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Easy algebra

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm so embarrassed that I cannot get this problem but here it is:

    2.07 = 1+B/(1-B)

    The answer is supposed to be .349 I think. But I can't figure out the steps involved in how to get it.



    3. The attempt at a solution

    I don't see why you can't just subtract 1 from both sides then B would equal -1.07
     
  2. jcsd
  3. Jun 4, 2012 #2
    It's supposed to be 2.07 = (1+B)/(1-B) if the answer is .349.
     
  4. Jun 4, 2012 #3
    right
     
  5. Jun 4, 2012 #4
    do you know how to solve the problem?
     
  6. Jun 4, 2012 #5
    2.07 = (1+b)/(1-b) => 2.07(1-b)=1+b => 2.07-2.07b = 1 + b => 1.07 = 3.07b => b=.349
     
  7. Jun 4, 2012 #6
    Thanks, got it. I'm glad that embarrassing problem is over.
     
  8. Jun 4, 2012 #7
    It happens, don't be embarrassed.
     
  9. Jun 4, 2012 #8
    But you have to find a different way. That is one of the biggest mistakes people do is to divide by 1-b. How do you know that B-1≠0?? You have to first find the domain and state that B≠1.
     
  10. Jun 4, 2012 #9

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, you don't. If B= -1, the right side of the equation does not exist so there is no equation to solve. Of course, the final result for B is NOT 0.
     
  11. Jun 4, 2012 #10
    Yes well that is obvious, but they have to state that. That's what I've been taught.
     
  12. Jun 4, 2012 #11

    Mark44

    Staff: Mentor

    I'm aware that the problem really is 2.07 = (1 + B)/(1 - B).

    You are apparently laboring under the false impression that you can subtract 1 from the right side to get -B.

    I'm guessing that your thought process went something like this:

    $$\frac{1 + B}{1 - B} = \frac{1}{1} + \frac{B}{-B} = 1 - B$$

    None of the expressions above is equal to any of the others. For the first equality, that is not how fractions work. For the second inequality, B/(-B) = -1, not -B.
     
  13. Jun 4, 2012 #12

    Ray Vickson

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    Science Advisor
    Homework Helper

    If you mean
    [tex] 2.07 = 1 + \frac{B}{1-B}[/tex] the solution is [itex] B = 107/207 \doteq 0.5169[/itex]. In this case you can, indeed, subtract 1 from both sides to get
    [tex] 1.07 = \frac{B}{1-B}.[/tex]
    If you mean
    [tex] 2.07 = \frac{1+B}{1-B}[/tex] the solution is [itex] B = 107/307 \doteq 0.3485. [/itex] In this case, subtracting 1 from both sides does not give you anything very useful.

    I guess you mean the second one, in which case you should USE BRACKETS, like this: 2.07 = (B+1)/(B-1).

    RGV
     
  14. Jun 4, 2012 #13
    There is also a trick to doing problems having this kind of mathematical form. You do (num-denom) divided by (num+denom) to both sides of the equation. In this problem you get

    B = (2.07 - 1)/(2.07 + 1) = 1.07 / 3.07 = 0.349
     
  15. Jun 4, 2012 #14

    Mark44

    Staff: Mentor

    Can you explain why this trick works? In particular, what you would need to do to go from this equation -- 2.07 = (1 + B)/(1 - B)

    to this equation --
    B = (2.07 - 1)/(2.07 + 1)
     
  16. Jun 4, 2012 #15
    Thanks, that clears things up for me
     
  17. Jun 6, 2012 #16
    If NL/DL = NR/DR

    Then you can prove that

    (c1 NL + c2DL)/(c3 NL + c4DL) = (c1 NR + c2DR)/(c3 NR + c4DR)

    where c1, c2, c3, and c4 are constants. See if you can prove this. It isn't very hard. IOW, whatever linear combinations of numerator and denominator you form on the left hand side of the equation, you form the exact same linear combinations of numerator and denominator on the right hand side.

    In our problem,

    NL = (1 + B)
    DL = (1 - B)
    NR = 2.07
    DR = 1

    so

    (NL - DL)/((NL + DL) = (NR - DR)/((NR + DR)

    or

    ((1 + B) - (1 - B))/((1 + B) + (1 - B)) = (2.07 - 1)/(2.07 +1)

    or

    (2B)/2 = B = 1.07 / 3.07
     
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