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

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The partial fractions can be found in a very easy way using the following method.

(3x-2)/(x-3)(x+1), the roots of the denominator are x = 3 and x = -1.

Representing as partial fractions we get the form A/(x - 3) + B(x + 1)

Let's find A, ignore the term (x - 3) and substitute the root x = 3 for all the other terms.

This gives us (3*3 - 2)/(3 + 1) = (9 - 2)/4 = 7/4

Similarly to find B, eliminate (x + 1) and substitute the root x = -1 in the other terms, we get (3*-1 - 2)/(-1 - 3) = 5/4

**Therefore the partial fractions are (7/4)/(x - 3) + (5/4)/(x + 1)**

(3x-2)/(x-3)(x+1)

so now for simplifying fractions like this we break this in 2 parts

(3x-2)/(x-3)(x+1)=a/(x-3) + b/(x+1)

or a(x+1)+b(x-3)=(3x-2)

now put x=3

a(3+1)+0=7

4a=7 or a=7/4

also now put x=-1

b(-4)=(-3-2)

b=5/4

so now put these all above these are your partial fractions.

We'll decompose the rational function in partial fractions:

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

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-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)

Comparing 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

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

**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.**