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Math Help - Rational Problems

  1. #31
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    m doesnt = -4/5
    yes


    then what am i supposed to cancel?
    all the (j + 3)'s cancel. and one of the (j + 4)'s in the top cancels with the (j + 4) in the bottom
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  2. #32
    Ari
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    Quote Originally Posted by Jhevon View Post
    yes

    all the (j + 3)'s cancel. and one of the (j + 4)'s in the top cancels with the (j + 4) in the bottom
    how?

    <br />
    " alt="\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)}
    " />

    it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top...
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  3. #33
    Ari
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    Quote Originally Posted by Jhevon View Post
    yes

    all the (j + 3)'s cancel. and one of the (j + 4)'s in the top cancels with the (j + 4) in the bottom
    how?

    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)}<br />


    it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top...
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  4. #34
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    how?

    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)}<br />


    it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top...
    there is a (j + 3) in the top of the first fraction as well as the top of the last fraction

     \frac {{\color{red} (j+3)}(j+4)}{{\color {red}(j+3)(j+3)}} \cdot \frac {{\color{red}(j+3)}(j-3)}{(j-4)(j+4)}
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  5. #35
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ari View Post
    how?

    <br />
\frac {(j+3)(j+4)}{(j+3)(j+3)} \cdot \frac {(j+3)(j-3)}{(j-4)(j+4)}<br />


    it doesnt make sense beacuse there are 2 (j+3) at the bottom and only 1 (j+3) at the top...
    the (j - 3) in the top of the second fraction should be (j + 4)
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