Expert: Scotto - 2/20/2009

Question
QUESTION: y=3x-1/2x

Attempt to solution:

y=(6x^2-1)/(2x)

x= (6y^2-1)/(2y)

2yx=6y^2-1

2yx-6y^2=-1


2y(x-3y)=-1

2y=-1/(x-3y)

( I get stuck her what do l do with the y in the denominator ? )

ANSWER: Here is how to find the inverse function.
Instead of expressing y in terms of x,
we will express x in terms of y.

The function starts out as y = 3x - 1/2x.

Our goal is to get x times some quantity that with no x in it.
Once this is done, we will divide both sides by that quantity.
It is very helpful to eliminate and fractions at first.

First, I will eliminate the denominator by multiplying by 2x,
giving 2xy = 3x – 1.  Now the fractions are gone.  
In the next step, we will put the x terms on the same side.

Now add 1 to both sides and subtract 2xy from both sides,
so we now have 1 = 3x – 2xy.  Note that the right has x terms
and the left has terms that are not (in this case, 1).

Once here, lets factor out the x on the right side, giving
1 = x(3-2y).  In this way, we have x alone.

Now that we have x all by itself, we can divide by the coefficient,
giving x = 1/(3-2y).  That is the inverse, x in terms of y.

If you want to reverse x and y, you would get y = 1/(3-2x).
I'm not sure why this would be done, but I just gave it to you.


---------- FOLLOW-UP ----------

QUESTION: So the inverse is y = 1/(3-2x) ? How did you solve for "y" ?

ANSWER: Multiply the entire equation by 2x to get 2xy = 3x – 1.

Add 1 to both sides, giving 2xy + 1 = 3x.

Subtract 2xy from both sides, giving 1 = 3x - 2xy.

Now that all of the terms with x are grouped together, factor out x.
The result is 1 = x(3-2y).

Divide both side by (3-2y), giving 1/(3-2y) = x.

Write the equation properly and say x = g(y) = 1/(3-2y).

==============================================================

Here is a summary:

2xy = 3x - 1; add 1, giving

2xy + 1 = 3x; subtract 2xy, giving

1 = 3x - 2xy; factor out x, giving

1 = x(3-2y); divide by 3-2y, giving

1/(3-2y) = x; x we can call g(y) at this point,

so be reversing the equation x = g(y) = 1/(3-2y).

Read the steps, one at a time.  You got it now?

To try it out, put in what y is { which is 1/(3-2x) } into
x=1/(3-2y), and you should get x = x after everything has
been worked out.


---------- FOLLOW-UP ----------

QUESTION: Hello Scotto,

The  function starts out 3x-1/(2x)so if we multiply both sides by two aren't we supposed to get :2x*(3x-1/(2x))=6x^2-1, not 3x-1 ?

Answer
Soory about that, but several people incorrectly put parenthesis in
the problem.  I have seen that most people turn parenthises into
spaces.  For example, 3 + 4*5 = 3 + 20, but 3+4*5 = 7*5 = 35.
This is not correct, but I have read many questions that way
and seemed to have got them right.  Thank you for putting them
in the problem correctly.

We has a (3x-1)/2x, so when we multiply by 2x, they cancel.
Don't forget the parenthesis when putting in addition and subtraction.  They're very important in that multiplication
and division come first if they are not there.  That we very
good that you put them around the 2x in the denominator.

Now if they weren't suppose to be there to start with,
the function must have been y = 3x - 1/2x, so then we get
y = (6x²-1)/2x, which then turns into 2xy = (6x²-1).
Putting this in proper format gives
6x² - 2yx - 1 = 0.  This gives us, by use of the quadratic equation,
x = (-b ±√(b²-4ac))/(2a)
x = (2y±√(4y²+24))/12 which can be reduced by cancelling 2, giving
x = (y±√(y²+6))/6.  Note that in the √, we divided by 2² to cancel.