partial fractions question

• September 14th 2010, 11:06 AM
porge111
partial fractions question
"(i) decompose the following expression into partial fractions and

6x + 7/ 3x (x +1)
• September 14th 2010, 12:09 PM
e^(i*pi)
Quote:

Originally Posted by porge111
"(i) decompose the following expression into partial fractions and

6x + 7/ 3x (x +1)

$\dfrac{6x+7}{3x(x+1)} = \dfrac{A}{3x} + \dfrac{B}{x+1}$ which, after algebraic manipulation, is equal to $6x+7 = A(x+1) + B(3x)$

Use subsitution to find the values of A and B (-1 and 0 would be a good start)
• September 14th 2010, 12:32 PM
masters
Quote:

Originally Posted by e^(i*pi)
$\dfrac{6x+7}{3x(x+1)} = \dfrac{A}{3x} + \dfrac{B}{x+1}$ which, after algebraic manipulation, is equal to

$6x+7 = A(x+1) + B(3x)$

Use subsitution to find the values of A and B (-1 and 0 would be a good start)

Or...you could multiply everything through, group the x terms and the constant terms:

$6x+7=Ax+A(1)+3Bx$

$6(x)+7(1)=(A+3B)x+A(1)$

For the two sides to be equal, the coefficients of the two polynomials must be equal. So you "equate the coefficients":

$A+3B=6$ and $A=7$

B is easy to find after that.
• September 14th 2010, 12:47 PM
wonderboy1953
Quote:

Originally Posted by e^(i*pi)
$\dfrac{6x+7}{3x(x+1)} = \dfrac{A}{3x} + \dfrac{B}{x+1}$ which, after algebraic manipulation, is equal to $6x+7 = A(x+1) + B(3x)$

Use subsitution to find the values of A and B (-1 and 0 would be a good start)

There's a formula way to solving as well (since the RHS denominators are linear).