Having trouble solving part two of an equation: Part one consisted of proving that 
$$\frac{x-1}{x-3} = 1+ \frac{2}{x-3}$$
I completed this and here is my working:
$$let \frac{x-1}{x-3}= LHS$$
$$RHS=\frac{x-3}{x-3} + \frac{2}{x-3}$$
$$\frac{2+(x-3)}{(x-3)}$$
$$\frac {x-3+2}{x-3}$$
$$\frac {x-1}{x-3} = LHS$$
Part two then asks me to hence solve:
$$\frac{x-1}{x-3}-\frac{x-3}{x-5} = \frac{x-5}{x-7}−\frac{x-7}{x-9}$$
I have attempted to sub in part one into part two but my answers always turn out to be incorrect. 
Any tips in the matter would be much appreciated.
I'd like to thank anyone who comments and answers in advance, your help is much appreciated. 

The hence suggests:
Let $x=x-2$ the first equation becomes 
$$ 
\frac{(x-2)-1}{(x-2)-3}  = 1+\frac{2}{(x-2)-3} \\
\frac{x-3}{x-5}=1+\frac{2}{x-5} 
$$
This process can be done with the other fractions in part 2. Notice that this will be of benefit because the $1$ will cancel because of the subtraction.
$$
1-\frac{2}{x-3}-\left(1-\frac{2}{x-5}\right)=1-\frac{2}{x-7}-\left(1-\frac{2}{x-9}\right)\\
-\frac{2}{x-3}+\frac{2}{x-5}=-\frac{2}{x-7}+\frac{2}{x-9}$$
Cancel the $2$ and simplify each side $$ \frac{-(x-5)+x-3}{(x-3)(x-5)} =\frac{-(x-9)+(x-7)}{(x-7)(x-9)}$$
Simplify  and note the numerators are equal. The fractions can therefore only be equivalent if the denominators are equal. Note that $x\neq7$ for example. $$ (x-3)(x-5)=(x-7)(x-9) \\
x^2-8x+15=x^2-16x+63 \\
8x=48\\
x=6
$$
