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Math Help - Problem with equation.

  1. #1
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    Problem with equation.

    Find value for A and B:

    A/2x+3 + B/x-4 = x+1/ 2x^2 -5x-12

    I found LCD

    A(x-4) + B(2x+3) / (2x+3)(x-4) = x+1/ (2x+3)(x-4)

    What do I do next???
    Last edited by mr fantastic; July 20th 2010 at 07:34 PM. Reason: Edited title.
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  2. #2
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    Quote Originally Posted by julia78 View Post
    Find value for A and B:

    A/2x+3 + B/x-4 = x+1/ 2x^2 -5x-12

    I found LCD

    A(x-4) + B(2x+3) / (2x+3)(x-4) = x+1/ (2x+3)(x-4)

    What do I do next???
    Learn to use parentheses! This should be

    A/(2x+3) + B/(x-4) = (x+1)/ (2x^2 -5x-12)

    and

    [A(x-4) + B(2x+3)] / [(2x+3)(x-4)] = (x+1)/ [(2x+3)(x-4)]

    I haven't checked to make sure your computations are correct, but in terms of general approach you're doing it right, now expand the numerator to

    Ax - 4A + 2Bx + 3B = (A + 2B) x + (3B - 4A)

    Set up system of linear equations

    A + 2B = 1

    3B - 4A = 1

    Now solve for A and B.
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  3. #3
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    Given that your problem is
    A/(2x+3) + B/(x-4) = (x+1)/ (2x^2 -5x-12)= (x+1)/(2x- 3)(x+ 4)
    you can multiply both sides by (2x+3)(x- 4) to get
    (x+4)A+ (2x- 3)B= x+ 1

    getting immediately what you get by adding the fractions on the left and the "setting the numerator equal" as undefined does.

    If that is to be true for all x, then there are several ways you can solve for A and B. One is to go ahead and multiply those out and combine "common terms" on the left: Ax+ 4A+ 2Bx- 3B= (A+ 2B)x+ (4A- 3B)= x+ 1 and then set "corresponding" coefficients equal, as undefined does, to get two equations to solve for A and B. Another way is to take x to be any two numbers to get two equations. If you take x= -4 and x= 3/2 you get particularly easy equations.
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