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Math Help - simplify check!

  1. #1
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    simplify check!

    simplify

    \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}

    my answer

     <br />
\frac {(x-3)(x-2)}{(x-2)^2}<br />

    correct? thanks
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    simplify

    \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}

    my answer

     <br />
\frac {(x-3)(x-2)}{(x-2)^2}<br />

    correct? thanks
    No. It would make it easier for us to check your work if you posted some working.

    CB
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    No. It would make it easier for us to check your work if you posted some working.

    CB
    sorry mate. ok here it is:

    \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}

    \frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}

    then cross multiply then mulitply denominators and start cancelling out.

     <br />
\frac {(x-3)(x-2)}{(x-1)^2}<br />
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  4. #4
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    Quote Originally Posted by jvignacio View Post
    sorry mate. ok here it is:

    \frac {x-3}{x^2-3x+2} - \frac {x-2}{x^2-4x+3}

    \frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}

    then cross multiply then mulitply denominators and start cancelling out.

     <br />
\frac {(x-3)(x-2)}{(x-1)^2}<br />
    I don't understand what you mean, but standard method of of adding fractions gives:

    \frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}= \frac{(x-3)(x-3)-(x-2)(x-2)}{(x-3)(x-2)(x-1)}

    and simplify

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    I don't understand what you mean, but standard method of of adding fractions gives:

    \frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}= \frac{(x-3)(x-3)-(x-2)(x-2)}{(x-3)(x-2)(x-1)}

    and simplify

    CB
    do i expand brackets to simplify or just cancel out the terms like top (x-3) and bottom (x-3) etc?
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    do i expand brackets to simplify or just cancel out the terms like top (x-3) and bottom (x-3) etc?
    How would you simplify \frac{3 \times 5 - 2}{3} .... ? Cancel the 3 .... \frac{\not{3} \times 5 - 2}{\not{3}} = \frac{5 - 2}{1} = 3 .... ?
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    How would you simplify \frac{3 \times 5 - 2}{3} .... ? Cancel the 3 .... \frac{\not{3} \times 5 - 2}{\not{3}} = \frac{5 - 2}{1} = 3 .... ?

     <br />
\frac{(x-3)-(x-2)}{(x-1)}<br />

    this okay?
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  8. #8
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    Quote Originally Posted by jvignacio View Post
     <br />
\frac{(x-3)-(x-2)}{(x-1)}<br />

    this okay?
    NO!

    Quote Originally Posted by mr fantastic View Post
    How would you simplify \frac{3 \times 5 - 2}{3} .... ? Cancel the 3 .... \frac{\not{3} \times 5 - 2}{\not{3}} = \frac{5 - 2}{1} = 3 .... ?
    Do you really think that \frac{3 \times 5 - 2}{3} is equal to 3!! I was trying to make a point. The point is that you can't do this. What you cancel has to be common to all of the numerator and all of the denominator before it can be cancelled!
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  9. #9
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    Quote Originally Posted by jvignacio View Post
    do i expand brackets to simplify or just cancel out the terms like top (x-3) and bottom (x-3) etc?

    You can only cancel something which is a factor of both the top and bottom of such an expressions and x-3 is not a factor of the top.

    What you do is expand the brackets at the top, collect like powers and factorise. Then if any of the factors on top are the same as a factor on the bottom you can canncel such facors.

    CB
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  10. #10
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    Quote Originally Posted by CaptainBlack View Post
    You can only cancel something which is a factor of both the top and bottom of such an expressions and x-3 is not a factor of the top.

    What you do is expand the brackets at the top, collect like powers and factorise. Then if any of the factors on top are the same as a factor on the bottom you can canncel such facors.

    CB
    for the top, i expanded it out and got \frac {-2x-5}{(x-3)(x-2)(x-1)}
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    I don't understand what you mean, but standard method of of adding fractions gives:



    \frac {x-3}{(x-2)(x-1)} - \frac {x-2}{(x-3)(x-1)}= \frac{(x-3)(x-3)-(x-2)(x-2)}{(x-3)(x-2)(x-1)}



    and simplify



    CB
    Quote Originally Posted by jvignacio View Post
    for the top, i expanded it out and got \frac {-2x-5}{(x-3)(x-2)(x-1)}
    The numerator should be -2x + 5.
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