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Difficult rational expressions thinking question

  1. Mar 22, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the Values of B and C given this:
    [tex]
    \frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}
    [/tex]
    2. Relevant equations

    The equation given:[tex]
    \frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}
    [/tex]


    3. The attempt at a solution

    My attempt:
    [tex]
    \frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}
    [/tex]
    [tex]
    \frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B(2x+1)+C(7x-3)}{(2x+1)(7x-3)}
    [/tex]
    [tex]
    3x-18=B(2x+1)+C(7x-3)
    [/tex]
    At this point I'm stuck and I think something I did was wrong.
    I'd appreciate any assistance that could be provided!
     
    Last edited: Mar 22, 2013
  2. jcsd
  3. Mar 22, 2013 #2

    SammyS

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    If your last equation
    [itex]
    3x-18=B(2x+1)+C(7x-3)
    [/itex]​
    is true for all values of x, then the initial rational equation
    [itex]\displaystyle \frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}
    [/itex]​
    will be true for all x except 3/7 and -1/2 .

    There are at least a couple of ways to solve the identity
    [itex]
    3x-18=B(2x+1)+C(7x-3)
    [/itex]​
    for B & C.

    One is to pick a couple of values for x, which will give you two equations in the two unknowns, B & C. One such value is x = -1/2 .

    Another method is to equate the coefficient of x on the left with the coefficient of x on the right; also equate the constant term on the left with the constant term on the right.
     
  4. Mar 22, 2013 #3
    Hi, Thank You so much for your response. I was just wondering if you could clarify what you meant by this:
    I apologize my math vocab is weak.
     
  5. Mar 22, 2013 #4
    He means that you can expand 3x−18=B(2x+1)+C(7x−3) as 3x-18=2Bx+B+7Cx-3C. Cleaning this up a bit, we get 3x-18=(2B+7C)x+(B-3C). Two polynomials are equal only if their corresponding coefficients are equal, we get 3x=2B+7C and -18=B-3C. This is a system of equations which you should be able to sove.
     
  6. Mar 22, 2013 #5
    But don't I have 3 variables still?
     
  7. Mar 22, 2013 #6
    I am sorry, there should be no x there. If 3x=(2B+7C)x, then 3=2B+7C. I apologize for not using latex, I am just beginning to learn as I am new to these forums.
     
  8. Mar 22, 2013 #7

    SammyS

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    If you expand the right side & collect terms, you get
    [itex]\displaystyle (2B+7C)x+(B-3C)\,,\ [/itex] right?​

    So, the coefficient of x on the left, which is 3 must equal the coefficient of x on the right, which is 2B+7C . Similarly, the constant term on the left must equal the constant term on the right.

    These must be equal because, the equation, [itex]\displaystyle \ 3x-18=B(2x+1)+C(7x-3)\ [/itex] is true for all possible values of x, thus this equation is an identity.
     
  9. Mar 22, 2013 #8
    Hi! Thank you both for responding.

    I'm with you guys so far until here:
    [itex]3x−18=(2B+7C)x+(B−3C){}[/itex]

    I don't quite understand why [itex]3x[/itex] must only equal [itex](2B+7C)x[/itex].

    After I try and solve this is what I get. Is everything I did okay because I don't know and method of verifying if this is right (not from a textbook)?

    [itex]-18=B-3C[/itex]
    [itex]B=-18+3C[/itex]

    [itex]3x=(2B+7C)x[/itex]
    [itex]3=2B+7C[/itex]
    sub in [itex]B=-18+3C[/itex]
    [itex]3=2(-18+3C)+7C[/itex]
    [itex]3=-36+6C+7C[/itex]
    [itex]39=13C[/itex]
    [itex]C=3[/itex]

    [itex]B=-18+3C[/itex]
    [itex]B=-18+3(3)[/itex]
    [itex]B=-18+9[/itex]
    [itex]B=-9[/itex]

    ∴B=-9 and C=3
     
  10. Mar 22, 2013 #9

    Dick

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    Yes, you solved it right. But [tex]\frac{(3x-18)}{(2x+1)(7x-3)}= \frac{B}{2x+1} + \frac{C}{7x-3}[/tex] actually leads to the equation [tex]3x-18=B(7x-3)+C(2x+1)[/tex] Which is not what you originally posted. The B and C are backwards.

    As to your first question Ax+B=Cx+D for all values of x means A=C and B=D. Can you show that?
     
  11. Mar 22, 2013 #10
    Ah, I plugged in the values of B and C and the expressions on both sides were equal. And that quote clarified quite a lot.

    Thank you so much to all of you for the help!
     
  12. Mar 23, 2013 #11

    SammyS

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    Sorry. I didn't check the details.
     
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