# Finding the inverse

1. Oct 16, 2011

### bael

1. The problem statement, all variables and given/known data
find a formula for f-1 when f is defined by
f(x) = 3x+5/x-4

2. Relevant equations

3. The attempt at a solution
Here is what I've done so far:
First I switch the x and the y so I get
x=3y+5/y-4
x(y-4)=3y+5
xy-4x=3y+5
xy=3y+4x+5
xy-3y=4x+5
y-3y=4x+5/x
-2y=4+5
I don't know what else to do. No matter what I try I always end up deleting a variable.

Last edited: Oct 16, 2011
2. Oct 16, 2011

### theclock54

Okay so first you switch the x's and y's.
Then you'll have:

x= (3y+5)/(y-4)

Let's multiply by y-4

which leaves... xy-4x = 3y+5

when you want to solve for a variable and there's two of them, you want to get them on the same side and factor it out.

so now what you do is subtract 5, (move it to the left) and move the subtract xy (move to right)

so you end up with

5-4x = 3y-xy

now lets factor out y

5-4x = y(3-x)

now we divide by (3-x)

3. Oct 16, 2011

### theclock54

*sorry I forgot to put the negative sign on 5...it should be (-5-4x)/(3-x) = y

4. Oct 16, 2011

### eumyang

@bael: First off, put in parentheses next time. What you wrote looks like this:
$$f(x) = 3x+\frac{5}{x}-4$$
The bolded is where your problem lies. You can't divide both sides by x like that. If you were to divide both sides by x, this would have been the result:
$xy-3y=4x+5$
$\frac{xy-3y}{x}=\frac{4x+5}{x}$
$\frac{xy}{x}-\frac{3y}{x}=\frac{4x}{x}+\frac{5}{x}$
$y-\frac{3y}{x}=4+\frac{5}{x}$
This is not the way to go.

What theclock54 posted isn't wrong (now that the correction was made), but from this step:
xy-4x = 3y+5

... I would instead add 4x to both sides and subtract 3y from both sides. I prefer to have the x terms before the constant terms in the numerator & denominator.
xy - 3y = 4x + 5

Then factor out the y, and divide both sides by (x - 3):
$y(x - 3) = 4x + 5$
$y = \frac{4x + 5}{x - 3}$

5. Oct 16, 2011

### bael

Thanks a lot, I understand it now.