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

Asked on by kikiri

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to write (3x-2)/(x-3)(x+1) as the sum of fractions.

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

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

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

equate the numeric terms and those with x

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

Now we have A + B = 3 and A - 3BĀ  = -2

=> A + B - A + 3B = 3 + 2

=> 4B = 5

=> B = 5/4

A = 3 - B = 3 - 5/4 = 7/4

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll decompose theĀ 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.

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