How do you find the inverse of the function (2x+1)/(3x-2)

$$y=\dfrac{2x+1}{3x-2}$$

First off note that this isn't defined at x=2/3 but other than that it is one to one.

$$(3x-2)y=2x+1$$

$$3xy - 2y = 2x+1$$

$$x(3y-2)=1+2y$$

$$x=\dfrac {1+2y}{3y-2}$$

now just replace x by f^-1(x) and y by x to obtain

$$f^{-1}(x)=\dfrac{2x+1}{3x-2}~~\forall x \neq \dfrac 2 3$$