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Math Help - Simplifying Algebraic Fractions

  1. #1
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    Simplifying Algebraic Fractions

    I've given this a try but I don't think I'm doing it right. If someone could explain what the best way to go about this that would be great. Thanks in advance.

    Question (simplify this fraction):
    (x/x-2)-(10/(x+3)(x-2))

    I tried to get a common denominator and expand out of the brackets but I have a feeling I'm going about this wrong.

    I did:

    (x^3+x^2-6x+10x-20)/(x^3+x^2-6x-2x^2-2x+12)
    (x^3+x^2+4x-20)/(x^3-x^2-8x+12)
    (4x-20)/(-8x+12)

    However I'm pretty sure I'm doing it wrong.

    Thanks!
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  2. #2
    MHF Contributor
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    Quote Originally Posted by hiyarearl View Post
    I've given this a try but I don't think I'm doing it right. If someone could explain what the best way to go about this that would be great. Thanks in advance.

    Question (simplify this fraction):
    (x/x-2)-(10/(x+3)(x-2))

    I tried to get a common denominator and expand out of the brackets but I have a feeling I'm going about this wrong.

    I did:

    (x^3+x^2-6x+10x-20)/(x^3+x^2-6x-2x^2-2x+12)
    (x^3+x^2+4x-20)/(x^3-x^2-8x+12)
    (4x-20)/(-8x+12)

    However I'm pretty sure I'm doing it wrong.

    Thanks!

    \frac{x}{x-2}-\frac{10}{(x+3)(x-2)}

     =\frac{x^2+3x-10}{(x+3)(x-2)}

     =\frac{(x+5)(x-2)}{(x+3)(x-2)}

    Do some canceling ..
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by hiyarearl View Post
    I've given this a try but I don't think I'm doing it right. If someone could explain what the best way to go about this that would be great. Thanks in advance.

    Question (simplify this fraction):
    (x/x-2)-(10/(x+3)(x-2))

    I tried to get a common denominator and expand out of the brackets but I have a feeling I'm going about this wrong.

    I did:

    (x^3+x^2-6x+10x-20)/(x^3+x^2-6x-2x^2-2x+12)
    (x^3+x^2+4x-20)/(x^3-x^2-8x+12)
    (4x-20)/(-8x+12)

    However I'm pretty sure I'm doing it wrong.

    Thanks!
    Mathaddict, you beat me, but I've added a step at the beginning that might be useful, so I'll post it anyway.

    \frac{x}{x-2}-\frac{10}{(x+3)(x-2)}

    Your common denominator will be (x+3)(x-2)

    \frac{x(x+3)-10}{(x+3)(x-2)}

    \frac{x^2+3x-10}{(x+3)(x-2)}

    \frac{(x+5)(x-2)}{(x+3)(x-2)}

    \frac{x+5}{x+3}
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  4. #4
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    Ah yes of course so:

    as when subtracting normal fractions we find the common denominator here effectively we have:

    \frac{x(x+3)}{(x-2)(x+3)}-\frac{10}{(x+3)(x-2)}

    When multiplied by (x+3) to achieve the common denominator.

    I understand now - thank you two very much for your help.


    Quote Originally Posted by masters View Post
    Mathaddict, you beat me, but I've added a step at the beginning that might be useful, so I'll post it anyway.

    \frac{x}{x-2}-\frac{10}{(x+3)(x-2)}

    Your common denominator will be (x+3)(x-2)

    \frac{x(x+3)-10}{(x+3)(x-2)}

    \frac{x^2+3x-10}{(x+3)(x-2)}

    \frac{(x+5)(x-2)}{(x+3)(x-2)}

    \frac{x+5}{x+3}
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