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Thread: Solve the equation

  1. #1
    Senior Member Mukilab's Avatar
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    Solve the equation

    $\displaystyle y-2x=3$
    $\displaystyle x^2+y^2=18$

    Following Mr Fantastic's previous advice I did

    $\displaystyle -9=4x^2-y^2$

    Sub into the second equation
    $\displaystyle
    9=4x^2+x^2$

    $\displaystyle \pm\sqrt{\frac{9}{5}}=x$
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  2. #2
    Senior Member Mukilab's Avatar
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    Did it again and got a different answer

    y^2=9+2x^2
    -y^2=x^2-18

    Add them together to eliminate y

    0=3x^2-9
    9=3x^2
    $\displaystyle \pm\sqrt{3}=x$
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  3. #3
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    Quote Originally Posted by Mukilab View Post
    $\displaystyle y-2x=3$
    $\displaystyle x^2+y^2=18$

    Following Mr Fantastic's previous advice I did

    $\displaystyle -9=4x^2-y^2$
    No $\displaystyle (y-2x)^2 \neq y^2-4x^2$

    $\displaystyle y=2x+3$

    $\displaystyle y^2=(2x+3)^2$

    Expand and substitute into the 2nd equation to get a quadratic with x as unknown
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  4. #4
    Senior Member Mukilab's Avatar
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    y=4x^2+12x+9

    Substituted
    0=5x^2+12x-9
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  5. #5
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    Yes
    Now you need to solve 0=5x^2+12x-9
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  6. #6
    Senior Member Mukilab's Avatar
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    yes but I get a horrendous equation along the lines of

    $\displaystyle \frac{-12\pm\sqrt{324}}{10}$

    This is a no calculator paper.

    Maybe I should just use that?
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  7. #7
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    Quote Originally Posted by Mukilab View Post
    yes but I get a horrendous equation along the lines of

    $\displaystyle \frac{-12\pm\sqrt{324}}{10}$

    This is a no calculator paper.

    Maybe I should just use that?
    Are you kidding us? SQRT(324) = 18
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