1. ## partial fractions question

"(i) decompose the following expression into partial fractions and

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

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

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

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

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

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

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:

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

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

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

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

B is easy to find after that.

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

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).