<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] 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 $0x=0$
$$0x=(0+0)x$$
$$0x=0x+0x$$
$$0x+(-0x)=0x+0x+(-0x)$$
$$0=0x+0$$
$$0=0x$$
</s>