<s>[INST] inverse of function Thanks for the help!  Here is the solution..
i had a problem:
$$f(x)=\frac{(\sqrt x+8)}{(5-\sqrt x)}$$
i had to find the inverse, so lets begin... 
1) first i write in terms of $y$
$$y=\frac{(\sqrt x+8)}{(5-\sqrt x)}$$
2) now try to get $x$ by itself
$$(5-\sqrt x)y=\sqrt x+8$$
3)Distribute y along the left hand side both sides
$$(5y-y\sqrt x)=\sqrt x+8$$
4)subtract 8 and from both sides and add $$ y\sqrt x $$ to both sides
$$ 5y-8=y\sqrt{x} + \sqrt x$$
5) factor right hand side 
$$ 5y-8=\sqrt{x} (y + 1)$$         
6)divide out $$ (y+1) $$ from right side
$$ ((5y-8)/(y+1))=\sqrt x $$
7) square both sides leaving
$$ x=((5y-8)/(y+1))^2$$
8) so the solution is $$f^-1(x)=((5x-8)/(x+1))^2$$
 [/INST] I have a problem:
$$f(x)= \frac{\sqrt{x}+8}{5-\sqrt{x}}$$
I have to find the inverse but my calculations are off. I have listed what I did below, could someone tell me where I went wrong? Thank You in advance.
1) Write in terms of y
$$y = \frac{\sqrt{x}+8}{5-\sqrt{x}}$$
2) Now try to get $x$ by itself
$$ y(5-\sqrt{x}) = \sqrt{x}+8 $$
3) Multiply the $y$ through the expression
$$ 5y-y\sqrt{x} = \sqrt{x}+8 $$
5) Subtract 8 from both sides
$$ 5y-y\sqrt{x} -8= \sqrt{x} $$
5) Add $y\sqrt{x}$ to both sides
$$ 5y - 8= \sqrt{x} +y\sqrt{x}$$
6) Factor out $\sqrt{x}$
$$ 5y - 8= \sqrt{x}(1+y)$$
7) Divide by $(1+y)$
$$ \frac{5y - 8}{(1+y)}= \sqrt{x}$$
8) Square both sides
$$ \frac{(5y - 8)^2}{(1+y)^2}= x$$
9) Replace $x$ with $y$
$$ f^{-1} = \frac{(5x - 8)^2}{(1+x)^2}$$
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