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

neela | Student

Let (3x-2/(x-3)(x+1) = A/(x-3)+B/(x+1) .The right side is the sum of two irreduces fractions We decide A and B, as follows:

Multiply both sides by  (x-3) and (x+1) . Then we get:

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

Put x = 3 in (1):

3*3-2  = A(3+1) +B*0

7  = 4A.

x = 7/4.

Now put x=-1 in (1):

3(-1)-2 = A*0+B(-1-3)

-3-2 = -4B.

4B = 5

B = 5/4 = 5/4.

Therefore A = 7/4 and B = 5/2

(3x-2)/(x-3)(x+1) = 7/(4(x-3)) + 5/(4(x+1))

giorgiana1976 | Student

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)

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)