1. ## [SOLVED]Partial Fraction concept

if a equation is 3x/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

will the next step is

= A(x-3) + B(x-1)(x-2) + C(x-1)

or

= A(x-3) + B(x-1)(x-3) + C(x-1)

thank you.

2. ## Re: Partial Fraction concept

Hi gilagila!

I'm afraid neither is correct.

To add all fractions together, you need to make the denominators equal.
For the first term with A this means:
A / (x-1) = A(x-2)(x-3) / [(x-1)(x-2)(x-3)].

3. ## Re: Partial Fraction concept

Originally Posted by ILikeSerena
Hi gilagila!

I'm afraid neither is correct.

To add all fractions together, you need to make the denominators equal.
For the first term with A this means:
A / (x-1) = A(x-2)(x-3) / [(x-1)(x-2)(x-3)].
oops...just noticed that need to multiple all the demoninator
if a equation is 3x/(x-1)(x-2)(x-3) = A/(x-1) + B/(x-2) + C/(x-3)

it should look like this, correct me if worng

3x = A(x-2)(x-3)/(x-1) + B(x-1)(x-3)/(x-2) + C(x-1)(x-2)/(x-3)

4. ## Re: Partial Fraction concept

You have the nominators correct now, but you should leave out the denominators.

You have multiplied the left hand side with (x-1)(x-2)(x-3) to give you 3x.
The same should be done on the right hand side, meaning there would be no denominators left.

5. ## Re: Partial Fraction concept

Originally Posted by ILikeSerena
You have the nominators correct now, but you should leave out the denominators.

You have multiplied the left hand side with (x-1)(x-2)(x-3) to give you 3x.
The same should be done on the right hand side, meaning there would be no denominators left.
oops, forgot to put in the denominator at left

3x/(x-1)(x-2)(x-3) = A(x-2)(x-3)/(x-1) + B(x-1)(x-3)/(x-2) + C(x-1)(x-2)/(x-3)

then only look like this

3x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)

6. ## Re: Partial Fraction concept

Originally Posted by gilagila
oops, forgot to put in the denominator at left

3x/(x-1)(x-2)(x-3) = A(x-2)(x-3)/(x-1) + B(x-1)(x-3)/(x-2) + C(x-1)(x-2)/(x-3)

then only look like this

3x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)
Ah well, it should be:
3x/[(x-1)(x-2)(x-3)] = A(x-2)(x-3)/[(x-1)(x-2)(x-3)] + B(x-1)(x-3)/[(x-1)(x-2)(x-3)] + C(x-1)(x-2)/[(x-1)(x-2)(x-3)]

but nevermind, you're result is correct.