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Math Help - 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 :

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

    Now, you may use that x^3+y^3=(x+y)(\ldots) and that 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
    Lord of certain Rings
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    IISc, Bangalore
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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 y = \sin \theta yields the equation \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 |x|,|y| \leq 1 first...
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  4. #4
    Junior Member
    Joined
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    An idea :

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

    Now, you may use that x^3+y^3=(x+y)(\ldots) and that 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
    Joined
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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 :

    \left\{<br />
\begin{array}{lc}<br />
x+y=3(x+y)-4(y^3+x^3) & (1)+(2)\\<br />
x-y=3(x-y)-4(y^3-x^3) & (1)-(2)<br />
\end{array}\right.<br />

    Simplifying and factoring gives :

     \left\{<br />
\begin{array}{l}<br />
-2(x+y)=-4(y^3+x^3)=-4(x+y)(y^2-xy+x^2) \\<br />
-2(x-y)=-4(y^3-x^3)=-4(y-x)(y^2+xy+x^2)<br />
\end{array}\right.<br />

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

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

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