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    How do you add/subtract 3 algebraic fractions?

    e.g. x + 3/(x-1) - 12/(x²+2x-3)

    Thank you...
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    i presume with that:
    x + 3/(x-1) - 12/(x²+2x-3)

    U take the (x-1) and divide it by (x^2+2x-3) to find its factor: ie. (x+3)


    and times (x+3) by both top and bottom to give u :

    x^2+6x+9 ```` 12
    ---------- - ----------
    (x^2+2x-3) (x^2+2x-3)



    then im sure you can do the rest....



    (i only presume this is right as im not sure i do the same topics as u, im on C1 and C2)
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    (Original post by Mushu)
    How do you add/subtract 3 algebraic fractions?

    e.g. x + 3/(x-1) - 12/(x²+2x-3)

    Thank you...
    Firstly, you want all fractions to be equilant, hence you have to have a common denominator, i.e. the denominator of all the fractions should be the same to add/subtract.

    Theres two ways of getting to the same denominator, by division of denominator or multiplication of denominator.
    Usually you would use multiplication.

    So you multiply all fractions by the denominators:

    x + 3/(x-1) - 12/(x²+2x-3)

    look at the first term, it is just x, so multiply by (x²+2x-3) and (x-1)

    x(x²+2x-3)(x-1)

    look at the second term, its already got our common denominator of (x-1) but not (x²+2x-3) so we multiply by (x²+2x-3)

    3(x²+2x-3)

    look at the third term, its already got our common denominator (x²+2x-3) so we just multiply by (x-1)

    - 12(x-1)

    Now put all these terms together to get

    x(x²+2x-3)(x-1) + 3(x²+2x-3) - 12(x-1)

    So whats missing, our common denominator which was ... (x-1)(x²+2x-3) which we multiplied every term by, so we have:

    x(x²+2x-3)(x-1) + 3(x²+2x-3) - 12(x-1)
    (x-1)(x²+2x-3)

    you have to now simplify by expnading:
    (x³+2x²-3x)(x-1)+3x²+6x-9 -12x+12
    (x-1)(x²+2x-3)

    You expand and simplify further to get:
    (x^4 + x^3 - 2x² -3x + 3)
    (x-1)(x²+2x-3)

    If required, this can be further simplified to:

    (x³+2x²-3)
    (x²+2x-3)

    then :
    x²+3x+3
    (x+3)
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    Thank you so much for that.

    I didnt understand this bit:


    (x^4 + x^3 - 2x² -3x + 3)
    (x-1)(x²+2x-3)

    If required, this can be further simplified to:

    (x³+2x²-3)
    (x²+2x-3)

    then :
    x²+3x+3
    (x+3)


    how did u further simplify the above?
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    (Original post by Mushu)
    Thank you so much for that.

    I didnt understand this bit:


    (x^4 + x^3 - 2x² -3x + 3)
    (x-1)(x²+2x-3)

    If required, this can be further simplified to:

    (x³+2x²-3)
    (x²+2x-3)

    then :
    x²+3x+3
    (x+3)


    how did u further simplify the above?
    So after expanding and simplifying we get
    (x^4 + x^3 - 2x² -3x + 3)
    (x-1)(x²+2x-3)

    Now the numerator has a factor of 1, so in the top, ifyou substitute x=1, you'll get 0, hence we can take out the factor of (x-1) in the top, by factorising to get:

    (x-1)(x³+2x²-3)
    (x-1)(x²+2x-3)

    Notice the (x-1) on both numerator and denominator can cancel to give:

    (x³+2x²-3)
    (x²+2x-3)

    If you further factorise the bottom to get (x-1)(x+3) and top to get (x-1)(x²+3x+3):

    (x-1)(x²+3x+3)
    (x-1)(x+3)

    Notice again the (x-1) can cancel giving:

    (x²+3x+3)
    (x+3)
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    Thank u vijay1
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    (Original post by Mushu)
    Thank u vijay1
    Anytime, no probs :rolleyes:
 
 
 
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