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Rational Expressions

  1. Sep 25, 2005 #1
    Right now in math class we are learning rational expressions. Since I am in an advanced math class, it seems like we learn a new lesson everyday. So if you don't understand something, you pretty much need to teach yourself. I don't really understand rational expressions, so can someone tell me if I am doing these two problems right(If I am doing wrong please tell me how to do them.)

    Problem: x/3 = 4/x+4

    Answer: 1) First I found a common denominator.
    3x+12(x/3) = 3x+12(4/x+4)
    2) So I got x+4 = x+3
    3) Then the answer would be x=-4, x=-3

    Am I right?
  2. jcsd
  3. Sep 25, 2005 #2


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    Watch out with the brackets, I suppose you mean

    [tex]\frac{x}{3} = \frac{4}{{x + 4}}[/tex]

    Now, multiply both sides with a common denominator to get rid of the denominators, so with for example [itex]3\left( {x + 4} \right)[/itex]
  4. Sep 25, 2005 #3


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    First off, you need to learn to use parentheses correctly. Whenever an addition or subtraction is supposed to happen before a multiplication or division, you need parentheses.

    For example:

    [tex]4/x+4 = \frac{4}{x} + 4[/tex]


    [tex]4/(x+4) = \frac{4}{x+4}[/tex]

    (3) is totally wrong -- if x+3 = x+4, then x=-4 and x=-3 are certainly not solutions.

    But... (2) is also totally wrong. You skipped a bunch of steps, so I don't know what you're doing wrong. Multiplication by 3x+12 was a reasonable idea, though. Could you post your work?
  5. Sep 25, 2005 #4
    See, I don't know my work. I have no idea what I am doing. This is as far as I got with the book. I am supposed to simplify and check
  6. Sep 25, 2005 #5


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    Well, try what I said. By multiplying both sides with 3, you lose the left denominator. Then, multiply both sides with x+4, this will get rid off the right denominator :smile:
  7. Sep 25, 2005 #6
    Another Question

    so would i then have 3(3) = x2 = 4 ?
  8. Sep 25, 2005 #7
    Never Mind, I think I'll just give up. It is way to hard for me to understand
  9. Sep 25, 2005 #8


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    I'll show you that first step. We multiply both sides with 3.
    At the LHS, the 3 will cancel out with the denominator, as we wanted.
    At the RHS, you can simplify it by multiplying it when the nominator.

    [tex]\frac{x}{3} = \frac{4}{{x + 4}} \Leftrightarrow 3 \cdot \frac{x}{3} = 3 \cdot \frac{4}{{x + 4}} \Leftrightarrow x = \frac{{12}}{{x + 4}}[/tex]

    Now, try losing the right denominator by multiplying both sides with (x+4) in the same way :smile:
  10. Sep 25, 2005 #9
    So would I do this?

    x+4 * x = 12/ x+4 * x+4

    Then I would get . . .

    x(x+4) = 12(x+4)


    If so, what do I do next?
  11. Sep 25, 2005 #10
    wait I made a mistake . . .

    It would be x(x=4) = 12
  12. Sep 25, 2005 #11
    I mean x(x+4) = 12

  13. Sep 25, 2005 #12


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    x/3 = 4/(x+4)


    x(x+4) = 12
  14. Sep 25, 2005 #13


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    Correct! :smile:

    Now bring everything to 1 side and you have a quadratic equation.
    Solve with the quadratic formula or by factoring.
  15. Sep 25, 2005 #14
    We are supposed to solve by factoring, so how do I do that?
  16. Sep 25, 2005 #15


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    So we have

    [tex]x\left( {x + 4} \right) = 12 \Leftrightarrow x^2 + 4x - 12 = 0[/tex]

    Personally, I would factor just by finding zeroes :smile:
    The divisors of the constant (-12) are 'possible candidates'...
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