<s>[INST] 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
 [/INST] 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 $\frac {1} {x-3}=-\frac {1} {3-x}$
</s>