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Thread: IM stuck on algebra!

  1. #1
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    Exclamation IM stuck on algebra!

    I have homework due tomorrow which I as usual forgot about and left till the last minute...its algebra which im hopeless at, any help would be much appreciated.

    1) -x < 4

    2) 12x > 2x^2

    3) x - 3 2x - 1
    ____ = _____

    x - 1 4x + 5

    4) 3(2x + 5)
    _________ = 6x + 1
    4

    5) find the x coordinates of the points of intersection of the two parabolas
    y = 3x^2 + 5x - 6 and y = -2x^2 - 3x + 18



    Any help wud be much appreciated
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  2. #2
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    Quote Originally Posted by Brownhash View Post
    I have homework due tomorrow which I as usual forgot about and left till the last minute...its algebra which im hopeless at, any help would be much appreciated.

    1) -x < 4

    2) 12x > 2x^2

    3) x - 3 2x - 1
    ____ = _____

    x - 1 4x + 5

    4) 3(2x + 5)
    _________ = 6x + 1
    4

    5) find the x coordinates of the points of intersection of the two parabolas
    y = 3x^2 + 5x - 6 and y = -2x^2 - 3x + 18



    Any help wud be much appreciated
    I'll post the solutions. If you can't follow them, say which steps don't make sense and I'll explain them...

    1. $\displaystyle -x<4$ gives $\displaystyle 0<x+4$ which means $\displaystyle -4<x$ or $\displaystyle x>-4$

    2. $\displaystyle 12x > 2x^2$
    $\displaystyle 6x > x^2$
    $\displaystyle 6>x$ if $\displaystyle x\geq 0$ else $\displaystyle 6<x$ if $\displaystyle x<0$

    3. $\displaystyle \frac{x-3}{x-1}=\frac{2x-1}{4x+5}$
    $\displaystyle (x-3)(4x+5)=(2x-1)(x-1)$
    $\displaystyle 4x^2+5x-12x-15=2x^2-2x-x+1$
    $\displaystyle 4x^2-7x-15=2x^2-3x+1$
    $\displaystyle 2x^2-4x-16=0$
    $\displaystyle 2(x^2-2x-8)=0$
    $\displaystyle x^2-2x-8=0$
    $\displaystyle (x-4)(x+2)=0$
    $\displaystyle x-4 = 0$ or $\displaystyle x+2=0$
    $\displaystyle x=4$ or $\displaystyle x=-2$

    4. $\displaystyle \frac{3(2x+5)}{4}=6x+1$
    $\displaystyle 3(2x+5)=4(6x+1)$
    $\displaystyle 6x+30=24x+4$
    $\displaystyle 30-4=24x-6x$
    $\displaystyle 26=18x$
    $\displaystyle \frac{26}{18}=x$
    $\displaystyle \frac{13}{9}=x$

    5. Intersection points between functions occur when the functions are equal. Since they are both equal to $\displaystyle y$, they are equal to each other...

    $\displaystyle 3x^2 + 5x - 6 = -2x^2 - 3x + 18$
    $\displaystyle 5x^2+8x-24=0$

    $\displaystyle x=\frac{-8\pm\sqrt{8^2-4\times5\times(-24)}}{2\times5}$ using the Quadratic Formula
    $\displaystyle =\frac{-8\pm\sqrt{64+480}}{10}=\frac{-8\pm\sqrt{544}}{10}=\frac{-8\pm4\sqrt{34}}{10}=\frac{-4\pm2\sqrt{34}}{5}$

    So $\displaystyle x=\frac{-4+2\sqrt{34}}{5}$ or $\displaystyle x=\frac{-4-2\sqrt{34}}{5}$
    Last edited by Prove It; Aug 19th 2008 at 03:04 AM.
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  3. #3
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    Thankyou

    I understand them now thanks for the help.
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