How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
 A: 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} $
A: 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 $\frac {1} {x-3}=-\frac {1} {3-x}$
A: 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.
A: We have $- \frac{1}{x-2} + \frac{1}{x-3} = \frac{1}{-1} \times \frac{1}{x-2} + \frac{-1}{-1} \times \frac{1}{x-3} = \frac{1}{(-1)\times (x-2)} + \frac{(-1)\times 1}{(-1)\times (x-3)} =  \frac{1}{-x+2} + \frac{-1}{-x+3}= \frac{1}{-x+2} - \frac{1}{-x+3} $. And we have the result by just nature of your own question.
A: $$
\frac{1}{x-a} = \frac{1}{-(a - x)} = - \frac{1}{a - x}
$$
A: This problem all boils down to the following relationship $$-1 = \frac{-1}{1}=\frac{1}{-1}$$
Part one is easy if you just express the division as a multiplication
$$x=\frac{-1}{1}\implies -1=1\cdot x\implies -1=x$$
For part two,
$$x=\frac{1}{-1}\implies1=-1\cdot x$$
$$1+(-1)+x=-1\cdot x+(-1)+x$$
$$x+0=-1\cdot x + 1\cdot x + (-1)$$
$$x=x((-1)+1)+(-1)$$
$$x=x((-1)+1)+(-1)$$
$$x=0x+(-1)$$
$$x=0+(-1)$$
$$x=-1$$
This assumes that $0x=0$
$$0x=(0+0)x$$
$$0x=0x+0x$$
$$0x+(-0x)=0x+0x+(-0x)$$
$$0=0x+0$$
$$0=0x$$
A: $$ -\frac{1}{x-2} = \frac{1}{-(x-2)} = \frac{1}{-x+2} = \frac{1}{2-x}$$ and 
$$ \frac{1}{x - 3} = \frac{1}{-3 + x} = \frac{1}{-(3 - x)} = -\frac{1}{3-x}$$
Thus $$ -\frac{1}{x-2} + \frac{1}{x - 3} = \frac{1}{2-x} - \frac{1}{3-x} $$
