How do you find the inverse of #f(x)=(2x+7)/(3x-1)#?

Let #y = f(x) = (2x+7)/(3x-1)#

#y = (2x+7)/(3x-1)#

I like to reduce the number of occurrences of #x# to one. Then it is clear how to isolate #x# on one side of the equation.

The #(2x ...) / (3x ...)# is going to make things a little messy -

unless we multiply through by #3# first, so let's do that:

Multiply both sides by #3# to get:

#3y = (3(2x+7))/(3x-1)#

#=(6x+21)/(3x-1)#

#=(6x-2+23)/(3x-1)#

#=(2(3x-1)+23)/(3x-1)#

#=2+23/(3x-1)#

Subtract #2# from both sides to get:

#3y - 2 = 23/(3x-1)#

Multiply both sides by #(3x-1)# to get:

#(3y-2)(3x-1) = 23#

Divide both sides by #(3y-2)# to get:

#3x-1 = 23/(3y-2)#

Add #1# to both sides to get:

#3x = 23/(3y-2)+1#

#=23/(3y-2)+(3y-2)/(3y-2)#

#=(23+(3y-2))/(3y-2)#

#=(3y+21)/(3y-2)#

#=(3(y+7))/(3y-2)#

Divide both sides by #3# to get:

#x = (y+7)/(3y-2)#

So #f^(-1)(y) = (y+7)/(3y-2)#

Replace #f(x)# with #y#.

#y=(2x+7)/(3x-1)#

Switch all occurrences of #x# with #y# and the #y# with #x#.

#x=(2y+7)/(3y-1)#

Solve for #y#. This may seem daunting, but just start by trying to eliminate fractions by cross-multiplying.

#x(3y-1)=2y+7#

Distribute on the left.

#3xy-x=2y+7#

Get all terms with #y# on one side of the equation. Move anything not including a #y# term to the other side of the equation.

#3xy-2y=x+7#

Factor a #y# term on the left.

#y(3x-2)=x+7#

Divide both sides by #(3x-2)# to solve for #y#.

#y=(x+7)/(3x-2)#

Since this is the inverse function, we can write is with #f^-1(x)# instead of #y#. All #f^-1(x)# means is the inverse of the function #f(x)#.

#f^-1(x)=(x+7)/(3x-2)#