How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$

Each of the terms was multiplied by $\frac{-1}{-1}$, which is really equal to $1$, so it's a "legal" thing to do:

$-\dfrac{1}{x - 2} + \dfrac{1}{x - 3}$

$ = -\dfrac{(-1)1}{(-1)(x - 2)} + \dfrac{(-1)1}{(-1)(x - 3)}$

$ = -\dfrac{-1}{2 - x} + \dfrac{-1}{3 - x}$

$ = \dfrac{1}{2 - x} - \dfrac{1}{3 - x} $

I am a grade 8 student, so I may not be able to explain really well.

First, I need to prove that $-\frac {1} {x-2}=\frac {1} {2-x}$

To prove, let's assume that "$x$" can be any number, for instance, I take $x$=8.

So by substituting,

\begin{align} -\frac {1} {x-2} & = -\frac {1} {8-2}\\ & = -\frac {1} {6} \end{align}

And same for this,

\begin{align} \frac {1} {2-8} & =\frac {1} {-6}\\ & = -\frac {1} {6} \end{align}

Therefore, we have proven that $-\frac {1} {x-2}=\frac {1} {2-x}$

I also need to prove that $\frac {1} {x-3}=-\frac {1} {3-x}$

So by substituting,

\begin{align} \frac {1} {8-3} & =\frac {1} {5}\\ \end{align}

and the same for this,

\begin{align} -\frac {1} {3-8} & =-\frac {1} {-5}\\ & = \frac {-1} {-5}\\ & = \frac {1} {5}\\ \end{align}

Therefore, we have proven that $\frac {1} {x-3}=-\frac {1} {3-x}$

By why it worked? The truth is, it is just having -1÷(-1)=1 (negative$\times$negative=positive)(And anything times 1 is the same number)

So, from $-\frac {1} {x-2}$ to $\frac {1} {2-x}$, they inserted both -1 for numerator and denominator as the following below.

\begin{align} -\frac {1} {x-2} & = \frac {-1} {x-2}\\ & = \frac {-1(-1)} {-1(x-2)}\\ & = \frac {1} {-x+2}\\ & = \frac {1} {2-x}\\ \end{align}

same goes to $\frac {1} {x-3}=-\frac {1} {3-x}$

It's very easy. It comes by using factorization and simplification rules in general. In the case of your question, we have

$- \frac{1}{x-2} = \frac{(-1)}{(-1)(-x+2)}= \frac{(-1)}{(-1)}\times \frac{1}{(-x+2)}= \frac{1}{(-x+2)}$

and in the case of $\frac{1}{x-3}$, by multiplying both denominator and numerator with $(-1)$, we have that

$\frac{1}{x-3} = \frac{(-1)}{(-1)} \times \frac{1}{x-3} = \frac{(-1)}{(-x+3)}= -\frac{1}{(3 - x)}$. So we will have

$- \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{(2 -x)} - \frac{1}{(3 - x)} $.

And we are done.