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

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Answer 1 · 2015-10-30T19:29:44

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

Writing #y=(1+3x)/(1-2x)#, we need to isolate #x#, which means to arrive to an expression of the form #x=f(y)#.

First of all, we multiply the whole expression by the denominator, obtaining

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

Expand the left term:

#y-2xy = 1+3x#

Bring every term involving #x# to the left, and everything else to the right:

#-3x-2xy = 1-y#

Factor our the #x# in the left term:

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

Divide everything by #(-3-2y)# to end:

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