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Simplify an expression

  1. Sep 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Simplify (1/3 + 1/3y)/(1/y + 1/3)

    2. Relevant equations
    (1/3 + 1/3y)/(1/y + 1/3) = x

    In this case I am using x to represent the simplified expression
    3. The attempt at a solution
    (1/3 + 1/3 * 1/y)/(1/y+1/3) = (2/3 * 1/y)/(1/y + 1/3) = (2/3y)/(1/y+1/3) = (2/3y)/(3/3y + y/3y) =
    2/3y * 3y/(3+y) = 6y/(9 + 3y^2)

    I obviously did something wrong but I don't see how. Every step I did correctly.
     
  2. jcsd
  3. Sep 3, 2015 #2

    mfb

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    Be careful with writing fractions. What does "1/3y" mean? Is it (1/3) y (the way a computer would interpret it) or 1/(3y)? Written on paper it is obvious, written in forum posts it is not. Based on your first step it has the second meaning here.

    You can use LaTeX for nice formulas:
    $$\frac{\frac{1}{3} + \frac{1}{3y}}{\frac{1}{y} + \frac{1}{3}}$$

    What did you do in the numerator in this step?
     
  4. Sep 3, 2015 #3
    I added like terms to get 2/3 * 1/y.
     
  5. Sep 3, 2015 #4

    andrewkirk

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    Why did you think that ##\frac{1}{3}+\frac{1}{3}\frac{1}{y}## is equal to ##\frac{2}{3}\frac{1}{y}##?

    Is that true if ##y=2##?
     
  6. Sep 3, 2015 #5
    Well I was taught in situations where you have multiple terms of the same type that you combine them. I thought I could do that in this step.
     
  7. Sep 3, 2015 #6

    andrewkirk

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    That is a very vague statement of what is a very specific rule, which is bound to lead to errors. What is 'like' about the two terms is that they both have a ##\frac{1}{3}## in them, not that they both have a ##\frac{1}{y}## in them (because they don't).

    What you do is factorise the sum of the 'like' terms, dividing each term by the common factor, then write the sum of the divided terms in brackets and then multiplying the bracket by the common factor.

    eg ##x+y+3x^2## has two terms with a common factor of ##x##, so we can write it as ##x(1+3x)+y##.

    In your numerator the common factor is ##\frac{1}{3}##. What happens if you apply that principle to this line?
     
  8. Sep 3, 2015 #7
    I get 1/3(1 + 1/y)
     
  9. Sep 3, 2015 #8

    mfb

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    That is correct.
     
  10. Sep 3, 2015 #9

    andrewkirk

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    OK. Now redo your calcs using that corrected numerator.

    Note however that the last step in your above calc is also incorrect: 2/3y * 3y/(3+y) = 6y/(9 + 3y^2)
     
  11. Sep 3, 2015 #10
    So if I did it correctly it would be like this:

    (1/3 + 1/3 * 1/y)/(1/y+1/3) = 1/3(1 + 1/y)/(1/y + 1/3) = (y+1)/(y+3)
     
  12. Sep 3, 2015 #11

    mfb

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    Now something went wrong with the denominator.
     
  13. Sep 3, 2015 #12
    Why? I simplified it as much as possible using Mathway and (y+1)/(y+3) was the solution it gave me.
     
  14. Sep 3, 2015 #13

    HallsofIvy

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    A good first step, rather than adding the fractions, would to be to multiply both numerator and denominator by 3y.
     
  15. Sep 3, 2015 #14

    andrewkirk

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    That solution is correct. The thing that was wrong in the denominator is in the last line of the OP, and you didn't re-make the error when you re-did the calc.
     
  16. Sep 4, 2015 #15

    mfb

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    Ah, I missed the additional step that was done there. Okay, ignore my previous post.
     
  17. Sep 4, 2015 #16

    SammyS

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    That looks good to me.
     
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