Write the fraction (3x-2)/(x-3)(x+1) as partial fractions.

We'll decompose the rational function in partial
quotients:

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

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)

We'll multiply both sides by (x-3)(x+1) and 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)

We'll compare and we'll
get:

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

We'll subtract 7/4 both
sides:

B = 3 - 7/4 => B = (12-7)/4 => B =
5/4

The fraction (3x-2)/(x-3)(x+1) = 7/4(x-3)
+ 5/4(x+1), where 7/4(x-3) and 5/4(x+1) are partial
fractions.