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Math Help - simplify by using a factor of 1 when possible.

  1. #1
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    simplify by using a factor of 1 when possible.

    [(7y+2)/(4y^2-12y-20)]+[(8)/(4y^2-24y-20)]
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  2. #2
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    Quote Originally Posted by mkpg24 View Post
    [(7y+2)/(4y^2-12y-20)]+[(8)/(4y^2-24y-20)]
    \frac{7y+2}{4y^2-12y-20}+\frac{8}{4y^2-24y-20}

    Not much to simplify here. And what do you mean by "use a factor of 1"?

    \frac{7y+2}{4(y^2-3y-5)}+\frac{8}{4(y^2-6y-5)}


    \frac{7y+2}{4(y^2-3y-5)}+\frac{2}{y^2-6y-5}

    I believe that's about it.
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  3. #3
    Moo
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    Hello,

    Multiplying by 1 means multiply by \frac aa

    While you're here :
    \frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}, do this :

    =\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}

    =\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}

    =\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}

    =\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}

    but we used the multiplication by 1 lol!
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  4. #4
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    Quote Originally Posted by Moo View Post
    Hello,

    Multiplying by 1 means multiply by \frac aa

    While you're here :
    \frac{7y+2}{4({\color{blue}y^2-3y-5})}+\frac{8}{4({\color{red}y^2-6y-5})}, do this :

    =\frac{7y+2}{4({\color{blue}y^2-3y-5})} \cdot \frac{\color{red}y^2-6y-5}{\color{red}y^2-6y-5}+\frac{8}{4({\color{red}y^2-6y-5})} \cdot \frac{\color{blue}y^2-3y-5}{\color{blue}y^2-3y-5}

    =\frac{(7y+2)(y^2-6y-5)+8(y^2-3y-5)}{4({\color{red}y^2-6y-5}) ({\color{blue}y^2-3y-5})}

    =\frac{7y^3-42y^2-35y+2y^2-12y-10+8y^2-24y-40)}{4(y^2-6y-5) (y^2-3y-5)}

    =\frac{7y^3-32y^2-71y-50}{4(y^2-6y-5) (y^2-3y-5)}

    but we used the multiplication by 1 lol!
    I guess I was confused about the word "simplify"
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