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Math Help - partial fractions question

  1. #1
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    partial fractions question

    "(i) decompose the following expression into partial fractions and
    (ii) check your answer by multiplying them out


    6x + 7/ 3x (x +1)
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  2. #2
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    Quote Originally Posted by porge111 View Post
    "(i) decompose the following expression into partial fractions and
    (ii) check your answer by multiplying them out


    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)
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    \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.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    \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).
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