Write as an addition of 2 irreducible fractions (3x-2)/(x-3)(x+1)

We'll write the ratio as a sum or difference of 2
irreducible ratios in this way:

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

We notice that the numerator of the original
ratio is a linear function and the denominator is a quadratic
function.

The irreducible ratios from the right side have
as numerators constant functions and as denominators, linear
functions.

We'll calculate LCD of the 2 ratios from the
right side.

The LCD is the same with the denominator from
the left side.

LCD =
(x-3)(x+1)

The expression will
become:

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

We'll remove the
brackets:

3x - 2 = Ax + A + Bx -
3B

We'll combine like terms form the right
side:

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

If the expressions from both sides are equal, then
the correspondent coefficients are equals:

3 =
A+B

-2 = A - 3B

We'll use the
symmetric property:

A+B = 3
(1)

A - 3B = -2 (2)

We'll
multiply (1) by 3:

3A+3B = 9
(3)

We'll add (3) to
(2):

3A+3B+A - 3B = 9-2

We'll
eliminate like terms:

4A =
7

We'll divide by 4:

A =
7/4

We'll substitute A in
(1):

A+B = 3

7/4 + B =
3

B = 3 - 7/4

B =
(12-7)/4

B =
5/4

(3x-2)/(x-3)(x+1) = 74/(x-3) +
5/4(x+1)