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. 

Start by realizing you can get a common denominator (and simplify once you have):
$$
-\frac{4}{(x-5)(x-3)}=-\frac{4}{(x-9)(x-7)}.
$$
Now take the reciprocal of both sides:
$$
-\frac{1}{4}(x-5)(x-3)=-\frac{1}{4}(x-9)(x-7).
$$
Now expand out the terms on the left- and right-hand sides:
$$
-\frac{x^2}{4}+2x-\frac{15}{4}=-\frac{x^2}{4}+4x-\frac{63}{4}.
$$
To make it easy to solve for $x$ (i.e., try to solve for when one of the sides equals $0$), subtract $-\frac{x^2}{4}+4x-\frac{63}{4}$ from both sides, obtaining
$$
12-2x=0 \Longleftrightarrow 2(6-x)=0 \Longleftrightarrow x=6.
$$
Thus, your equality is true when $x=6$.
