Question

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

Accepted Answer

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 - 3B  = -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)

Answer

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.

Math

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

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