Write as partial fractions . (5x-4)/(x-2)(x+3)

We'll decompose the fraction into elementary irreducible
fractions:

(5x-4)/(x-2)(x+3) = A/(x-2) +
B/(x+3)

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

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

LCD = (x-2)(x+3)

We'll multiply
both sides by (x-2)(x+3) and the expression will become:

(5x-4) =
A(x+3) + B(x-2)

We'll remove the
brackets:

5x - 4 = Ax + 3A + Bx -
2B

We'll combine like terms form the right
side:

5x - 4 = x(A+B) +
(3A-2B)

Comparing we'll get:

5 =
A+B

-4 = 3A-2B

We'll use the symmetric
property:

A+B = 5 (1)

3A-2B = -4
(2)

We'll multiply (1) by 2:

2A+2B = 10
(3)

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

2A+2B+3A - 2B
= 10-4

We'll eliminate like terms:

5A =
6

We'll divide by 5:

A =
6/5

We'll substitute A in (1):

A+B =
5

6/5 + B = 5

We'll subtract 6/5 both
sides:

B = 5 - 6/5 => B = (25-6)/5 => B =
19/5

The given fraction written as partial fractions
is: (5x-4)/(x-2)(x+3) = 6/5(x-2) + 19/5(x+3), where
6/5(x-2) and
19/5(x+3) are partial
fractions.