Finding the inverse function of $f(x)=\frac{3x+1}{2-7x}$

Question

Find the inverse function of: $$f(x)=\frac{3x+1}{2-7x}$$ I did the question and when I checked my answer with the key it was wrong, can someone please show me how to properly do this problem? I followed all the steps and the answer I came up with was: $$\frac{2x-1}{7x+3}$$ but the correct answer is supposed to be: $$\frac{-(1-2x)}{7x-3}$$

Answer 1

$$y=\frac{3x+1}{2-7x}$$

$$2y-7xy=3x+1$$

$$3x+7xy=2y-1$$

$$x(3+7y)=2y-1$$

This means the inverse of $f(x)$ can be written

$$g(y)= \frac{2y-1}{7y+3}$$

I think you are correct.

Answer 2

Say $y=f(x)$ Therefore the inverse function is $x=f^{-1}(y)$ $$y=\frac{3x+1}{2-7x}$$ or, $$y(2-7x)=(3x+1)$$ or, $$2y-7xy=3x+1$$ or, $$2y-1=x(3+7y)$$ or, $$x=\frac{2y-1}{3+7y}$$ or, $$f^{-1}(y)=\frac{2y-1}{3+7y}$$ or, $$f^{-1}(x)=\frac{2x-1}{3+7x}$$

Hence $y=\frac{2x-1}{3+7x}$ is the required inverse function. So your answer is correct.