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I hope you'll forgive such a basic question but I'm relearning maths after a long layoff. Working on the OCW MIT 18.01 Calculus course, but I'm getting hung up on manipulating fractional equations. One question asks you to find the inverse function of:

$$f(x) = \frac{x-1}{2x+3}$$

The approach I took (which the answer sheet confirms) is to solve this equation for $x$.

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

Specifically the answer sheet says to Crossmultiply and solve for x, getting:

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

I cross multiply, which gives $x-1 = y(2x +3)$. I can see that this is not too far away from the answer, but I'm now stuck on what to do next.

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    $\begingroup$ Expand the rhs and group the $x$ together. $\endgroup$ – Claude Leibovici Aug 17 '16 at 10:18
  • $\begingroup$ Googling the term "Distributive Property" may be productive. [e.g. $a(b+c) = ab + ac$] $\endgroup$ – John Joy Aug 18 '16 at 18:51
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Simply take all the terms with $x$ on it to LHS side and all other to RHS. Then, try to factoring x out on LHS, and divivide RHS by the factor left to get a solution for $x$. This results in:

$x- 1 = y(2x+3) \rightarrow x - 2yx = 3y +1$

$(1 - 2y)x = (3y + 1) \rightarrow x = \frac{3y +1}{1-2y}$

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Your result $$x−1=y(2x+3) $$ is equivalent to $$x−1=2xy+3y $$ which is $$x-2xy=3y+1$$ $$x(1-2y)=3y+1$$ so $$x=\frac{3y+1}{1-2y}$$

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$$x-1=y(2x+3)\iff x-1-2yx=3y\iff (1-2y)x=3y+1$$

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$x-1=2xy+3y \rightarrow x-2xy=3y+1 \rightarrow x(1-2y) = 3y+1 \rightarrow x=\frac{3y+1}{1-2y}$

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Expand the term at the Right hand side so you have: $$x-1=2xy+3y$$ Now isolate the terms with $y$ and you have $$x-2xy=3y+1$$ Then: $$x(1-2y)=3y+1$$ And from here dividing by $(1-2y)$ both sides you have the thesis.

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