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I am having a bit of trouble solving for x when trying to find $f^{-1}$. I have $y=\frac{x+5}{x-4}$ How can I get x on one side? I tried multiplying both sides by the denominator but I am still left with an $x$ on both sides...

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    That was a good start! After that move the terms containing an $x$ to the other side. Remember to swap signs when appropriate.2012-09-03

4 Answers 4

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Multiply with the denominator:

$y(x-4)=x+5$ $xy -4y=x+5$ Take all x and bring them to the left side, all other terms to the right side. $xy -x = 5+4y$ Single out x: $x(y-1)=5+4y$ Divide by the factor: $x=\frac{5+4y}{y-1}$

Note: $x\ne4$, because of the first equation.

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Note that $x+5 = 1\cdot(x-4) + 9$. Just as with numbers, this means that $x-4$ divides $x+5$ once with a remainder of 9. So you can write $(x+5)/(x-4)$ as $1+\frac{9}{x-4}$. It should be easy to get $x$ on one side now.

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    Usually when we are working with rational **functions**; we delete the roots of its denominator of our domain. So, what you did here is someway legal.2012-09-03
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You just need to expand out the brackets and re-factor:

$ \begin{array}{rcl} y & = & \frac{x+5}{x-4}\\ y(x-4) & = & x+5\\ yx-4y&=&x+5\\ yx-x &=&4y+5\\ x(y-1)&=&4y+5\\ x&=&\frac{4y+5}{y-1} \end{array} $

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First multiply out to get rid of the fraction: $y(x-4)=x+5$, or $yx-4y=x+5$. Now collect the $x$ terms on one side and everything else on the other: $yx-x=5+4y$. Factor the lefthand side: $(y-1)x=5+4y$. Now just divide both sides by $y-1$, and you’ll have $x$ in terms of $y$.