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

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