Find a formula for the inverse function?

If you want to find an inverse function of

#y=(3x+2)/(7x-6)#

first you can multiply both sides by the denominator:

#y(7x-6)=3x+2# ##

#7xy-6y=3x+2#

Now we can rearrange the equation so that the new unknown #x# is on the left side:

#7xy-3x=6y+2#

#x*(7y-3)=6y+2#

#x=(6y+2)/(7y+3)#

Now we can change the variables back. This causes that #x# becomes the argument and #y# the function value (the usual naming convention)

The inverse function is:

#y=(6x+2)/(7x+3)#

Given: #h(x) = (3x+2)/(7x-6)#

Substitute #h^-1(x)# for every instance of #x#:

#h(h^-1(x)) = (3(h^-1(x))+2)/(7(h^-1(x))-6)#

One of the two parts of the definition of an inverse is that #h(h^-1(x)) = x#, therefore, the left side becomes x:

#x = (3(h^-1(x))+2)/(7(h^-1(x))-6)#

Multiply both sides by #(7(h^-1(x))-6)#:

#x(7(h^-1(x))-6) = 3(h^-1(x))+2#

Distribute the x:

#7x(h^-1(x))-6x = 3(h^-1(x))+2#

Add 6x to both sides:

#7x(h^-1(x)) = 3(h^-1(x)) + 6x+2#

Subtract #3(h^-1(x))# from both sides:

#7x(h^-1(x)) - 3(h^-1(x)) = 6x+2#

Factor out #h^-1(x)#:

#(7x - 3)h^-1(x) = 6x+2#

Divide both sides by #(7x - 3)#:

#h^-1(x) = (6x+2)/(7x - 3)#

To verify that this is truly the inverse, we must test that #h(h^-1(x)) = x# and #h^-1(h(x)) = x#.

Start verification with #h(h^-1(x))#:

#h(h^-1(x)) = (3((6x+2)/(7x - 3))+2)/(7((6x+2)/(7x - 3))-6)#

Multiply the right side by 1 in the form of #(7x - 3)/(7x - 3)#:

#h(h^-1(x)) = (7x - 3)/(7x - 3)(3((6x+2)/(7x - 3))+2)/(7((6x+2)/(7x - 3))-6)#

This has the effect of remove the embedded fractions and distributing the denominator over the constants:

#h(h^-1(x)) = (3(6x+2)+2(7x - 3))/(7(6x+2)-6(7x - 3))#

Use the distributive property to perform 4 multiplications:

#h(h^-1(x)) = (18x+6+14x - 6)/(42x+14-42x + 18)#

Combine like terms:

#h(h^-1(x)) = (32x)/(32)#

#h(h^-1(x)) = x#

Finish the verification with #h^-1(h(x))#:

#h^-1(h(x))= (6((3x+2)/(7x-6))+2)/(7((3x+2)/(7x-6)) - 3)#

Multiply by 1 in the form of #(7x-6)/(7x-6)#:

#h^-1(h(x))= (7x-6)/(7x-6)(6((3x+2)/(7x-6))+2)/(7((3x+2)/(7x-6)) - 3)#

Eliminate the embedded fractions and distribute the numerator among the constants:

#h^-1(h(x))= (6(3x+2)+2(7x-6))/(7(3x+2) - 3(7x-6))#

Use the distributive property to perform 4 multiplications:

#h^-1(h(x))= (18x+12+14x-12)/(21x+14 - 21x+18)#

Combine like terms:

#h^-1(h(x))= (32x)/(32)#

#h^-1(h(x))= x#

It is verified that #h^-1(x) = (6x+2)/(7x - 3)#