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+0 Answers


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

13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details