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# Algebraic Fractions watch

1. How do you add/subtract 3 algebraic fractions?

e.g. x + 3/(x-1) - 12/(x²+2x-3)

Thank you...
2. 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)
3. (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)
4. 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?
5. (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)
6. Thank u vijay1
7. (Original post by Mushu)
Thank u vijay1
Anytime, no probs

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