \left\{
    \begin{array}{lc}
    1=2(y^2-xy+x^2) & (3)\\
    -1=2(y^2+xy+x^2) & (4)\\
    \end{array}\right. $

    This is where you stopped. You can take a look at $\displaystyle (3)+(4)$, it'll give the solutions. (we're looking for solutions in $\displaystyle \mathbb{R}$ aren't we ?)
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