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Going from x(y) to y(x), please help

  1. Nov 9, 2012 #1
    going from x(y) to y(x), please help :)

    If we have a function y=1/2x+½ how come if we isolate x as a function of y on a calculator we get x=2y-1??

    i get the 2*y but not the -1
    What algebra is needed to get from y(x) to x(y) in this case?
     
  2. jcsd
  3. Nov 9, 2012 #2
    Re: going from x(y) to y(x), please help :)

    Multiply the original equation by 2 and then subtract 1 from both sides.
    y=½x+½
    2y=x+1
    x=2y-1
     
  4. Nov 9, 2012 #3
    Re: going from x(y) to y(x), please help :)

    Woaa, i see :D
    So normal algebra by isolating x instead of y, does not work when there are two variables?
     
  5. Nov 9, 2012 #4
    Re: going from x(y) to y(x), please help :)

    How come i would get x=y/½ -½ if i use the regular add, subtract, multiply and divide operations?
     
  6. Nov 9, 2012 #5

    SammyS

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    Re: going from x(y) to y(x), please help :)

    According to the rules for order of operations,

    y = 1/2x+½

    is equivalent to

    y = (1/2)x + ½ .
     
  7. Nov 9, 2012 #6
    Re: going from x(y) to y(x), please help :)

    Hmm. I don't see the difference between the two steps (except for the brackets?)
     
  8. Nov 9, 2012 #7
    Re: going from x(y) to y(x), please help :)

    What SammyS did is a perfectly normal operation.

    You could also do it like this, keeping in mind that you'll get the same thing and that the other way is even easier.

    y = (1/2)x + 1/2

    y - 1/2 = (1/2)x

    Now multiply both sides by two (note that this is the exact same thing as "dividing both sides by 1/2")

    2y-1 = x

    To show why dividing by 1/2 is the same thing:

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

    Note that 1/2 divided by 1/2 is 1, so y divided by 1/2 is 2. Dividing by a fraction is equivalent to multiplying by the reciprocal of the fraction.
     
  9. Nov 9, 2012 #8
    Re: going from x(y) to y(x), please help :)

    Ahh Now i see!
    Thank you for the help, now I can continue on my double integral of 3-D shapes (I often forget the simple algebra so i'm trying to strengthen it by doing it all by hand) :)
     
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