# partial fractions question

• Sep 14th 2010, 10:06 AM
porge111
partial fractions question
"(i) decompose the following expression into partial fractions and

6x + 7/ 3x (x +1)
• Sep 14th 2010, 11:09 AM
e^(i*pi)
Quote:

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)
• Sep 14th 2010, 11:32 AM
masters
Quote:

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.
• Sep 14th 2010, 11:47 AM
wonderboy1953
Quote:

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).