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Thread: System of non-linear equations

  1. #1
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    System of non-linear equations

    I read this problem on a site:

    $\displaystyle x^2-xy+y^2=21$
    $\displaystyle x^2+xy-8y^2=0$

    I believe it's not possible to isolate either x or y in either equation using standard methods, right? It was suggested to actually use the quadratic formula to solve in terms of a single variable. I also read that the second equation can be factored, but I can't seem to figure out how the factoring was done. Apparently it factors into:

    (x+4y)(x–2y)
    Last edited by mr fantastic; May 4th 2009 at 08:14 PM. Reason: Fixed one of the equations
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  2. #2
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    Quote Originally Posted by Phire View Post
    I read this problem on a site:

    $\displaystyle x^2-xy+y^2=21$
    $\displaystyle x^2+2xy-8y^2=0$

    I believe it's not possible to isolate either x or y in either equation using standard methods, right? It was suggested to actually use the quadratic formula to solve in terms of a single variable. I also read that the second equation can be factored, but I can't seem to figure out how the factoring was done. Apparently it factors into:

    (x+4y)(x2y)
    The Quadratic Formula looks to be easiest, though Completing the Square would be OK too.

    Do it on the 2nd equation.

    If $\displaystyle x^2 + 2xy - 8y^2 = 0$ then

    $\displaystyle x = \frac{-2y \pm \sqrt{(2y)^2 - 4(1)(-8y^2)}}{2(1)}$

    $\displaystyle = \frac{-2y \pm \sqrt{36y^2}}{2}$

    $\displaystyle = \frac{-2y \pm 6y}{2}$

    $\displaystyle = \frac{4y}{2}\textrm{ or }\frac{-8y}{2}$

    $\displaystyle = 2y\textrm{ or }-4y$.


    Substitute these values into the first equation and you should be able to solve for y, and then back-substitute to find x.
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  3. #3
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    Quote Originally Posted by Phire View Post
    I read this problem on a site:

    $\displaystyle x^2-xy+y^2=21$
    $\displaystyle x^+2xy-8y^2=0$

    I believe it's not possible to isolate either x or y in either equation using standard methods, right? It was suggested to actually use the quadratic formula to solve in terms of a single variable. I also read that the second equation can be factored, but I can't seem to figure out how the factoring was done. Apparently it factors into:

    (x+4y)(x2y)
    $\displaystyle x^2 + 2xy - 8y^2 = 0
    $

    $\displaystyle (x+4y)(x-2y) = 0$

    $\displaystyle x = -4y$ or $\displaystyle x = 2y$

    go back to the first equation ...

    $\displaystyle x^2 - xy + y^2 = 21$

    sub in $\displaystyle -4y$ for $\displaystyle x$ ...

    $\displaystyle 16y^2 + 4y^2 + y^2 = 21$

    $\displaystyle 21y^2 = 21$

    $\displaystyle y = \pm 1$ ... $\displaystyle x = \mp 4$


    sub in $\displaystyle 2y$ for $\displaystyle x$ ...

    $\displaystyle 4y^2 - 2y^2 + y^2 = 21$

    $\displaystyle 3y^2 = 21$

    $\displaystyle y = \pm \sqrt{7}$ ... $\displaystyle x = \pm 2\sqrt{7}$
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  4. #4
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    Thanks, but can you guys tell me how you were able to factor the 2nd equation? I can't seem to do that. I tried:

    $\displaystyle x^2 + 2xy-8y^2$

    $\displaystyle x(x+2y)...$ Already don't know what to do

    So then I try:

    $\displaystyle x^2 +xy+xy-8y^2$

    $\displaystyle x(x+y)y(x+8y)$ No good...
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  5. #5
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    Quote Originally Posted by Phire View Post
    Thanks, but can you guys tell me how you were able to factor the 2nd equation? I can't seem to do that. I tried:

    $\displaystyle x^2 + 2xy-8y^2$

    $\displaystyle x(x+2y)...$ Already don't know what to do

    So then I try:

    $\displaystyle x^2 +xy+xy-8y^2$

    $\displaystyle x(x+y)y(x+8y)$ No good...
    Post #2 tells you how to find the roots. And if you know the roots then you know the factors. But getting the roots is all you need.

    Alternatively:

    $\displaystyle x^2 + 2xy - 8y^2 = (x^2 - 2xy) + (4xy - 8y^2) = x(x - 2y) + 4y(x - 2y) = (x + 4y)(x - 2y)$.
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