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
Best Answer
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} $