# how can I solve for $x$?

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. – Jyrki Lahtonen Sep 3 '12 at 6:36

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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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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For doing $x+5=1.(x-4)+9$, assuming $x\neq 4$ is essential. :-) – Babak S. Sep 3 '12 at 6:38
Just as with numbers, one needs to be careful when dividing by zero. Simply by paying careful attention to the statement of the problem we can see that $x=4$ could be problematic. – shoda Sep 3 '12 at 17:18
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. – Babak S. Sep 3 '12 at 17:35

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 $y$.

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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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