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

$\begin{array}{rl}\frac{1}{{x}^{2}-5x+6}& =\frac{1}{(x-2)(x-3)}=\frac{1}{-3-(-2)}(\frac{1}{x-2}-\frac{1}{x-3})=-\frac{1}{x-2}+\frac{1}{x-3}\\ & =\frac{1}{2-x}-\frac{1}{3-x}=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{2}^{n+1}}{x}^{n}-\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{3}^{n+1}}{x}^{n}=\sum _{n=0}^{\mathrm{\infty}}(\frac{1}{{2}^{n+1}}-\frac{1}{{3}^{n+1}}){x}^{n}\end{array}$

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

$\begin{array}{rl}\frac{1}{{x}^{2}-5x+6}& =\frac{1}{(x-2)(x-3)}=\frac{1}{-3-(-2)}(\frac{1}{x-2}-\frac{1}{x-3})=-\frac{1}{x-2}+\frac{1}{x-3}\\ & =\frac{1}{2-x}-\frac{1}{3-x}=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{2}^{n+1}}{x}^{n}-\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{3}^{n+1}}{x}^{n}=\sum _{n=0}^{\mathrm{\infty}}(\frac{1}{{2}^{n+1}}-\frac{1}{{3}^{n+1}}){x}^{n}\end{array}$