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I have solved it as follows:

$\displaystyle x=\frac{5y-3}{2y+1}$

$5y-3=2xy+x$

$5y-2xy=3+x$

$y(5-2x)=3+x$

$\displaystyle y=\frac{3+x}{5-2x}.$

$\displaystyle {f^{-1}}(x)=\frac{3+x}{5-2x}$

That is my answer. But On the screen, there appeared another answer due to a slightly different method of isolating the second stroke:

$2xy+x=5y-3$

$2xy-5y=-x-3$

$y(2x-5)=-3-x$

$\displaystyle y=\frac{-3-x}{2x-5}$

$\displaystyle {f^{-1}}(x)=\frac{-3-x}{2x-5}$.

As you can see there are two answers now. So which is the real answer?

user36339
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    They're the same answer. The only difference is the numerator and denominator of the second one are both multiplied by $-1$ – Brenton Jun 07 '17 at 21:06
  • Please use Mathjax to format math text. This link may be useful: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Dave Jun 07 '17 at 21:07
  • So you mean that the first answer (which is mine) can be multiplied by -1, yes? But there is not any track of multiplication by -1 in the second answer. – user36339 Jun 07 '17 at 21:15
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    @YoqubjonJo'rayev Brenton said both numerator and denominator multiplied by $-1$, not the whole fraction multiplied by $-1$. – peterwhy Jun 07 '17 at 21:18
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    both are correct, both are equivalent. When you multiply your answer by $1 = \dfrac{-1}{-1}$, you'll see for yourself. – amWhy Jun 07 '17 at 21:33
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    I cannot understand why this question was downvoted; There is no requirement that a new user with rep =1, to know from the start how to use mathjax. Not a reason for a downvote. (We can expect more, in terms of formatting, when an asker has been around for at least a little while. – amWhy Jun 07 '17 at 21:39

2 Answers2

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$\displaystyle y = \frac{5x-3}{2x+1}$

$\displaystyle (2x+1)y=5x-3$

$\displaystyle 2xy+y=5x-3$

$\displaystyle x(2y-5)=-y-3$

$\displaystyle x(5-2y)=y+3$

$\displaystyle \boxed{x=\frac{y+3}{5-2y}}$

Hope this helped. Have a nice day :D

Both answers are correct. Multiply by $\displaystyle \frac{-1}{-1}$ to get from one answer to the other.

Saketh Malyala
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    Why didn't you change y to x because here the inverse must be found? – user36339 Jun 07 '17 at 21:23
  • What I gave was just a hint, finding x in terms of y. Once you have found that, you can simply swap x and y. – Saketh Malyala Jun 07 '17 at 21:24
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    That's not a "hint" (I refer to your answer) it's a full-blown answer with all the workings, but the failure to note that the inverse function, is $$f^{-1}(x)= \frac{x+3}{5-2x}$$ – amWhy Jun 07 '17 at 21:29
  • amWhy, I also got this answer. Is this answer above similar to the {f^{-1}}=(-3-x)/(2x-5) (this is the teacher's answer)? If yes, why? Because they don't seem to be so. – user36339 Jun 07 '17 at 21:36
  • Also, Saketh, the OP's work has been correct all along; (and the OP one-upped you by knowing how to express an inverse function correctly. – amWhy Jun 07 '17 at 21:39
  • Yoqubjon The answer would be fine if the answerer would have replaced $y$ with $x$ and designated it as the inverse function $f^{-1}(x)$. Your answer is indeed correct, and more correct than this answer. What I list in the comments (the formula f^{-1} is correct, as is the teacher's (see my comment below your question.) $$f^{-1}(x)= \frac{x+3}{5-2x} = \frac{-1}{-1}\cdot \frac{x+3}{5-2x} = \frac{-x-3}{2x-5}$$ – amWhy Jun 07 '17 at 21:45
  • Thanks a lot for help! Now I have fathomed it! – user36339 Jun 07 '17 at 21:49
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This method is not applicable always , but sometimes , it helps very much when you are having confusion whether two quantities are equal or not

Just set $x=1$ or any natural value in both expression, for simpler calculation , and see whether both are giving same results or not

In your case :

putting $x=1$ in Expression $1st$=$${{3+x}\over {5-2x}}={{3+1}\over {5-2(1)}}={4\over 3}$$

similarly putting $x=1$ in expression $2nd$=

$${{-3-x}\over {2x-5}}={{-3-1}\over {2(1)-5}}={-4\over -3}={4\over 3}$$

Both expressions are resulting same at $x=1$ , so both should be equivalent

But as I said, it doesn't works always , but most of times it is useful in comparing equality of two quantities

Atul Mishra
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