Algebra
posted by Audrey .
Which would be the inverse of this:
Let f(x)=(x+3)(2x5)
f^ (1) x = (5x  3)/ (2x + 1)
f^ (1) x = (5x + 3)/ (2x  1)

Could you check if there wasn't a typo, namely,
"Let f(x)=(x+3)/(2x5)... "
If that's the case, it's the second response.
To double check,
let g(x)=f^{1}(x)=(5x+3)/(2x1)
Evaluate
f(g(x))
= ((5x+3)/(2x1)+3)/((2(5x+3))/(2x1)5)
= x 
Let y = f(x)
y = (x + 3)/ (2x  5)
Switch x and y.
x = (y + 3)/ (2y  5)
Multiply both sides by (2y  5).
2xy  5x = y + 3
Subtract 3 from both sides.
2xy  5x  3 = y
Subtract 2xy from both sides.
5x  3 = 2xy + y
Factor the right side.
5x  3 = y (2x + 1)
Divide both sides by (2x + 1).
(5x  3)/ (2x + 1) = y
(5x  3)/ (2x + 1) = y
Now we replace y with the inverse function notation: f^ (1) x.
f^ (1) x = (5x + 3) / (2x  1)
This my work to reflect the answer. Is it correct? Thanks! 
Yes, the calculation is correct.
Note: You may not have noticed that you omitted the division sign in the initial post. 
Thanks for pointing that out!

You're welcome!