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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?

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    multiply by denominator $3-4x$ both sides, collect terms of x together and divide by the factor that x is multiipled by – 2017-02-01

5 Answers 5

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$$y=\frac{2x+1}{3-4x}$$

Chase the denominator away.

$$y(3-4x)=2x+1$$

Expand.

$$3y-4xy=2x+1$$

Move all $x$ to the right-hand side.

$$3y-1=2x+4xy$$

Factor.

$$3y-1=(2+4y)x$$

Divide.

$$\frac{3y-1}{2+4y}=x$$

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Good start. You have: $$y(3-4x)=2x+1$$ You can expand this to obtain: $$3y-4xy=2x+1$$ And you can gather all the $x$ terms together: $$-4xy-2x=1-3y$$ $$4xy+2x=3y-1$$ Now, you can factor out the $x$ on the left hand side.

Can you continue? If not, feel free to ask.

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    Jesus,, I was at the right track after all. How embarrassing. Ty for pointing it out! – 2017-02-01
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$$y=\frac{2x+1}{3-4x}=-\frac{2x+1}{4x-3}=-\frac{1}{2}-\frac{5}{8x-6}$$ 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 $$f^{-1}(x)=\frac{3x-1}{4x+2}$$

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    Ty! will check your answer as solved when the time expires. Yes I know matrices, if you have the time it would be awesome if you could also show that solution! – 2017-02-01
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    @Salviati I'm not sure either, but it involves the fact that $$ \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)^{-1} = \frac{1}{ad-bc} \left( \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right), ad-bc \ne 0 $$ Damn if I can remember the rest of it. I thought I knew, but I forgot. – 2017-02-01
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    Right! I remember now, think I can solve it on my own now that way. Ty! – 2017-02-01
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After your first step:

$y(3-4x) = 2x+1\quad | -2x$
$3y-4xy-2x = 1\quad | -3y$
$x(-4y-2) = 1-3y\quad | : (-4y-2)$
$x = \frac{1-3y}{-4y-2} = \frac{3y-1}{4y+2}$

From the first to second line, I also multiplied out on the left hand side. From the second to third line, I factored out x.

This can of course be solved quicker, I tried to do it in steps that are easy to follow.

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We have to express $y $ in terms of $x $. That is our ultimate goal. We thus get on rearranging, $$3y-4xy =2x+1 \Rightarrow 3y-1 =x (2+4y)$$ $$x = \frac {3y-1}{2+4y} $$ This is our inverse function. It can also be expressed as $f^{-1}(x) = \frac {3x-1}{2+4x} $. Hope it helps.