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

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