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 .
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).