I need a bit of help trying to find the inverse of this function:

y=(7x+3)/(5x+7) ==> x=(7y^{-1}+3)/(5y^{-1}+7)

It'd be preferred if you also walked through solving the problem, as that is the main issue I have, I just don't know what to do to isolate the y^{-1} .

Guest Sep 5, 2017

#1**+2 **

From x=(7y^{-1}+3)/(5y^{-1}+7) multiply both sides by (5y^{-1}+7) to get;

x(5y^{-1}+7) = 7y^{-1}+3 Now distribute the x in to the parentheses on the left hand side:

5xy^{-1} + 7x = 7y^{-1}+3 Subtract 5xy^{-1} from both sides:

7x = 7y^{-1}+3 - 5xy^{-1} Subtract 3 from both sides:

7x - 3 = 7y^{-1} - 5xy^{-1} Factor the right hand side:

7x - 3 = (7 - 5x)y^{-1} Divide both sides by 7 - 5x:

(7x - 3)/(7 - 5x) = y^{-1}

or y^{-1} = (7x - 3)/(7 - 5x)

.

(Note that the original function is not defined at x = -7/5, so there is no inverse at this point; and the inverse function isn't defined at x = +7/5).

Alan
Sep 5, 2017