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. 
 A: 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 $x=6$.
A: 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. $$ (x-3)(x-5)=(x-7)(x-9) \\
x^2-8x+15=x^2-16x+63 \\
8x=48\\
x=6
$$
A: Because there is a difference of two each time between numerator and denominator, we can use the rearrangement you have from part 1 on each fraction in part 2. 
Then, after a little simplification, you can cross-multiply your fractions and get a constant in the numerator on each side. This allows you to invert the fractions and solve.
