So I just got back from a Calculus test and I have some trouble figuring out one of the questions, it states:
"Calculate the inverse of the function $y=\dfrac{2x+1}{3-4x}$."
What first came into my mind was to eliminate the denominator somehow. But I quickly realized that it would be rather difficult science the numerator is also of degree one. I then got desperate and just tried to move the denominator and got this expression witch did not help me at all. $$y(3-4x) = 2x+1$$ After several more equally stupid moves I gave up. Even now with the help of Wolfram Alpha I can not figure out how to solve it. Can anyone explain?
$$y=\frac{2x+1}{3-4x}=-\frac{2x+1}{4x-3}=-\frac{1}{2}-\frac{5}{8x-6}$$ So $$y+\frac{1}{2}=-\frac{5}{8x-6} \iff \frac{2}{2y+1}=\frac{6-8x}{5}$$
So $$\frac{10}{2y+1}=6-8x \iff x=\frac{3y-1}{4y+2}$$ A simpler solution is possible using matrices, I believe. The answer is thus $$f^{-1}(x)=\frac{3x-1}{4x+2}$$