1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse func.

  1. Aug 1, 2011 #1
    1. The problem statement, all variables and given/known data


    y = 3
    ___
    2(x-3)

    2. Relevant equations



    3. The attempt at a solution

    Ive tried several times, I cant get the steps down. They say use reverse bedmass, and I dont understand.

    Here's what i did.

    y= 3/2(x-3)

    x= 3/2(y-3)

    x= 3/2y-6
    x+6=3/2y

    x+6
    ___ = 3
    2

    Which is already wrong, as you can see. Can someone please tell me the step so i can stop being stuck on this for 2 hours? =)
     
  2. jcsd
  3. Aug 1, 2011 #2
    "x= 3/2y-6"

    You were correct up to here. If it helps, write parenthesis in the denominator.

    x=[itex]\frac{3}{(2y-6)}[/itex]

    What you did was just cancelled out the 6 in the denominator which you cannot do. Our objective is to get y by itself. Do that by multiplying 2y-6 to both sides. That should get rid of the fraction and leave you with.

    x(2y-6)=3

    Try getting y by itself from here.
     
  4. Aug 1, 2011 #3
    Ok...

    x = 3 / (2y-6)

    x(2y-6) 3
    _______ = _____
    (2y-6) (2y-6)

    2xy-6x 3
    ______=_____
    2y-6 2y-6

    Pretty sure im already wrong. .. .
     
  5. Aug 1, 2011 #4
    When I mean multiply both sides you don't do it like this:

    [itex]\frac{2y-6}{2y-6}[/itex] x=[itex]\frac{3}{2y-6}[/itex][itex]\frac{2y-6}{2y-6}[/itex]

    [itex]\uparrow\uparrow\uparrow\uparrow[/itex] That is a no no.

    You do it like this:

    x([itex]\frac{2y-6}{1}[/itex])=[itex]\frac{3}{2y-6}[/itex]([itex]\frac{2y-6}{1}[/itex])

    Notice how the (2y-6) in the right side would cancel out. That would leave us with this.

    x(2y-6)=3

    Remember this: If you do anything to one side, you MUST do it to the other side of the equation.
     
  6. Aug 1, 2011 #5
    So... the x still factors into 2y-6 on both sides..? your eq would give 2xy-6x/x on the left side..
     
  7. Aug 1, 2011 #6
    What do you mean "factors on both sides)? My equation would not leave you with a fraction (though the answer would be a fraction). Try distributing the x term to (2y-6). Then just solve for y.

    *Update: I see what you did. No that is wrong. Remember your algebraic properties:

    a*[itex]\frac{b}{c}[/itex]=[itex]\frac{ab}{c}[/itex]
     
  8. Aug 1, 2011 #7
    So, i did it like this..


    2x(y-3) = 3

    2x*y -6x =3

    2x*y = 6x+3

    y= 6x+3 /2x

    y= 3x +3/2x

    The answer is 2/3x +3 , however. Why is the x on the bottom?
     
  9. Aug 1, 2011 #8
    Starting from here:
    "2xy = 6x+3"

    We would divide 2x to get y by itself.

    [itex]\frac{2xy}{2x}[/itex]=[itex]\frac{6x+3}{2x}[/itex]

    I see you split the fraction on the right side to divide 6x/2x, however you forgot to cancel out the x. However you cannot divide 3/(2x), so that is why 2x is in the denominator. The answer would look like this.

    y=[itex]\frac{3}{2x}[/itex]+3


    I do not see how the right answer is: "2/3x +3" (assuming this is (2/(3x))+3) the 3 is supposed to be in the numerator and 2x is supposed to be in the denominator.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inverse func.
  1. Inverse Functions (Replies: 2)

  2. Inverse function (Replies: 3)

  3. Inverse proportion (Replies: 4)

Loading...