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Isolating a variable when variable is in both sides of equation

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    y=(2-x)/(5x)

    Solving for x=f(y)

    I can get as far as 5xy=(2-x)

    Tried 5xy=-1(x+2)

    All attempts at solving this have given me the wrong answer, and I end up with x on both sides of the equation.

    Stuck and not sure what to do.

    Correct answer is x=2/(5y+1) , but I don't know how to get there.

    Thanks for any help!
     
  2. jcsd
  3. Nov 12, 2012 #2

    symbolipoint

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    Start with what you know and go step by step.

    From the given equation:
    1. Multiply both members by 5x.
    2. Add the additive inverse of you-know-what to both sides.
    3. Can you then see to combine two terms of the variable, x ?
    4. ...And if you did, then you can finish, finding x as a formula of y.
     
  4. Nov 12, 2012 #3

    HallsofIvy

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    This the same as 5xy= 2- x, you don't need the parentheses. If you want x only on the left, then add x to both sides.

     
  5. Nov 12, 2012 #4
    The best way to solve this problem would be to separate the division part, that is to say 2/(5x)- x/(5x) = 2/(5x) - 1/5, then multiply all by 5x to get
    5xy=2-x. Now add x to both sides to get 5xy+x = 2. Now do you notice something? We can factor this to x(5y+1)=2, divide by 5y+1 to get x=2/(5y+1)



    Bonaparte
     
  6. Nov 12, 2012 #5
    Aha! Of course...thats what I was missing! I was so close just forgot to factor the x out.

    Thanks very much - this problem was really getting under my skin!
     
  7. Nov 12, 2012 #6

    Mentallic

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    Separating the division is both an unnecessary step and makes things harder.
     
  8. Nov 12, 2012 #7

    SammyS

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    An alternate method:
    [itex]\displaystyle y=\frac{2-x}{5x}[/itex]

    [itex]\displaystyle 5y=\frac{2-x}{x}[/itex]

    [itex]\displaystyle 5y=\frac{2}{x}-1[/itex]

    [itex]\displaystyle 5y+1=\frac{2}{x}[/itex]

    [itex]\displaystyle x=\frac{2}{5y+1}[/itex]​
     
  9. Nov 12, 2012 #8

    symbolipoint

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    NOT necessarily, in case it helps the o.p. to understand.
     
  10. Nov 12, 2012 #9

    SammyS

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    Good point.

    In this case, breaking up the rational expression (separating the division) leaves only one term with x in it -- a fairly uncomplicated term at that. This makes it rather easy to solve for x.
     
  11. Nov 12, 2012 #10

    Mentallic

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    I wouldn't think to explain to a student that's working with functions and factorizing that (a+b)/c is the same as a/c+b/c, but I suppose I shouldn't make that assumption.
     
  12. Nov 12, 2012 #11
    Mentallic, not that you too seperated the division part, just later.
    That was from (2-x)/x to 2/x-x/x. You are right that it is not necessary, but when he gets to much more complicated problems, I personally find it helpful to seperate. Where you seperated you could have multiplied by x both sides and continue, you are perfectly right that it is not "necessary".

    Bonaparte
     
  13. Nov 13, 2012 #12

    symbolipoint

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    Actually, when working the problem forward again myself, Mentallic is right, that splitting the rational expression into a sum (or a difference) should not be necessary. The problem can be solved in four steps (slightly differently than #7, SammyS). Myself, I found no use in splitting the expression into sum or difference.
     
  14. Nov 13, 2012 #13
    I'm tired of repeating myself, of course it is not necessary, I disagree it makes things harder however. I will rather spend a extra 30 seconds to separate then have the O.P. not understand and explain. Please use your time more efficiently and help others.

    Thanks, Bonparte
     
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