Q: How to solve this algebra expression? I'm attempting work out why : 
$$\frac{(6x/x) - (9/x)}{(x/x) - (1/x)}=6$$
So far I have : 
$$\frac{(6x/x) - (9/x)}{(x/x) - (1/x)} =$$
$$\frac{(6x/x) - (9/x)}{(1) - (1/x)} =$$
Specifically how to simplify ? : 
$${(6x/x) - (9/x)}$$
Update : 
This is a limit question, reason I did not include full equation as I was just confused how it simplified, I should really have included full equation and have taken note. 
Simplify : 
$$\lim_{x\rightarrow \infty} \frac{(6x/x) - (9/x)}{(x/x) - (1/x)}$$
src : 


A: We have to assume $x \ne 0$ and $x \ne 1$.  But if we do that we have:
$\frac{(6x/x) - (9/x)}{(x/x) - (1/x)}=$
$\frac{6(x/x) - 9(1/x)}{(x/x) - (1/x)}=$
$\frac{6 - 9(1/x)}{1 - (1/x)}=$
$\frac{[6 - 9(1/x)]*x}{[1 - (1/x)]*x}=$
$\frac{6x - 9}{x-1}= 3\frac{2x - 3}{x-1}$
which does not equal any constant.
I suspect the $9$ was supposed to be a $6$ in which case we'd have:
$\frac{(6x/x) - (6/x)}{(x/x) - (1/x)}= \frac{6[(x/x) - (1/x)]}{(x/x) - (1/x)}=6\frac{(x/x) - (1/x)}{(x/x) - (1/x)}= 6$
===
Ah... if this was a limit problem, that changes everything...
$\lim_{n\rightarrow \infty} \frac{(6x/x) - (9/x)}{(x/x) - (1/x)}= $ (as we may assume $x > 0$ we have...)
$\lim_{n\rightarrow \infty} \frac{6 - 9/x}{1 - 1/x}= $ (as we may assume x > 1, 6 -9/x and 1-1/x are continuous and 1- 1/x $\ne$ 0 we have...)
$\frac{\lim_{n\rightarrow \infty} 6 - 9/x}{\lim_{n\rightarrow \infty} 1 - 1/x}= $
$\frac{6 - 9\lim_{n\rightarrow \infty} 1/x}{1 - \lim_{n\rightarrow \infty} 1/x}= $ (and as $\lim_{n\rightarrow \infty} 1/x = 0$ we have...)
$\frac {6-9*0}{1-0} = \frac{6}{1} = 6$.
===
It's interesting to note that:
$\lim_{n\rightarrow 0} \frac{(6x/x) - (9/x)}{(x/x) - (1/x)}= $
$\lim_{n\rightarrow 0} \frac{6 - 9/x}{1 - 1/x}= $
$\lim_{n\rightarrow 0} \frac{6x - 9}{x - 1}= $
$-9/-1 = 9$.
and 
$\lim_{n\rightarrow 1^\pm} \frac{(6x/x) - (9/x)}{(x/x) - (1/x)}= $
$\lim_{n\rightarrow 1^\pm} \frac{(6- 9x)}{1 - x}= $
$\frac 6{\lim_{n\rightarrow 1^\pm}1-x} = \mp \infty$.
===
for all other values of $x$ (not $0$ or $1$) we have $\frac{(6x/x) - (9/x)}{(x/x) - (1/x)}=3\frac{2x - 3}{x-1}$ which can be many different values but never $6$.

A: Assume a solution(s) exists. $x=0,1$ is obviously not in the domain, and we can do the following
$$
(\mathrm{EQ}) \iff \frac{1/x}{1/x}\frac{6x - 9}{x-1} = 6 \iff 6x-9 = 6x-6 \iff 9 = 6 
$$
which is a contradiction. Thus no solution exists.

Update:
To start with, in limits, you cannot algebraically work with the limit and the expression in the limit, like the way you seem to want to. Also the suggested solution in your textbook is somewhat odd to me.
Since the denominator and numerator have finite (non-zero in denominator) limits, then the limit of the fraction can be taken "separately/algebraically"
$$
\frac{6x - 9}{x - 1} = \frac{6 - 9/x}{1 - 1/x} \to 6, \quad x \to \infty
$$
