Provide all details to decompose the fraction (x-4)/(x-2)(x-3) in partial fractions?

We'll write the model of decomposing such a fraction into
partial irreducible fractions;

(x-4)/(x-2)(x-3) = A/(x-2) +
B/(x-3)

To compute the sum of the 2 fractions, they must
have the same denominator. Since it is not the case, we'll create the same denominator
to both elementary fractions.

(x-4) = A(x-3) +
B(x-2)

We'll remove the
brackets:

x-4 = Ax - 3A + Bx -
2B

We'll combine like
terms:

x-4 = x(A+B) - 3A
-2B

Comparing, we'll
get:

According to this:

A+B=1
(1)

- 3A -2B = -4 (2)

We' ll
multiply A+B=1 by 2, to eliminate the unknown B.

2A + 2B=2
(3)

Now, we'll add (3) to
(2):

2A + 2B - 3A -2B =
2-4

We'll eliminate like
terms:

-A = -2

A=2 =>
A+B=1 <=> 2+B=1 => B=1-2 =>
B=-1

The result of
decomposition into partial fractions is: (x-4)/(x-2)(x-3) = 2/(x-2) -
1/(x-3).